Calculus and Above
Calculus Questions? Ask a Mathematician for Answers ASAP
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Calculus and Above
Calculus Questions? Ask a Mathematician for Answers ASAP
Connect oneonone with {0} who will answer your question
I would like to work on this. When do you need an answer by?
I would like it by Thursday or Friday but to include all work applied to get the answer.
Here is what I came up with. Your questions are welcome. Please copy and paste in your browser
https://www.box.com/s/n4ioj7zg48bp8gcv1odt
So the questions can't be solved using normal probability distribution, estimates and sample sizes, and or hypothesis testing? I have no knowledge of Wilcoxon test so I am not sure how to follow or if its what is being asked about. Surely I wouldn't be asked questions about something we haven't covered yet.
we could assume that the sample is a simple random sample of 10 items form a normal population of teenagers who take the supplement. In part A the standard deviation is known so we can apply a Z test with the standard deviation given In part B we can apply a t test since the standard deviation is not given and we use the sample standard deviation. The vitamin company objects to the analysis since normality was assumed given such a small sample. The FDA then gives the Wilcoxen Test which does not assume normality to counter the latter objection. I believe this is how you should present your analysis I will send you the t test for part B standard deviation unknown.
Here is the analysis using the T test for part B
https://www.box.com/s/bu85g7vl28nondcw94tm
hey I have some questions if you would like to answer them
1.The Ultrahostile Air Lines (UAL ?) is notorious for its costcutting substandard customer service. Consequently, only 85% of the ticket holders will show up for a flight. The Air Lines workhorse, the Boring 878, has 234 seats. UAL, of course, will make extra money by overbooking. However, they have to worry about the FAA fines, and they have to keep a safety margin, say, that there is at least 95% probability that they can accommodate all those who show up for the Boring 878 flight. How many seats can they actually sell in this case?
You already know that the exact solution is based on the binomial distribution. However, is there an approximate method to carry out the analysis so, how would you do it?
2. In this recession, yours truly, XXXXX XXXXX the Outrageous Products Enterprise, would like to make extra money to support my frequent filetmignonanddoublelobstertail dinner habit. A promising enterprise is to massproduce tourmaline wedding rings for brides. Based on my diligent research, I have found out that women's ring size normally distributed with a mean of 6.0, and a standard deviation of 1.0. I am going to order 5000 tourmaline wedding rings from my reliable Siberian source. They will manufacture ring size from 4.0, 4.5, 5.0, 5.5, 6.0, 6.5, 7.0, 7.5, 8.0, 8.5, 9.0, and 9.5. How many wedding rings should I order for each of the ring size should I order 5000 rings altogether? (Note: It is natural to assume that if your ring size falls between two of the above standard manufacturing size, you will take the bigger of the two.)
3. An amusement park is considering the construction of an artificial cave to attract visitors. The proposed cave can only accommodate 36 visitors at one time. In order to give everyone a realistic feeling of the cave experience, the entire length of the cave would be chosen such that guests can barely stand upright for 98% of the all the visotors.
The mean height of American men is 70 inches with a standard deviation of 2.5 inches. An amusement parl consultant proposed a height of the cave based on the 36guestatatime capacity. Construction will commence very soon.
The park CEO has a second thought at the last minute, and ask yours truly XXXXX XXXXX proposed height is appropriate. What would be the propsed height of the amusement park consultant? And do you think that it is a good recommendation? If not, what should be the appropriate height? And why?
4. It is commonly accepted that the mean temperature of human is 98.6^{o}F. Yours truly XXXXX XXXXX better to do but measured the temperatures of 26 colleagues 1 to 4 times daily to get a total of 123 measurements. The collected data yielded a sample mean of 98.4^{o}F and a sample standard deviation of 0.7^{o}F. Is the mean temperature of his colleagues less than 98.6^{o}F at the 0.01 significance level? Justify your answer with the proper statistics.
5. My brother wants to estimate the proportion of Canadians who own their house. What sample size should be obtained if he wants the estimate to be within 0.02 with 90% confidence if
a. he uses an estimate of 0.675 from the Canadian Census Bureau?
b. he does not use any prior estimates? But in solving this problem, you are actually using a form of "prior" estimate in the formula used. In this case, what is your "actual" prior estimate? Please expalin.
Saturday will be good
ok thank you
ok here it goes...
92 (12)In a randomized controlled trial in Kenya, insecticidetreated bednets were tested as a way to reduce malaria. Among 343 infants using bednets, 15 developed malaria. Among 294 infants not using bednets, 27 developed malaria (based on data from “Sustainability of Reductions in Malaria Transmission and Infant Mortality in Western Kenya with Use of InsecticideTreated Bednets,” by Lindblade, et al., Journal of the American Medical Association, Vol. 291, No.21) Use a 0.01 significance level to test the claim that the incidence of malaria is lower for infants using bednets. Do the bednets appear to be effective?
(30)Use the sample data in Exercise 29 with a 0.05 significance level to test the claim that the percentage of makes who answer “yes” is less than the percentage of females who answer “yes.”
(38)Equivalence of Hypothesis Test and Confidence Interval Two different simple random samples are drawn from two different populations. The first sample consists of 20 people with 10 having a common attribute. Compare the results from a hypothesis test of p1 = p2 (with a .05 significance level) and a 95%confidence interval estimate of p1 – p2.
93 (18) Refer to the sample data given in Exercise 17 and use a .005 significance level to test the claim that the mean braking distance of fourcylinder cars is greater than the mean braking distance of sixcylinder cars.
A simple random sample of 13 fourcylinder cars is obtained, and the braking distances are measured. The mean braking distance is 137.5ft and the standard deviation is 5.8ft. A simple random sample of 12 sixcylinder cars is obtained and the braking distances have a mean of 136.3ft with a standard deviation 9.7ft (based on Data Set 16 in Appendix B). Construct a 90% confidence interval estimate of the difference between the mean braking distance of fourcylinder cars and the mean braking distance of sixcylinder cars. Does there appear to be a difference the means?
(32)White Blood cell counts are helpful for assessing liver disease, radiation, bone marrow, failure, and infectious diseases. Listed below are white blood cell counts found in simple random samples of males and females (based on data from the Third National Health and Nutrition Examination Survey).
A. Use a .01 significance level to test the claim that females and males have different mean white blood cell counts.
B. Construct a 99% confidence interval of the difference between the mean white blood cell count of females and males. Based on the result, does there appear to be a difference?
Females: 8.90 6.50 9.45 7.65 6.40 5.15 16.60 5.75 11.60 5.90 9.30 8.55 10.80 4.85 4.90 8.75 6.90 9.75 4.05 9.05 5.05 6.40 4.05 7.60 4.95 3.00 9.10
Male: 5.25 5.95 10.05 5.45 5.30 5.55 6.85 6.65 6.30 6.40 7.85 7.70 5.30 6.50 4.55 7.10 8.00 4.70 4.40 4.90 10.75 11.00 9.60
94 (6) Use a .05 significance level to test the claim that paired sample data come from a population for which the mean difference is µ_{d }=0 Find (a) ̅d , (b) s_{d}, (c) the test static, and (d) the critical values. Forecast Temperatures Listed below are predicted high temperatures that were forecast before different days (based on Data Set 11in Appendix B.)
Predicted high temperature forecast three days ahead 79 86 79 83 80
Predicted high temperature forecast five days ahead 80 80 79 80 79
(12) Assume that the paired sample data are simple random samples and that the differences have a distribution that is approximately normal. Listed below are the costs (in dollars) of flights from New York (JFK) to San Francisco for US Air, Continental, Delta, United, American, Alaska, and Northwest. Use a .01 significance level to test the claim that flights scheduled one day in advance cost more than flights scheduled 30 days in advance. What strategy appears to be effective in saving money when flying?
