Oh, ok. Well that's not too unusual, but it changes things a little bit. Before anybody commits to anything, can you give me a little more information about what this might entail? How much time do you have to complete it? What course material do you expect to be covered?

I can take a break between parts You just tell me and I will stop after a part, but once I start a part I have 3 hours to complete it. This is Intermediate Statistics.

Do you recall what the last topic covered was? I'm trying to get some kind of feel for what portion of the whole statistics spectrum might be in play. And it wouldn't be a good thing for you if we were to find out too late that this includes material that I'm not familiar with.

Also, other than the actual limit for each section, is there a deadline for when you need to have all four parts completed? A 12-hour stretch to do this all in one shot might be a bit unreasonable.

Question 1Assume that you are using a significance level of α = 0.05 to test the claim that p1 = p2. Use the given sample sizes and the number of hits to find the z test statistic for hypothesis testing. A report on the nightly news program said that 10 108 households with dogs were stolen and 20 208 without dogs were stolen.A) z = -0041B) z = -0.102C) z = 0.000D) z = -0173

B. z = -0.102 Note: Be careful about copying and pasting information. Sometimes parts of the problem statement are not text, and they tend to disappear. The "10 108 households" part looks odd, but I was able to figure out what it meant. If you can take screenshots and post those, issues like this might be avoided.

2. Build confidence interval indicated by the difference between the two population means. Suppose that the two samples are independent simple random samples selected from populations with normal distribution. Do not assume that the population standard deviations are equal. A researcher was interested in comparing the resting pulse rate of people who exercise regularly and pulse rates of people who do not exercise regularly. She obtained independent simple random samples of 16 people who do not exercise regularly and 12 people who exercise regularly. Pulse rates at rest (beats per minute) were recorded and summary statistics are as follows."DO NOT EXERCISE REGULARLY _X1 = 72.3 BEATS / MIN S1 = 10.9 BEATS / MIN N1 = 16Exercise regularly _X2 = 68.1 BEATS / MIN S2 = 8.2 BEATS / MIN N2 = 12 Construct a confidence interval of 95% for μ1 - μ2, the difference between the average pulse frequency of people who do not exercise regularly and pulse frequency of average people who exercise regularly.-3.22 Beats / min <μ1 - μ2 <11.62 beats / min-3.55 Beats / min <μ1 - μ2 <11.95 beats / min-3.74 Beats / min <μ1 - μ2 <12.14 beats / min-4.12 Beats / min <μ1 - μ2 <14.72 beats / min

Suppose the data are normally distributed and the number of observations is greater than fifty. Find the critical value z used to test the null hypothesis. α = 0.05 for a two-tailed test.A) ± 1.96B) -1.96C) ± 1.64D) 1.96

Use the given data to find the equation of the regression line. Rounding the final values to three significant digits, if necessary.X 1 3 5 7 9 Y 143 116 100 98 90A) Y = -150.7 + 6.8xB) Y = 150.7 - 6.8xC) Y = 140.4 - 6.2xD) Y = -140.4 + 6.2x

Use a computer program to find the multiple regression equation. Can the equation be used for prediction? An anti-snuff group used the data in the table to link carbon monoxide various brands of cigarettes and their contentART CO NIC 15 1.2 16 15 1.2 16 17 1.0 16 6 0.8 9 1 0.1 1 8 0.8 8 10 0.8 10 17 1.0 16 15 1.2 15 11 0.7 9 18 1.4 18 16 1.0 15 10 0.8 9 7 0.5 5 18 1.1 16 CO = carbon monoxide TAR = tar NIC = nicotinea) CO = 1.37 - 5.53TAR + 133NIC; Yes, because the R2 is high b) CO = 1.37 + 5.50TAR - 138NIC; Yes, because the p value is high c) CO = 1.3 + 5.5TAR - 1.3NIC; Yes, because the R2 is low d) CO = 1.25 + 1.55TAR - 5.79NIC; Yes because the p is low

Customer:replied 1 year ago.

