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Confidence intervals and tests for these data use the t distribution with degrees of freedom

(a) 9.29.

(c) 15.

(e) 30.

(b) 14.

(d) 16.

Short Answer

Expert verified

(b) 14

Step by step solution

01

Given Information

Need to find in which answer is correct.

02

Explanation

Given:

n=16

The degrees of freedom is the sample size n decreased by 2 (because the least-squares regression equation involved the estimation of 2 parameters):

localid="1650543862751" df=n-2=16-2=14

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Most popular questions from this chapter

The school board in a certain school district obtained a random sample of 200residents and asked if they were in favor of raising property taxes to fund the hiring of more statistics teachers. The resulting confidence interval. for the true proportion of residents in favor of raising taxes was (0.183,0.257). The margin of error for this confidence interval is

(a) 0.037

(b) 0.183

c) 0.220

(d) 0.257

(e) 0.740

The cell that contributes most to the chi-square statistic is

(a) men who developed a rash.

(b) men who did not develop a rash.

(c) women who developed a rash.

(d) women who did not develop a rash.

(e) both (a) and (d).

Question refers to the following situation:

Could mud wrestling be the cause of a rash contracted by University of Washington students? Two physicians at the University of Washington student health center wondered about this when one male and six female students complained of rashes after participating in a mud-wrestling event. Questionnaires were sent to a random sample of students who participated in the event. The results, by gender, are summarized in the following table.

Some Minitab output for the previous table is given below. The output includes the observed counts, the expected counts, and the chi-square statistic.

A residual plot from the least-squares regression is shown below. Which of the following statements is supported by the graph

(a) The residual plot contains dramatic evidence that the standard deviation of the response about the population regression line increases as the average number of putts per round increases.

(b) The sum of the residuals is not 0. Obviously, there is a major error present.

(c) Using the regression line to predict a playerโ€™s total winnings from his average number of putts almost always results in errors of less than \(200,000.

(d) For two players, the regression line under predicts their total winnings by more than\)800,000.

(e) The residual plot reveals a strong positive correlation between average putts per round and prediction errors from the least-squares line for these players.

The bodyโ€™s natural electrical field helps wounds heal. If diabetes changes this field, it might explain why people with diabetes heal more slowly. A study of this idea compared randomly selected normal mice and randomly selected mice bred to spontaneously develop diabetes. The investigators attached sensors to the right hip and front feet of the mice and measured the difference in electrical potential (in millivolts) between these locations. Graphs of the data for each group reveal no outliers or strong skewness. The computer output below provides numerical summaries of the data26.

The researchers want to know if there is evidence of a significant difference in mean electrical potentials between normal mice and mice with diabetes. Carry out a test using a 5%level of significance and report your conclusion.

An old saying in golf is โ€œYou drive for show and you putt for dough.โ€ The point is that good putting is more important than long driving for shooting low scores and hence winning money. To see if this is the case, data from a random sample of 69of the nearly 1000players on the PGA Tourโ€™s world money list are examined. The average number of putts per hole and the playerโ€™s total winnings for the previous season is recorded. A least-squares regression line was fitted to the data. The following results were obtained from statistical software.

The correlation between total winnings and the average number of putts per hole for these players is

(a)-0.285

(b)-0.081

(c)0.007

(d)0.081

(e) 0.285

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