Flight scheduled one day in advance 456 614 628 1088 943 567 536
Flight scheduled 30 days in advance 244 260 264 264 278 318 280
95 (10) Test the given claim. Assume that both samples are independent simple random samples from populations having normal distributions. Braking Distances of Car. A simple random sample of 13 fourcylinder cars is obtained, and the braking distances are measured. The mean braking distance is 137.5ft and the standard deviation is 5.8ft. A random sample of 12 sixcylinder cars is obtained and the braking distances have a mean of 136.3ft with a standard deviation 9.7ft (based on Data Set 16 in Appendix B). Use a .05 significance level to test the claim that braking distances of fourcylinder cars and braking distances of sixcylinder cars have the same standard deviation.
(16) Listed below are body mass indexes (BMI) for Miss America winners from two different time periods. Use a .05 significance level to test the claim that winners from both time periods have BMI values with the same amount of variation.
BMI (from recent winners):
19.5 20.3 19.6 20.2 17.8 17.9 19.1 18.8 17.6 16.8
BMI (from the 1920s and 1930s):
20.4 21.9 22.1 22.3 20.3 18.8 18.9 19.4 18.4 19.1
Sunday night 11pm
oh sorry (30) A Pew Research Center Poll asked subjects "Is there solid evidence that the earth is getting warmer>" 69% of 731 male respondents answered "yes" and 70% of 770 female respondents answered "yes".
(38)Equivalence of Hypothesis Test and Confidence Interval Two different simple random samples are drawn from two different populations. The first sample consists of 20 people with 10 having a common attribute. The second sample consists of 2000 people with 1404 of them having the same common attribute. Compare the results from a hypothesis test of p1 = p2 (with a .05 significance level) and a 95%confidence interval estimate of p1 – p2.
Hey for those last questions my professor did not like the use of software and gave me little to no points. I have 2 assignments where he did all the work and answered the problem but when I did what he did he wasn't satisfied either. So I guessing I am missing something because no matter how the questions are answered I never get credit for it. If you want to take a look at what he did and figure out what he expects different from me when I resubmit please let me know
This is for the last one. This is how he did the problems.
9212. Bednets to Reduce Malaria In a randomized trial in Kenya, insecticidetreated bednets were tested as a way to reduce malaria. Among 343 infants using bednets, 15 developed malaria. Among 294 infants not using bednets, 27 developed malaria (based on data from "Sustainability of Reductions in Malaria Transmission and Infant Mortality in Western Kenya with Use of InsecticideTreated Bednets," by Lindblade et al., Journal of the American Medical Association, Vol. 291, No. 21). Use a 0.01 significance level to test the claim that the incidence of malaria is lower for infants using bednets. Do the bednets appear to be effective?
Let us summarize the information:
Bednets  No Bednet 
n_{1} = 343  n_{2} = 294 
x_{1} = 15  x_{2} = 27 
Hypothesis:
H_{o}: p_{1} = p_{2}
H_{1}: p_{1} < p_{2}
Test statistics:
Pvalues: This is a lefttailed test. Again, we are using Z table  Normal Distribution.
Or we read off, from Table A2, the area under the curve at z = 2.44, and it is 0.0073. Definitely, this value is well below the α = 0.01 value, and it is in the rejection area.
Critical value: We can also look at it from the point of critical zvalue. At 0.01 significance level, this lefttail critical zvalue is 2.3263.
Again, we see that z = 2.44 is in the rejection region, the solid blue region!
Therefore, we reject the null hypothesis.
There is sufficient evidence to support the claim that the incidence of malaria is lower for infants who use the bednets. And the bednets appear to be effective.
9230. Global Warming Survey Use the sample data in Exercise 29 with a 0.05 significance level to test the claim that the percentage of males who answer "yes" is less than the percentage of female who answer "yes."
Let us summarize the relevant information in Exercise 29:
Male  Female 
n_{1} = 731  n_{2} = 504 
x_{1} = 770  x_{2} = 539 
Hypothesis:
H_{o}: p_{1} = p_{2}
H_{1}: p_{1} < p_{2}
Test statistics:
Note that this is a lefttailed test. Therefore, we are talking about the scenario depleted in the following figure (Z table  Normal Distribution) for a 0.05 significance level.. The critical value in this case is z = 1.645. Well, you see the same picture in Figure 84 on page 399.
A value of z = 0.4430 is to the left of the critical value of z = 1.645, well within the blue zone in the above figure, the rejection zone. Therefore, there is not sufficient evidence to support the claim that the percentage of male who answer "yes" is less than the percentage of female who answer "yes." That is, we fail to reject the null hypothesis.
9238. Equivalence of Hypothesis Test and Confidence Interval Two different simple random samples are drawn from two different populations. The first sample consists of 20 people with 10 having a common attribute. The second sample consists of 2000 people with 1404 of them having the same common attributes. Compare the results with a hypothesis test of p_{1} = p_{2} (with a 0.05 significance level) and a 95% confidence interval estimate of p_{1}  p_{2}.
Well, it is time to organize again....
Sample #1  Sample #2 
n_{1} = 20  n_{2} = 2000 
x_{1} = 10  x_{2} = 1404 
Hypothesis:
H_{o}: p_{1} = p_{2}
H_{1}: p_{1} < p_{2}
First, let us try out the hypothesis test........
Test statistics:
Pvalue:
If we go to Table A2, and look for the area under the curve to the left of 1.92, we get 0.0281. And, Z table  Normal Distribution gives us a very close result.
The pvalue of 0.0281 is less than the specified α = 0.05 (95% confidence level). Therefore, we will reject the null hypothesis.
Now, let us try the confidence interval estimate:
Well, what is our z_{α/2} for 95% confidence level? Go to Table A2, we get a value of 1.96.
And, the margin of error, E, is
The corresponding confidence interval estimate is
Well, in the resultant confidence interval, zero is included. In this case, we fail to reject the null hypothesis.
What is going on here? We arrive at different conclusions, actually contradicting conclusions, when we use hypothesis test and confidence interval estimate!
We consider the two conclusions to be contradictory only if we consider the hypothesis test and the confidence interval estimate to be equivalent tests. Yes, the are closely related to each other, but they are not equivalent.
The hypothesis test is a much more stringent test than the confidence interval estimates. In the hypothesis test, we are asking a very narrow question if p_{1} = p_{2}. On the other hand, the confidence interval estimate is a much more forgiving test in that we are asking if p_{1}  p_{2} lies with a confidence interval. Therefore, even if the null hypothesis is rejected in the hypothesis test, it is possible that it is not rejected in the confidence interval estimate.
In these two exercises, assume that the two samples are independent random samples selected from normally distributed populations. Do not assume that the population standard deviations are equal, unless your instructor stipulates otherwise.
9318 Hypothesis Test for Breaking Distances of Cars Refer to the sample data given in Exercise 17 and use a 0.05 significance level to test the claim that the mean braking distance of fourcylinder cars is greater than the mean braking distance for sixcylinder cars.
Let us make a table again....
 FourCylinder Cars  SixCylinder Cars 
n  13  12 
 137.5  136.3 
s  5.8  9.7 
And let us go on with the hypothesis test.....
Hypothesis:
H_{o}: μ_{1} = μ_{2}
H_{1}: μ_{1} > μ_{2}
Test statistics:
Since this is a righttailed test, we have to determine our t_{α} value. Per Triola's prescription that we should take the simple and conservative estimate of df, i.e., the smaller of the n_{1} 1 (12) and n_{2} 1 (11), we have, for α = 0.05,
t_{α} = 1.796 based on our T Distribution Calculator. Well, how about with Table A3? what value will we get? Surprisingly, if we look across the row of df = 11, for α = 0.05 in a onetail test, we also get t_{α} = 1.796!