Find the unexplained variation to the data below. Paired data show test scores and hours of preparation for 5 students randomly selected. The equation of the regression line is = 44.8447 + 3.52427 x. Find the variation does not explcada. x hours of preparation | 2 May 9 June 10 ____________________ | _______________________ | and test scores | 64 48 72 73 80A) 511 724B) 87.4757C) 599.2D) 96 103

Going back to the one about the cigarettes, can you double-check the column headings? I'm not getting any of the given answer choices for the regression equation. I suspect that maybe the column labels are in the wrong order.

7. Find standard error of the estimate for the data below. Build the prediction interval for an individual and indicated. Paired data show test scores and hours of preparation for 5 students randomly selected. The equation of the regression line is = 44.8447 + 3.52427 x and standard error of estimate is SE = 5.40. Find the prediction interval 99% for test score of a person who spent 7 hours preparing for the exam. x hours of preparation | 2 May 9 June 10 ____________________ | _______________________ | and test scores | 64 48 72 73 80A) 58 <y <82B) 62 <y <78C) 35 <y <104D) 32 <y <104

For the "test scores" problem, it seems like there is missing data, or mangled data. What do "May" and "June" have to do with this problem? It seems that there should be a set of study times, and a set of test scores. Can you please clarify?

Find standard error of the estimate for the data below. Build the prediction interval for an individual and indicated. Paired data show test scores and hours of preparation for 5 students randomly selected. The equation of the regression line is Y = 44.8447 + 3.52427 x and standard error of estimate is SE = 5.40. Find the prediction interval 99% for test score of a person who spent 7 hours preparing for the exam. x hours of preparation 5 2 9 6 10 ____________________ | _______________________ | and test scores | 64 48 72 73 80A) 58 <y <82B) 62 <y <78C) 35 <y <104D) 32 <y <104

the other one for the cigarette CO TAR NIC 15 1.2 16 15 1.2 16 17 1.0 16 6 0.8 9 1 0.1 1 8 0.8 8 10 0.8 10 17 1.0 16 15 1.2 15 11 0.7 9 18 1.4 18 16 1.0 15 10 0.8 9 7 0.5 5 18 1.1 16

Customer:replied 1 year ago.

It comes out like that 15 1.2 16

Customer:replied 1 year ago.

try your best...

Customer:replied 1 year ago.

Use the information given to calculate the coefficient of determination. A regression equation is obtained for a set of paired data. It was found that the total variation is 24.488, explained variation is 15,405, and the unexplained variation is 9,083. Calculate the coefficient of determinationA) 0629B) 0590C) 1590D) 0.371

Customer:replied 1 year ago.

9.Dados the linear correlation coefficient r and the sample size n, determine the críiticos r values and use them to establish whether the given r represents a significant or linear correlation. Use a significance level of 0.05. r = 0.75, n = 9A) Critical values: r = ± 0.666, there is a significant linear correlationB) Critical values: r = 0.666, there is a significant linear correlationC) Critical values: r = - 0.666, there is a significant linear correlationD) Critical values: r = ± 0.666, significant correlation Linal

Customer:replied 1 year ago.

Use the given data to find the best predicted value of the response variable. Six pairs of data gives r = 0.789 and the regression equation What is the best predicted for x = 5 value?A) 22.0B) 18.0C) 18.5D) 19.0

The data itself was fine, it was the headings that were wrong. For the cigarette question, the solution is a) CO = 1.37 - 5.53TAR + 133NIC; Yes, because the R2 is high For the one about unexplained variation, the solution is B) 87.4757 For the one about the 99% predication interval, the answer choices are not intervals. Is there something wrong there? For the one about the coefficient of determination, the solution is A) 0.629 For the one about the linear correlation coefficient, the solution is A) Critical values: r = ± 0.666, there is a significant linear correlation For the last one, what is the regression equation?

