Accepted Answer

Dear XXXXXXXX,

Great questions! Here are your answers. Btw, can I ask a quick question? Are these for an online class or something? I have seen these questions several times lately.

Here are your answers:

Part I T/F & Multiple Choice (20 points)

1. T

2. F

3. F

4. F

5. F

6. F

7. F

8. The symbols didn't come out. But the answer is pM<=pF.
      
9. The symbols didn't come out, but the answer is H0: mu <= 6.5 vs. Ha: mu > 6.5

10. It's sigma. The symbols didn't come out.

Part II. Short Answers & Computational Questions

1. A machine produces 3-inch nails. A sample of 10 nails was selected and the lengths determined. The results are as follows:
2.89     2.95     3.00     3.05    ; 2.99     2.96     3.10     3.06     3.00     3.12
Use these results to test H0: ? = 3 and Ha: ? ? 3 at ? = 0.01. Give the critical region, the computed test statistic, and the conclusion. (6 pts)

Hypotheses didn't come out. Here's what I think they should be:

H0: mu=3
Ha: mu not equal to 3.
alpha = 0.01.

The critical value is 3.690.

The mean of the data is 3.012, and the standard deviation is 0.071. So the test statistic is

t=(3.012 - 3)/(0.071/sqrt(10)) = 0.535.

So, the test statistic does not exceed the critical value, so we do not reject the null hypothesis.

2. A machine produces 3-inch nails. A sample of 100 nails is selected, and it is found that 25 are shorter than 3 inches. Find the 95% confidence interval on the proportion of all such nails that are shorter than 3 inches (4 pts)

25/100 +/- 1.96*sqrt(.25*.75/100) = (0.165, 0.335)

3. A sample of size n = 14 is selected from a normal population to construct a 95% confidence interval estimate for a population mean. The interval was computed to be (7.82 to 9.64). Determine the sample standard deviation. (4 pts)

the sample average was (9.64+7.82)/2 = 8.73

So the sample standard deviation is calculated by solving the following education.

7.82 = 8.73 - 2.160(s/sqrt(14)) --> 0.91 = 2.160*s/sqrt(14) --> s=1.576

4. Determine the p-value for each of the following hypothesis-testing situations: (4 pts)
a. H0: p = 0.25 and Ha: p ? 0.25; z test value = 1.84 -- assuming that the alternative is not-equal. p=0.066

b. H0: ? ? 13.5 and Ha: ? < 13.5; t test value = -1.94 d.f. = 10 -- p=0.041

5. An insurance company states that 90% of its claims are settled within 5 weeks. A consumer group selected a random sample of 100 of the company’s claims and found 75 of the claims were settled within 5 weeks. Is there enough evidence to support the consumer group’s claim that fewer than 90% of the claims were settled within 5 weeks? Test using the traditional approach with ? = 0.05 (6 pts)

H0: p>=0.9
Ha: p<0.9

critical value = -1.96

test statistic: (.75-.9)/sqrt(.75*.25/100) = -3.464

Since the test statistic exceeds the critical value, we reject the null hypothesis. The rate of settling claims is not 90%.

6. A random sample of 51 observations was selected from a normally distributed population. The sample mean was = 88.6, and the sample variance was s2 = 38.2. Does the sample show sufficient reason to conclude that the population standard deviation is not equal to 8 at the 0.05 level of significance? Use the p-value method. (6 pts)

H0: sigma = 8
Ha: sigma not equal to 8

critical value: 68.67

The test statistic is: s^2(n-1)/sigma^2 = 38.2(50)/64 = 29.84375.

The p-value is 0.98.

Since the p-value greatly exceeds the alpha value, there is not enough evidence to reject the null hypothesis. There is no reason to thing that the population standard deviation is not 8.

7. A researcher wishes to compare two different groups of students with respect to their mean time to complete a particular task. The time required is determined for each group. The data summary is given below. Test the claim at ?= 0.05, that there is no difference in variance. Give the critical region, test statistic value, and conclusion for the F test. (6 points)
n1=100, s1=20
n2=120, s2=34

Um... I'm going to need the sample averages to do this, I think.


8. The number of sit-ups that a person can do in one minute, both before and after a physical fitness course was recorded. The data is shown below for 10 randomly selected participants. Is there sufficient evidence to conclude that a significant amount of improvement took place? Assume variable is normally distributed. Test using ? = 0.01. (6 pts)

Before    ; 30     23     25    &nb sp;29     27     25     30     45     34     26
After     31     27     25     35&nb sp;    34     35     32     52     50     43

Paired T-Test and CI: Before, After

Paired T for Before - After

             N   Mean StDev SE Mean
Before      10 29.40   6.35     2.01
After       10 36.40   9.12     2.88
Difference 10 -7.00   5.87     1.86


T-Test of mean difference = 0 (vs not = 0): T-value=-3.77 P-Value = 0.002

Since alpha>p-value, we reject the null hypothesis. The difference is not 0.


9. You are given the following data. Test the claim that there is a difference in the means of the two groups. Use ? = 0.10. (6 pts)

Group A        Group B

There is no data here. Go ahead and add it and I'll help!

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Let me know if you have questions about this. I'd be glad to clarify any points!

Expert:	Greg
Pos. Feedback:	99.6 %
Accepts:
Answered:	2/9/2009

Graduate Student

I teach math and statistics at a college level as a graduate teaching assistant