Now what?
OK, t = 0.37 is definitely well outside the critical region, right of t_{α} = 1.796. We do not have sufficient evidence to reject the null hypothesis that the mean braking distance for fourcylinder cars is greater than that for sixcylinder cars.
How about the pvalue test?
Oh, yes, what about the pvalue?
Well, the pvalue for the test statistics t = 0.3716 is 0.3586. It is much greater than our α = 0.05.
In either case, we do not have sufficient evidence to reject the null hypothesis.
9332 Sex and Blood Cell Counts White blood cell counts are helpful for assessing liver disease, radiation, bone marrow failure, and infectious diseases. Listed below are white blood cell counts found in simple random samples of male and females (based on data from the Third National Health and Nutrition Examination Survey).
Female: 8.90 6.50 9.45 7.65 6.40 5.15 16.60 5.75 11.60
5.90 9.30 8.55 10.80 4.85 4.90 8.75 6.90 9.75
4.05 9.05 5.05 6.40 4.05 7.60 4.95 3.00 9.10
Male: 5.25 5.95 10.05 5.45 5.30 5.55 6.85 6.65 6.30
6.40 7.85 7.70 5.30 6.50 4.55 7.10 8.00 4.70
4.40 4.90 10.75 11.00 9.60
a. Use a 0.01 significance level to test the claim that females and males have different mean white blood cell counts.
b. Construct a 99% confidence interval of the difference between the mean white blood cell count of females and males. Based on the result, does there appear to be a difference?
a. Let us make a table again.... Oh, I use ProStat for the means and standard deviations.......
 Females  Males 
n  27  23 
 7.443  6.787 
s  2.892  1.971 
And we are ready to get on with the hypothesis test.....
Hypothesis:
H_{o}: μ_{1} = μ_{2}
H_{1}: μ_{1} ≠ μ_{2}
Test statistics:
Pvalue:
OK, let us settle down with a pvalue for this test statistics........ Oh, yes, we take the lesser of n  1 for the degrees of freedom.
Remember that this is a twotailed test, we have to count both end of the tcurve. Based on our cheating tool, it is p = 2 x 0.1767 = 0.3534.
Now, let us recall our significance level. We have α = 0.01. Definitely, our pvalue is much larger than the assigned significance level. Therefore, we do not have enough evidence to warrant rejection of the null hypothesis that females and males have the same mean white blood cell counts.
Unfortunately, because of the information content, or rather the lack of it, in Table A3, we cannot easily obtained the pvalue for the test statistics in our current problem. Nonetheless, we can obtain a good estimate of it, though. Let us take a look across the row for degrees of freedom 22. Our test statistics is 0.9482, and it is to the right of 1.321. We can conclude that for a twotailed test, the corresponding pvalue must be greater than 0.02. And, again, this value from Table A3 will give us enough information to make the same conclusion........
Well, how about the critical value method?
Now, the moment of truth....... via the critical value method.......
So far so good, but we need a significance level to test our hypothesis against!
What are the critical tvalue in a twotailed test for α = 0.01? Again, per Triola, we take the smaller of the df = n 1 based on our given data set. Remember, because we have a "≠" in the alternative hypothesis, this is a twotail test, this is a twotailed test.
Miraculously, we obtained the same result, t_{α/2} = ±2.819 from Table A3!
For α = 0.01, our test statistics is far away from the critical region. Again, we come up with the same conclusion: we do not reject the null hypothesis that there is no difference in the mean white blood cell counts for females and male.
b. How about a 99% confidence interval?
Let's obtain the margin of error for a 99% confidence interval.....
And, hence, the 99% confidence interval is
Since the confidence interval includes a zero, we come up with the same conclusion as before that we do not have enough evidence to warrant rejection of the null hypothesis that females and males have the same mean white blood cell counts.
Calculations with Paired Sample Data. In this exercise, assume that you want to use a 0.05 significance level to test the claim that the paired sample data come from a population for the mean difference is μ_{d} = 0. Find (a) d, (b) s_{d}, (c) the test statistic, and (d) the critical values.
946 Forecast Temperatures Listed below are the predicted high temperatures that were forecast before different days (based on Data Set 11 in Appendix b).
Predicted high temperature forecast three days ahead 79 86 79 83 80
Predicted high temperature forecast five days ahead 80 80 79 80 79
a. Definitely, it is table time.....
Flight scheduled one day in advance  79  86  79  83  80 
Flight scheduled 30 days in advance  80  80  79  80  79 
Difference d  1  6  0  3  1 
The mean of the difference is
b. And now comes the messy job of finding the sample standard deviation of the difference d.
Difference, x 


1  2.8  7.84 
6  0.5  17.64 
0  1.8  3.24 
3  1.2  1.44 
1  0.8  0.64 

 Σ = 30.8 
c. Test statistics:
d. Note the key statement in the question:to test the claim that the paired sample data come from a population for the mean difference is μ_{d} = 0. We still have the null hypothesis as:
H_{o}: μ_{d} = 0
but the alternate hypothesis would be:
H_{1}: μ_{d} ≠ 0
It is, therefore a twotailed test. The scenario of the critical value is
You find the same result of t = ±2.776 from Table A3. It would be along the row for df = 4, and the column "Area in Two Tails, 0.05.
9412 Are Flights Cheaper When Scheduled Earlier? Listed below are the costs (in dollars) of flights from New York (JFK) to San Francisco for US Air, Continental, Delta, United, American, Alaska, and Northwest. Use a 0.01 significance level to test the claim that flights scheduled one day in advance cost more than flights scheduled 30 days in advance. What strategy appears to be effective in saving money when flying?
Flight scheduled one day in advance 456 614 628 1088 943 567 536
Flight scheduled 30 days in advance 244 260 264 264 278 318 280
Well, let us go on with our table making business.....
Flight scheduled one day in advance  456  614  628  1088  934  567  536 
Flight scheduled 30 days in advance  244  260  264  264  278  318  280 
Difference d  212  354  364  824  656  249  256 
Oh, I think I can do away with the tedious calculations by using ProStat, I obtained
Now, we are in a position to carry on with our hypothesis testing business....
Hypothesis:
H_{o}: μ_{d} = 0
H_{1}: μ_{d} > 0
Test statistics:
Because we have a ">" in the alternative hypothesis, this is a righttailed test. Well, for df = 7  1 = 6, we have
And, of course, we also get t_{α} = 3.143 in Table A3!
Now what? Well, since the test statistics is well inside the critical region, we reject the null hypothesis. In deed, we have sufficient evidence to warrant rejection of the null hypothesis. In human language, it is much cheaper to schedule flights well in advance!!!
Let us try the pvalue.....
Well, since t = 4.73 is so far in the rejection zone, we cannot even see the red area under the curve. Anyway, based on our T Distribution Calculator, it is 0.0016.
Well, this pvalue is less than our significance level of α = 0.01. Therefore, we arrive at the same conclusion as before: reject the null hypothesis.
It is unfortunate that we cannot extract our pvalues from Table A3. However, if we scan across the row for df = 6, we can deduce that the pvalue must be smaller than 0.005. This result is compatible with our applet result. In any case, we come to the same conclusion if we depend on on Table A3.
9420 Heights of Winners and RunnersUp Listed below are the heights (in inches) of candidates who won presidential elections and the heights of the candidates who were runnersup. The data are in chronological order, so the corresponding heights from the two lists are matched. For candidates who won more than once, only the heights from the first election are included, and no elections before 1900 are included.