Find standard error of the estimate for the data below. Build the prediction interval for an individual and indicated. Paired data show test scores and hours of preparation for 5 students randomly selected. The equation of the regression line is Y = 44.8447 + 3.52427 x and standard error of estimate is SE = 5.40. Find the prediction interval 99% for test score of a person who spent 7 hours preparing for the exam. x hours of preparation 5 2 9 6 10 ____________________ | _______________________ | and test scores | 64 48 72 73 80A) 58 <y <82B) 62 <y <78C) 35 <y <104D) 32 <y <104

I don't know what to do with that one. It seems to be asking for the standard error of the estimate, and then it gives a value of 5.40, which then doesn't match with any of the answer choices. And then it also asks for a prediction interval, but the answer choices are not in the form of intervals.

Find standard error of the estimate for the data below. Paired data show test scores and hours of preparation for 5 students randomly selected. The equation of the regression line is Y = 44.8447 + 3.52427 x. Find standard error of the estimate. x hours of preparation X 5 2 9 6 10 ____________________ | _______________________ | Y 64 48 72 73 80A) 4.1097B) 7.1720C) 5.3999D) 13 060

Find the total variation for the data below. Paired data show test scores and hours of preparation for 5 students randomly selected. The equation of the regression line is Y = 44.8447 + 3.52427 x. Find the total variation. x hours of preparation X 5 2 9 6 10_________________ | _______________________ | and test scores Y 64 48 72 73 80A) 599.2B) 498 103C) 511,724D) 87.4757

Customer:replied 1 year ago.

Find the variance explained for the following data. Paired data show test scores and hours of preparation for 5 students randomly selected. The equation of the regression line is = 44.8447 + 3.52427 x. Find the explcada variation. x hours of preparation X 5 2 9 6 10____________________ | _______________________ | and test scores Y 64 48 72 73 80A) 498 103B) 511 724C) 87.4757D) 599.2

Customer:replied 1 year ago.

Use a computer program to obtain multiple regression equation. Use the estimated equation to find the predicted value. A health specialist collected the data in the table to see if the pulses can be explained by exercise and smoking. For exercise, he assigned 1 to if and 2 to no. For smoking, if assigned 1 to 2 to no. Then use the results to predict the pulse rate of a person whose value was 1 year and whose value was smoking PULSO EXERCISE 1. SMOKING97 2 2 88 1 2 69 1 2 67 1 2 83 1 2 77 1 2 66 2 2 78 2 2 73 1 1 67 1 1 55 1 2 82 1 1 70 1 2 55 1 2 76 1 2A) 70 beats / minB) 74 beats / minC) 81 beats / minD) 77 beats / min

Use a computer program to find the multiple regression equation, R2, R2 and p-value adjusted. An anti-snuff group used the data in the table to link carbon monoxide various brands of cigarettes and content of tar and nicotine. Note: CO = carbon monoxide TAR = tar NIC = nicotine CO TAR NIC 15 1.2 16 15 1.2 16 17 1.0 16 6 0.8 9 1 0.1 1 8 0.8 8 10 0.8 10 17 0.1 16 15 1.2 15 11 0.7 9 18 1.4 18 16 1.0 1.5 10 0.8 9 7 0.5 5 18 1.1 16A) 0976, 0921, 0002B) 0.943, 0.934, 0.000C) 0.931, 0.902, 0.000D) 0861, 0900, 0015

Use the results shown to answer the question. A collection of data pairs is the number of years students have studied Spanish and their scores on a test of Spanish language skills. a computer program was used to obtain the linear least squares regression, and computer output shown below. Along with the data of paired samples, the program also gave a value x 2 (years of study) to be used to predict the test score.The regression equation isScore = 31.55 + 10.9 yearsPredictor Coef. T PConstant 31.55 6.36 4.96 0,00010.90 1,744 6.25 0,000 yearsS = 5,651 R2 R2 = 83.0% (Adjusted) = 82.7%predicted valuesEst IP 95% 95%3,168 53.35 (42.72, 63.98) (31.61,75.09)SD: standard deviationWhat percent of the total variation in test scores can be explained by the linear relationship between the years of study and test scores?A) 82.7%B) 17.0%C) 91.1%D) 83.0%