Won Presidency  RunnerUp  
71  74.5  74  73  69.5  71.5  75  72  73  74  68  69.5  72  71  72  71.5 
70.5  69  74  70  71  72  70  67  70  68  71  72  70  72  72  72 
a. A wellknown theory is that winning candidates tend to be taller than the corresponding losing candidates. Use a 0.05 significance level to test that theory. Does height appear to be an important factor in winning the presidency?
b. If you plan to test the claim in Part (a) by using a confidence interval, what confidence level should be used? Construct a confidence interval using the confidence level, then interpret the result.
Oh, well, let there be yet another table.... A summary table this time......
Won Presidency  71  74.5  74  73  69.5  71.5  75  72  70.5  69  74  70  71  72  70  67 
Runnerup  73  74  68  69.5  72  71  72  71.5  70  68  71  72  70  72  72  72 
Difference d  2  0.5  6  3.5  2.5  0.5  3  0.5  0.5  1  3  2  1  0  2  5 
Using ProStat, I obtained
a. Now, we will now carry on with our hypothesis testing business....
Hypothesis:
H_{o}: μ_{d} = 0
H_{1}: μ_{d} > 0
Test statistics:
Because we have a ">" in the alternative hypothesis, this is a righttailed test. For df = 16  1 = 15, and α = 0.05, we have
And again, we also get t_{α} = 1.753 from Table A3.
Well, our test statistics is not in the critical region. We do not reject the null hypothesis.
Basically, at 0.05 significance level, we have enough evidence to conclude that there is no difference in height between candidates who won the presidency and those who lost the race!
Oh, well, do you want to see the pvalue?
Definitely, this pvalue is greater than our significance level of α = 0.05. We arrive at the same conclusion as before.....
What about Table A3? Again, if we scan across the row for df = 15, for a onetailed test, we can see that our test statistics will yield a pvalue much greater than 0.10. We should be able to arrive at the same conclusion.......
b. Confidence interval? OK, let us use the same confidence level as we use before, α = 0.05. But confidence interval is a twotailed test. Therefore, we should have 0.05 at each tail, which will lead us to a confidence interval of 90%. From Table A3, the t_{α/2} is ±1.753. Now, we have to determine the margin of error, which is
And our confidence interval is
Well, the confidence interval limits do contain 0, which indicates that there is not a significant height difference between winners and losers of presidential races.
9510 Breaking Distances of Cars A random sample of 13 fourcylinder cars is obtained, and the breaking distances are measured and found to have a mean of 137.5 ft and a standard deviation of 5.8 ft. A random sample of 12 sixcylinder cars is obtained and the breaking distance s have a mean of 163.3 ft and a standard deviation of 9.7 ft (based on Data Set 16 in Appendix B). Use a 0.05 significance level to test the claim that braking distances of fourcylinder cars and braking distances of sixcylinder cars have the same standard deviation.
First of all, let us recall the table from Exercise 9318..........
 FourCylinder Cars  SixCylinder Cars 
n  13  12 
 137.5  136.3 
s  5.8  9.7 
Now, we are dealing with Fdistribution. Let us do a strategic switch for the sake of computational simplicity, and let n_{1} = 12 and n_{2} = 13, σ_{1} = 9.7 and σ_{2} = 5.8
Hypothesis:
H_{0}: σ_{1} = σ_{2}
H_{1}: σ_{1} ≠ σ_{2}
_{ }Test statistics:
Let us see what we get with our F Distribution Calculator ......... For a twotailed test with α = 0.05, numerator df = n_{1}  1 = 11, and denominator df = n_{2}  1 = 12, we have
0.2916 and 3.321 as the two critical values. By the way, because we can only have the area to the right of a Fvalue, the "area" under the curve we are interested in is actually the white area.
Definitely, our test statistics is not in the critical regions. Therefore, there is not enough evidence to warrant the rejection of the null hypothesis that the standard deviation of the fourcylinder cars and the sixcylinder cars are the same.
Now, can we get anywhere with Table A5 from page 756759? Well, we have to go to page 757 first. Across the row for denominator df = 12, we have the average the two values, 3.376 and 3.2773, for numerator df = 10 and 12 respectively. The resultant Fvalue is 3.3255, which is practically the same value from the applet.
How about pvalue? Let us go back to our applet again......
Remember that this is a twotailed test. Therefore, we have to double the value read out from above. And our pvalue is 2 x 0.0455 = 0.091, which is greater than our α = 0.05. We arrive at the same conclusion.....
9516 BMI for Miss America Listed below are body mass indexes (BMI) for Miss America winners from two different periods. Use a 0.05 significance level to test the claim that winners from both time periods have BMI values with the same amount of variation.
BMI (from recent winners): 19.5 20.3 19.6 20.2 17.8 17.9 19.1 18.8 17.6 16.8
BMI (from 1920s and 1930s): 20.4 21.9 22.1 22.3 20.3 18.8 18.9 19.4 18.4 19.1
Of course, let us organize..... Oh, I use ProStat again........
 n 
 s  s^{2} 
BMI (from recent winners)  10  18.76  1.186  1.4071 
BMI (from 1920s and 1930s)  10  20.16  1.479  2.1871 
Now, here we go again.... OK, let us assign index 1 to the Miss America's of 1920s and 1930s, and index 2 to the recent Miss America's.
Hypothesis:
H_{0}: σ_{1} = σ_{2}
H_{1}: σ_{1} ≠ σ_{2}
_{ }Test statistics:
We are going to continue our cheating spree with technology...... Remember that it is, again, a twotailed test because we have the "≠" sign. At a significance level α = 0.05, and both numerator and denominator df = 10  1 = 9, we have
The critical values turn out to be 0.2484 and 4.026.
Again, Table A5 on page 756 gives us a critical value of 2.026 for α/2 = 0.025, which is the same as the result from our applet.
Anyway, we the test statistics is not in the critical region. Therefore, there is not enough evidence to warrant rejection of the null hypothesis that the BMI of Miss America winners in the two given time periods have the same amount of variation.
Oh, yes, the dreaded pvalue.......
Again, since this is a twotailed test, we have to double what we read out from our applet. We have p = 2 x 0.2621 = 0.5242. Definitely, it is greater than the significance level of α = 0.05. Again, our previous conclusion stands.......
This is for the one before that.
Chapter 8
Finding Critical Values. In this exercise, assume that the normal distribution applies and find the critical z value.
8222. α = 0.01; H_{1} is p > 0.5.
Since we have p > 0.5, the critical region is in the righthand side of the normal curve. We have a righttailed case.
First, let us illustrate with our Z table  Normal Distribution.
We see that the critical zvalue is z = 2.3263 ~ 2.33. The blue shaded area has the area of α = 0.01. And this is the area (α = 0.01) that we are testing against when we use the pvalue approach later.
Since we have cheated. We have some ideas of what we are looking for now. In Table A2, because it gives the area to the left of the zscore, we have to look for the zscore with an area of 0.99. Well, the closest value of the area we can find is 0.9901, which corresponds to a zscore value of 2.33.
Note: The p > 0.5 in the alternative hypothesis has nothing to do with either the pvalue (to be introduced later) or the critical zscore value. It is just a statement.
Finding Test Statistics. In this exercise, find the value of the test statistics z using
8228. Seat Belts The claim is that more than 75% of adults always wear a seat belt in the front seat. A Harris Poll of 1012 adults resulted in 870 who say that they always wear a seat belt in the front seat.
In this case, we have
Meanwhile, the key word in our case here is "more than." Therefore, it should be safe to set p = 0.75 and q = 1  p = 0.25. Therefore, the test statistics is
Testing Claims About Proportions. In these two exercises, test the given claim. Identify the hypothesis, alternative hypothesis, test statistic, Pvalue or critical value(s), conclusion about the null hypothesis, and final conclusion that addresses the original claim. Use the Pvalue method unless your instructor specifies otherwise. Use the normal distribution as an approximation to the binomial distribution (as described in Part 1 of this section).
8318. Gender Selection for Boys The Genetics and IVF Institute conducted a clinical trial of the YSORT method designed to increase the probability that a baby is a boy. As of this writing, among the babies born to parents using the YSORT method, 172 were boys and 39 were girls. Use the sample data with a 0.01 significance level to test the claim with this method, the probability of a baby being a boy is greater than 0.5. Does the YSORT method of gender selection appear to work?
Now, let us get on with the business with the hypotheses and test statistics.......
p = 0.5
q = 1  p = 0.5
n = 172 + 39 = 211
α = 0.01
Null Hypothesis: H_{0} : p = 0.5
Alternative Hypothesis: H_{1} : p > 0.5
Test statistics:
Pvalues:
Holy cow! With a test statistics of 9.16, we really do not have to go to our favorite applet, Z table  Normal Distribution, to figure out that the area to the right of z = 9.16 is zero for all practically purpose.
Well, what about taking a look at Table A2? In Table A2, we have the area to the left of the critical zvalue up to 3.50, which is 0.9999. We will assume that for critical zvalue of 9.16 will have an area very much less than 1  0.9999 = 0.0001 to the right of the critical zvalue. We can safely conclude that our Pvalue is 0.
But what now? Since our Pvalue is practically zero, it is very much less than our α = 0.01, we can reject our null hypothesis that the chance of getting a boy with this YSORT method is 0.5. Based on the alternate hypothesis, we can conclude that the YSORT method of gender selection works.
Now that we have our answer already. It may be wise to see if we will get the same result with the critical value method.....
Critical value:
Well, we have the same old procedure......
The approximate critical value is z = 2.33. The test statistics of 9.16 is well within the rejection region.
Therefore, we can conclude that there is sufficient evidence to reject the hull hypothesis that the chance of getting a boy is 50%. And, in human language, it means that the YSORT method appears to work very well!
8328. Bias in Jury Selection In the case of Casteneda v. Partida, it was found that during a period of 11 years in Hidalgo County, Texas, 870 people were selected for grand jury duty, and 39% of them were Americans of Mexican Ancestry. Among the people eligible for grand jury duty, 79.1% were Americans of Mexican ancestry. Use a 0.01 significance level to test the claim that the selection process is biased against Americans of Mexican Ancestry. Does the jury selection system appear to be fair?
OK, let see what we have.......
p = 0.791
q = 1  p = 0.209
n = 870
α = 0.01
Null Hypothesis: H_{0} : p = 0.791
Alternative Hypothesis: H_{1} : p < 0.791
Test statistics:
Pvalue:
Based on Table A2, any zvalues less than 3.50, the area to the left of the normal curve is taken as 0.0001. And we do not even have to try our applet, a zvalue of 29.1 will give us a flat zero!
Definitely, our pvalue of less than 0.0001 is much less than our α = 0.01, and we are in the rejection region!
We reject our null hypothesis that the probability of an American of Mexican ancestry being selected to serve in a grand jury is 79.1% . In fact, it is much less....... Therefore, we can conclude that the selection process is biased.
How about the fairness of the jury selection system? You tell me........ With a jury pool of 79.1% are of Americans of Mexican ancestry, the demographic of the jury should reflect that proportion. But it is only 39%, which is about half of what it should be. It is not even close. It is way off! Definitely, the jury selection in this case is very far from being fair, as it is reflected in the test statistics. It is so far out in the left field (no pun intended), 29.1 standard deviation from the mean!!!
How about the critical value method? In this case, the test statistics is so extreme that we are certain that it is well within the critical region. In fact, if you look at Table A2, for α = 0.01, the critical zscore is 2.33. Remember that this is a lefttailed test.
Note: Based on the result of the hypothesis testing, it is "obvious" that there is considerable bias in the jury selection process. The real challenging question is: Do we see that coming right at the beginning, even before the statistical analysis?
I would stick my neck out to say that statistics should be a quantitative verification of your hunch. Yes, most physics professors told their students that they should know the answer before they start. What? This sounds like yet another riddle. But it is not......... What these physics professors are saying is that we should develop a physical intuition of should the answer roughly be. So, if we get a strange answer, at least we know that we are wrong.
Now, let us get back to our problem above. We are told that about 80% of the residents are MexicanAmerican ancestry, whereas only about 40% of the jurors are of MexicanAmerican ancestry. You would expect roughly 80% of the jurors are of MexicanAmerican ancestry based on common sense (our understanding of random sampling). Definitely, we should have smelled something fishy right at the beginning. Our statistical analysis is just a quantitative nail to the coffin: We have backed up our hunch with numbers!
Testing Hypothesis. In this exercise, test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, Pvalue or critical value(s), conclusion about the null hypothesis, and final conclusion that addresses the Original claim. Use the Pvalue method unless your instructor specifies otherwise.
8412. Loaded Die When a fair die is rolled many times, the outcomes of 1, 2, 3, 4, 5, and 6 are equally likely, so the mean of the outcomes should be 3.5. The author drilled holes into a die and loaded it by inserting lead weights, then rolled it 16 times to obtain a mean of 2.9375. Assume that the standard deviation of the outcome is 1.7078, which is the standard deviation for a fair die. Use a 0.05 significance level to test the claim that the outcomes from the loaded die have a mean different from the value of 3.5 expected with a fair die. Is there anything about the sample data suggesting that the methods of this section should not be used?
Well, what do we have here.......
σ = 1.7078
n = 16
α = 0.05
Null Hypothesis: H_{0} : μ = 3.5
Alternative Hypothesis: H_{1} : μ ≠ 3.5
Test statistics:
Pvalue:
If we go to Table A2, and look for the area under the curve to the left of z = 1.32, we get 0.0934. Since this is a twotailed test, we double the "area under the curve." This is the Pvalue, which is 0.1864. Definitely, this Pvalue is greater than our α, i.e., p > α. Therefore, we fail to reject the null hypothesis that the mean outcomes of the loaded die is not 3.5 at 95% confidence level.
What? Whenever in doubt, try another route....... Let's take a look at the criticalvalue method....
Criticalvalue:
What is the critical zscore value for a twotailed test at 95% confidence level? It is z = ±1.96 in Table A2.......
Definitely, our test statistics of z = 1.32 is outside the critical region shaded blue above. Therefore, the test statistics is NOT in the rejection zone!
Therefore, based on the above statistical analysis, the loaded die behaves like a fair die statistically!!! Based on the above statistical test, Dr. Triola is getting away with murder!!!
Now, here is where common sense, or statistical sense in our situation, comes in......... We know that the die is loaded, and the mean of the outcomes is about 13% less than that of a fair die. And, what does the statistical test tells us? It tells us that we cannot tell it is a loaded one! Something has gone awry........
I would doublecheck my calculations, and they check out fine. I have perform the test and calculation as I am told by Dr. Triola. It is time to question our methodology. It may be our starting assumption of a normal distribution is in question......
First of all, we have 16 trials in our data set. It is less than the 30 sample data points that is required. In this case, our assumption of a normal distribution is in doubt.
And, most importantly, the distribution of the outcomes is NOT a normal distribution! We have to keep in mind that the distribution of the outcomes of a fair die is a UNIFORM distribution, in which each face has equal chance to show up! We are not dealing with a normal distribution in this case. What have we been doing?
We have apply the WRONG method to analysis our problem. Yes, we have ask the wrong question, or a meaningless question. Therefore, we will get a meaningless answer!!!
8420. California Speeding Listed below are recorded speeds (in mi/h) of randomly selected cars traveling on a section of Highway 405 in Los Angeles (based on data from Sigalert). That part of the highway has a posted speed limit of 65 mi/h. Assume that the standard deviation is 5.7 mi/h and use a 0.01 significance level to test the claim that the sample is from a population with a mean that is greater than 65 mi/h.
68 68 72 73 65 74 73 72 68 65 65 73 66 71 68 74 66 71 65 73
59 75 70 56 66 75 68 75 62 72 60 73 61 75 58 74 60 73 58 75
First of all, we have to determine the sample mean of the given data. It is 68.375 mi/h.
OK, let us see what we get ......
σ = 5.7
n = 40
α = 0.01
Now, let us get on with the hypothesis testing.
Null Hypothesis: H_{0} : μ = 65 mi/h
Alternative Hypothesis: H_{1} : μ > 65 mi/h
Test statistics:
Pvalue:
Remember that this is a righttailed test. We are looking at the region to the right of z = 3.74.
The total shaded area, if you can actually see it, is 0.000092 ≈ 0.0001, which is our pvalue. Definitely, this pvalue is very much less than our α = 0.01. Therefore, there is sufficient evidence to reject our null hypothesis.
What is the botXXXXX XXXXXne? The null hypothesis is rejected. We are left with the alternative hypothesis that the sample data are from a population with a mean that is greater than 65 mi/h. Oh, yes, the Californians are speeding!!!
What if you have to use Table A2?
Let's go for it. In Table A2, the maximum zvalue is 3.50! Well, to the right of this maximum zvalue, the area under the curve is 0.0001. Therefore, we can safely assume that the area under the curve to the right of z = 3.74 is less than 0.0001. We end up with practically the same pvalue.
Critical value:
Well, we can refer back to Problem 8318 for the critical zvalue, which is 2.33. Definitely, our test statistics of 3.74 is well to the right of 2.33 in the region zone. Yes, we come up with the same conclusion, of course.
Testing Hypotheses. In the following exercise, assume that a simple random sample has been selected from a normally distributed population and test the given claim. Unless specified by your instructor, use either the traditional method or Pvalue method for testing hypotheses. Identify the null hypothesis and alternative hypothesis, test statistics, Pvalue (or range of Pvalues), critical value(s), and state the final conclusion that addresses the original claim.
8528. BMI for Miss America The trend of thinner Miss America winners has generated charges that the contest encourages unhealthy diet habits among young women. Listed below are body mass indices (BMI) for recent Miss America winners. Use a 0.01 significance level to test the claim that recent Miss America winners are from a population with a mean BMI less than 20.16, which was the BMI for winners from 1920's and 1930's. Do the recent winners appear to be significantly different from those in the 1920's and 1930's?
19.5 20.3 19.6 20.2 17.8 17.9 19.1 18.8 17.6 16.8
Well, first of all, let see what is the mean and sample standard deviation of the above data set.........
Really, I have to rely on "technology" to save time. We come up with a sample mean of 18.76 and a sample standard deviation of 1.186
OK, let's see what we have here......
s = 1.186
n = 10 (Remember that the degrees of freedom is n  1 = 9)
α = 0.01
Then, let's get on with the hypothesis test:, a lefttailed test.
Null Hypothesis: H_{0} : μ = 20.16
Alternative Hypothesis: H_{1} : μ < 20.16
Test statistics:
We are now in a position to take a look at the pvalue. Anyway, let us make good use of the T Distribution Calculator. Since this is a lefttailed test, we are interested in the shaded area to the left of t = 3.73.
And, again, if you can ever see it, the shaded area under the tcurve is 0.0023. Definitely, this pvalue is much smaller than our assumed α = 0.01.
Yes,, we have enough evidence to reject the null hypothesis that the recent Miss America winners have BMI equals to a population with a mean BMI index of 20.16.
We are left with the alternative hypothesis!!! The recent Miss America winners have smaller BMI indices. Well, what's wrong with the more rubenseques Miss America winners in the 1920's and 1930's?
Of course, we can go to Table A3 for our pvalue too. If we look across the row with degrees of freedom of 9, the maximum tvalue is 3.250, which corresponds to a pvalue of 0.005. We can infer that our pvalue is definitely less than 0.005. Again,. we come up with the same conclusion that Miss America winners in the 1920's and 1930's were more rubensesques........
8534. Using the Wrong Distribution When testing a claim about a population mean with a simple random sample selected from a normally distributed population with unknown σ, the Student t distribution should be used for finding critical values and/or a pvalue. If the standard normal distribution is incorrectly used instead, does that mistake make you more or less to reject the null hypothesis, or does it not make a difference? Explain.
To address this question, we may want to go back and review Problem 7434. Basically, we observed that a tdistribution is wider (fatter) than a normal distribution. And the tdistribution will get slimmer and slimmer with increasing sample size until it becomes a normal distribution with n greater than, say, 700.
That said, what we mean is that the corresponding critical value of a tdistribution is further away from the origin that that of a normal distribution. Therefore, by using the normal distribution instead of a tdistribution, it will be more likely to reject a null hypothesis.
Finding Test Components. The this exercise, find the test statistics and critical value(s). Also, use Table A4 to find limits containing the Pvalue, then determine whether there is sufficient evidence to support the given alternative hypothesis.
868. Precipitation Amounts H_{1}: σ ≠ 0.25, α = 0.01, n = 26, s = 0.18.
Alright, let's try χ^{2}. What we have is
Null Hypothesis: H_{0} : σ = 0.25
Alternative Hypothesis: H_{1} : σ ≠ 0.25
n = 26
s = 0.18
α = 0.01
Test statistics:
Critical values:
Well, remember that this is a twotailed test. Therefore, we have to split the α into two, 0.005 in the left extreme and 0.005 in the right extreme of the χ^{2}curve.
Based on Table A4, the corresponding critical values for α = 0.01 is 10.520 and 46.928. This is basically what we get from our Chi Square Calculator.
In simple language, the test statistics is not in the critical region/rejection zone.
Pvalue:
Let us take advantage of our Chi Square Calculator at this point. But, remember that it is a pain in the neck that we can only obtain the area to the right of our χ^{2}value.......
It is the white area that we are interested in, which is (1  0.977) = 0.023.
But remember! This is a twotailed test. Therefore, our pvalue is 2 x 0.023 = 0.046.
OK, what about Table A4? If we scan across the row for degrees of freedom of 25, our critical value of 12.96 falls in between 11.524 and 13.120, which corresponds to areas of 0.01 and 0.025 to the left. Based on Table A4, we have 2 x 0.01 < p < 2 x 0.025, even though our pvalue should be much closer to 0.05 but still less than 0.05. This pvalue is compatible to our finding with our Chi Square Calculator.
Conclusion:
There is not enough evidence to reject the null hypothesis. In other words, that there is not enough evidence to support the alternative hypothesis.
Testing Claims About Variation. In the following exercises, test the given claim. Assume that a simple random sample is selected from a normally distributed population. Use either the Pvalue method or the traditional method of testing hypothesis unless your instructor indicates otherwise.
8618. Vitamin Supplement and Birth Weight Listed below are birth weights (in kilograms) of male babies born to mothers on a special vitamin supplement (based on data from the New York State Department of Health). Test the claim that this sample comes from a population with a standard deviation equal to 0.470 kg, which is the standard deviation for male birth weights in general. Use a 0.05 significance level. Does the vitamin supplement appear to affect the variation among birth weights?
3.73 4.37 3.73 4.33 3.39 3.68 4.68 3.52
3.02 4.09 2.47 4.13 4.47 3.22 3.43 2.54
Well, first of all, we have to determine the sample variance of the above listed sample. We have s^{2} = 0.432.
Again, it is time for χ^{2}.
Null Hypothesis: H_{0} : σ = 0.470 kg
Alternative Hypothesis: H_{1} : σ ≠ 0.470 kg
α = 0.05
Test statistics:
Critical value:
Remember that since we have a ≠ case, this is a twotailed test. We are looking at the regions to the left and the right with an area of 0.05 with a degree of freedom of 15 (= 16  1). Yes, it is an area of 0.025 to the left and another 0.025 to the right. Now, let us cheat by using the Chi Square Calculator. Because of the set up of the calculator, we have to look at 0.975 and 0.025 on the right.
Yes, the critical χ^{2} values are 6.262 and 27.49. Of course, we may look up Table A4, we look at degrees of freedom 15, and Area in One Tail of 0.975 and 0.025, we have χ^{2} = 6.262 and 27.488.
Well, the χ^{2} test statistics is in the critical region, and it is to the right of 27.488. Therefore, we have enough evidence to reject the null hypothesis that the sample comes from a population with a standard deviation equals to 0.470 kg.
How about the pvalue method?
We have to be mindful that this is a twotailed test, and our pvalue should be double of the blueshaded area. Therefore, our pvalue should be 0.0292. Obviously, this pvalue is less than our α, which is 0.05. Again, we have the same conclusion as before: reject the null hypothesis!
But, again, what about Table A4? Well, if we look across the row for degrees of freedom of 15, our test statistics is closest to a value of 30.578, which corresponds to an α = 0.01. In fact, the actual area under the curve should be a little bit greater than 0.01. Doubling that, we have our pvalue a little bit greater than 0.02, which is still small than our α. Well, again, our pvalue method and the criticalvalue method yield the same conclusion......
the test statistic is 15*s2^2/(.47^2)=12.67 You will reject H0 if you are greater than 27.488 or less than 6.262. You are looking for an unusual occurrence in the tails of the distribution to reject H0. The conclusion is do not reject H0 there is insufficient evidence to reject the H0 that the standard deviation is .47kg. the pvalue is 2*chisquare(0,12.67,15 d.f)=.7445>.05 hence do not reject H0since pvalue >.05
If you think of this intuitively the test statistic is
chi=(n1)S^2)/(sigma ^2). It is looking at the ratio of the S^2(sample)/(sigma^2 of the population). If this ratio is very large or very small you are led to the conclusion that the data does not suggest the sigma of the population is .47 hypothesized.
The pvalue is a measure of how likely of getting a test statistic of 12.67 if the true value of the sigma of thepopulation is .47. A small pvalue leads you to the conclusion that the H0 is very unlikely. In this case the pvalue is too large and hence we do not reject H0.
I don't understand I sent you like 20 questions with the answers included. I was just looking for a different way to answer each question but you sent me back one or two answers?
I did not notice 20 questions. I will look at all of them and get back to you. One thing I noticed however is in some cases you did not tell how the test statistic was computed.
ok never mind about those questions for now. I need help with a whole list of new questions. Please show work as I said before and please do not use software to answer the questions. I know its a lot but let me know how fast you can get this done if at all. Thank you
Refer to the following frequency distribution for Questions 1, 2, 3, and 4.
The frequency distribution below shows the distribution for suspended solid concentration (in ppm) in river water of 50 different waters collected in September 2012.
Concentration (ppm) Frequency
20 – 29 1
30 – 39 7
40 – 49 8
50 – 59 11
60 – 69 11
70 – 79 7
80 – 89 3
90 – 99 2
1. What percentage of the rivers had suspended solid concentration less than 40? (5 pts)
2. Calculate the mean of this frequency distribution. (10 pts)
3. In what class interval must the medXXXXX XXXXXe? Explain your answer. (You don’t have to find the median) (5 pts)
4. Assume that the largest observation in this dataset is 98. Suppose this observation were incorrectly recorded as 988 instead of 98. Will the mean increase, decrease, or remain the same? Will the median increase, decrease or remain the same? Explain your answers.
(5 pts)
Refer to the following information for Questions 5 and 6.
A coin is flipped three times. Let A be the event that the outcome of the first flip is a heads. Let B be the event that the outcomes of second and third flips are both tails..
5. What is the probability that the outcomes of the second and third flips are both tails, given that the first one is a heads? (10 pts)
6. Are A and B independent? Why or why not? (5 pts)
Refer to the following data to answer questions 7 and 8. Show all work. Just the answer, without supporting work, will receive no credit.
A random sample of song playing times in seconds is as follows:
231 220 213 230 293
7. Find the standard deviation. (10 pts)
8. Are any of these playing times considered unusual in the sense of our textbook? Explain. Does this differ with your intuition? Explain. (5 pts)
Refer to the following situation for Questions 9, 10, and 11.
The fivenumber summary below shows the grade distribution of two STAT 200 quizzes.
 Minimum
 Q1  Median  Q3
 Maximum 
Quiz 1
 12  40  60  95  100 
Quiz 2
 20  35  50  80  100 
For each question, give your answer as one of the following: (a) Quiz 1; (b) Quiz 2; (c) Both quizzes have the same value requested; (d) It is impossible to tell using only the given information. Then explain your answer in each case. (5 pts each)
9. Which quiz has greater interquartile range in grade distribution?
10. Which quiz has the greater percentage of students with grades 80 and over?
11. Which quiz has a greater percentage of students with grades less than 60?
12. A random sample of 225 SAT scores has a mean of 1522. Assume that SAT scores have a population standard deviation of 300. Construct a 95% confidence interval estimate of the mean SAT scores. (15 pts)
Refer to the following information for Questions 13 and 14.
There are 1000 students in the senior class at a certain high school. The high school offers two Advanced Placement math / stat classes to seniors only: AP Calculus and AP Statistics. The roster of the Calculus class shows 95 people; the roster of the Statistics class shows 86 people. There are 43 overachieving seniors on both rosters.
13. What is the probability that a randomly selected senior is in exactly one of the two classes (but not both)? (10 pts)
14. If the student is in the Calculus class, what is the probability the student is also in the Statistics class? (10 pts)
Refer to the following information for Questions 15, 16, and 17.
A box contains 5 chips. The chips are numbered 1 through 5. Otherwise, the chips are identical. From this box, we draw one chip at random, and record its value. We then put the chip back in the box. We repeat this process two more times, making three draws in all from this box.
15. How many elements are in the sample space of this experiment? (5 pts)
16. What is the probability that the three numbers drawn are all different? (10 pts)
17. What is the probability that the three numbers drawn are all odd numbers? (10 pts) Questions 18 and 19 involve the random variable x with probability distribution given below.
X 2 3 4 5 10
()Px 0.1 0.2 0.3 0.2 0.2
18. Determine the expected value of x. (10 pts)
19. Determine the standard deviation of x. (10 pts)
Consider the following situation for Questions 20 and 21.
Airline overbooking is a common practice. Due to uncertain plans, many people cancel at the last minute or simply fail to show up. Capital Air is a small commuter airline. Its past records indicate that 85% of the people who make a reservation will show up for the flight. The other 15% do not show up. Capital Air decided to book 11 people for today’s flight. Today’s flight has just 10 seats.
20. Find the probability that there are enough seats for all the passengers who show up. (Hint: Find the probability that in 11 people, 10 or less show up.) (10 pts)
21. How many passengers are expected to show up? (5 pts)
22. Given a sample size of 65, with sample mean 726.2 and sample standard deviation 85.3, we perform the following hypothesis test. 0:750Hμ=
1:750Hμ<
What is the conclusion of the test at the 0.10α=level? Explain your answer. (20 pts)
Refer to the following information for Questions 23, 24, and 25.
The IQ scores are normally distributed with a mean of 100 and a standard deviation of 15.
23. What is the probability that a randomly person has an IQ between 85 and 115? (10 pts)
24. Find the 90th percentile of the IQ distribution. (5 pts)
25. If a random sample of 100 people is selected, what is the standard deviation of the sample mean?
(5 pts)
26. Consider the hypothesis test given by 01:530:530.HHμμ=≠
In a random sample of 81 subjects, the sample mean is found to be 524.x=Also, the population standard deviation is 27.σ=
Determine the Pvalue for this test. Is there sufficient evidence to justify the rejection of 0H at the 0.01α= level? Explain. (20 pts)
27. A certain researcher thinks that the proportion of women who say that the earth is getting warmer is greater than the proportion of men.
In a random sample of 250 women, 70% said that the earth is getting warmer.
In a random sample of 220 men, 68% said that the earth is getting warmer.
At the 0.05 significance level, is there sufficient evidence to support the claim that the proportion of women saying the earth is getting warmer is higher than the proportion of men saying the earth is getting warmer? Show all work and justify your answer. (25 pts)
Refer to the following data for Questions 28 and 29.
:
X 0 – 1 1 2 3
Y 2 – 2 5 4 7
28. Is there a linear correlation between x and y at the 0.05 significance level? Justify your answer. (10 pts)
29. Find an equation of the least squares regression line. Show all work; writing the correct equation, without supporting work, will receive no credit. (15 pts)
30. The UMUC Daily News reported that the color distribution for plain M&M’s was: 40% brown, 20% yellow, 20% orange, 10% green, and 10% tan. Each piece of candy in a random sample of 100 plain M&M’s was classified according to color, and the results are listed below. Use a 0.05 significance level to test the claim that the published color distribution is correct. Show all work and justify your answer. (25 pts)
Color Brown Yellow Orange Green Tan
Number 45 13 17 7 18
________________________________________________________________________
yes please
well I need those and have some other questions I will need help on by Sunday
yeah he didn't give me one point. I can use a calculator or STATDISK
yes
its been awhile but I can get familiar
I am having trouble reading 22 26and data for 28. please help
Here is what I came up with after 3 hours of work.
a bonus would be greatly appreciated
https://www.box.com/s/nf41kksfq6rx7a30qlyg
So what I see right away is question 29 specifically says no credit will be given without supporting work. Number 8 is not answered.
ok I'll be waiting on that but thanks a lot and I have a few more questions like I said before but I need to have that back at the end of today. I can give them now or later if you want. But in the meantime I will go ahead and tip. Thanks a lot
These are the next set of questions
102 (16) Heights of Presidents and RunnersUp Theories have been developed about the heights of winning candidates for the U.S. presidency and the heights of candidates who were runnersup. Listed below are heights (in inches) from the recent president elections. Is there a linear correlation between the heights of candidates who won and the heights of the candidates who were runnersup?
Winner 69.5 73 73 74 74.5 74.5 71 71
Runnerup 72 69.5 70 68 74 74 73 76
(26) One classic application of correlation involves the association between the temperature and the number of times a cricket chirps in a minute. Listed below are the numbers of chirps in 1 min and the corresponding temperatures in ^{o}F (based on data from The Song of Insects by George W. Pierce, Harvard University Press). Is there a linear correlation between the number of chirps in 1 min and the temperature?
Chirps in 1 min 882 1188 1104 864 1200 1032 960 900
Temperature (^{o}F) 69.7 93.3 84.3 76.3 88.6 82.6 71.6 79.6
103 (16) ) Heights of Presidents and RunnersUp Find the best predicted height of runnerup Goldwater, given that the height of the winning presidential candidate Johnson in 75 in. Is the predicted height of Goldwater close to his actual height of 72 in?
Winner 69.5 73 73 74 74.5 74.5 71 71
Runnerup 72 69.5 70 68 74 74 73 76
(26) Crickets and Temperature Find the best predicted temperatures in (^{o}F ) at a time when a cricket chirps 3000 times in one minute. What is wrong with this predicted value?
Chirps in 1 min 882 1188 1104 864 1200 1032 960 900
Temperature (^{o}F) 69.7 93.3 84.3 76.3 88.6 82.6 71.6 79.6
104 (16) Global Warming Listed below are concentrations (in parts per million) of CO_{2} and temperatures (in ^{o}C) for different years (based on data from the Earth Policy Institute).
CO_{2} 314 317 320 326 331 339 346 354 361 369
Temperature 13.9 14.0 13.9 14.1 14.0 14.3 14.1 14.5 14.5 14.4
112 (8) Conduct the hypothesis test and provide the test statistic, critical value and/or Pvalue, and state the conclusion. Flat Tire and Missed Class A classical tale involves four carpooling students who missed a test and gave as an excuse a flat tire. On the makeup test, the instructor asked the students to identify the particular tire that went flat. If they really didn’t have flat tire, would they be able to identify the same tire? The author asked 41 other students to identify the tire they would select. The results are listed in the following table (except for one student who selected the spare). Use a 0.05 significance level to test the author’s claim that the results fit a uniform distribution. What does the result suggest ability of the four students to select the same tire when they really didn’t have a flat?
Tire Left Front Right front Left rear Right rear
Number selected 11 15 8 6
(24)
113 (18) Global Warming Survey A Pew Research poll was conducted to investigate opinions about warming. The respondents who answered yes when asked if there is solid evidence that the earth is getting warmer were then asked to select a cause of global warming. The results for two age brackets are given in the table below. Use the 0.01 significance level to test the claim that the age bracket is independent of the choice for the cause of global warming. Do respondents from both age brackets appear to agree, or is there a substantial difference?
Human activity Natural patterns Don’t know or refused to answer
Under 30 108 41 43 65 and over 121 71 43
(22) Injuries and Motorcycle Helmet Color A casecontrol (or retrospective) study was conducted to investigate a relationship between the colors of helmets worn by motorcycle drivers and whether they are injured or killed in a crash. Results are given in the table below (based on data from “Motorcycle Rider Conspicuity and Crash Related Injury: CaseControl Study,” by Wells et. al., BMJ USA, Vol. 4) Test the claim that injuries are independent of helmet color. Should motorcycle choose helmets with a particular with a color? If so, which color appears best
Color of Helmet
Black White Yellow/Orange Red Blue
Controls (not injured) 491 377 31 170 55
Case (injured or killed) 213 112 8 70 26
122 (14) Car Emissions Listed below are measured amounts of greenhouse gas emissions from cars in three different categories (from Data Set 16 in Appendix B). The measurements are in tons per year, expressed as CO2 equivalents. Use a 0.05 significance level to test the claim that the different car categories have the same mean amount of greenhouse gas emissions. Based on the results, does the number of cylinders appear to affect the amount of greenhouse gas emissions?
Four cylinder 7.2 7.9 6.8 7.4 6. 6.6 6.7 6.5 6.5 7.1 6.7 5.5 7.3
Six cylinder 8.7 7.7 7.7 8.7 8.2 9.0 9.3 7.4 7.0 7.2 7.2 8.2
Eight cylinder 9.3 9.3 9.3 8.6 8.7 9.3 9.3
26. Consider the hypothesis test given by
H 0 : µ = 530
H1 : µ ≠ 530.
In a random sample of 81 subjects, the sample mean is found to be population standard deviation is σ = 27.
x = 524. Also, the
Determine the Pvalue for this test. Is there sufficient evidence to justify the rejection of
H 0 at the α = 0.01 level? Explain. (20 pts)
22. Given a sample size of 65, with sample mean 726.2 and sample standard deviation 85.3, we perform the following hypothesis test.
H 0 : µ = 750
H1 : µ < 750
What is the conclusion of the test at the α = 0.10 level? Explain your answer. (20 pts)
Refer to the following data for Questions 28 and 29.
:
x 0 – 1 1 2 3
y 2 – 2 5 4 7
28. Is there a linear correlation between x and y at the 0.05 significance level? Justify your answer. (10 pts)
Uh its getting down to crunch time, please let me know if you can not make this deadline.
This completes the project. Please note the revisions to problems 9,10,11,22,26,28,29 thank you for the bonus. nice to work with you
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