Q. 45 45. Paying for college College f... [FREE SOLUTION]

91影视

The Practice of Statistics for AP

David Moore,Daren Starnes,Dan Yates

$Math Studyset 91影视 Explanations$ Math

4th Edition 路 809 pages

ISBN: 9781319113339

Chapter 10: Q. 45 (page 654)

45. Paying for college College financial aid offices expect students to use summer earnings to help pay for college. But how large are these earnings? One large university studied this question by asking a random sample of $1296$ students who had summer jobs how much they earned. The financial aid office separated the responses into two groups based on gender. Here are the data in summary form:
(a) How can you tell from the summary statistics that the distribution of earnings in each group is strongly skewed to the right? A graph of the data reveals no outliers. The use of two-sample t procedures is still justified. why?
(b) Construct and interpret a $90 %$ confidence interval for the difference between the mean summer earnings of male and female students at this university.
(c) Interpret the $90 %$ confidence level in the context of this study.

Short Answer

Expert verified

(a) The standard deviation is nearly identical to the mean. At least $30$ in sample size is required.

(b) The mean difference in earnings has a $90 %$ of probability is between $413.62$ and $634.64$ .

(c) The confidence interval reveals that on average $90 %$ of the samples had confidence intervals reflecting the mean difference in men and women's earnings.

Step by step solution

Part (a) Step 1: Given information

The data in summary form:

Group	n	$\overset{炉}{x}$	$s_{x}$
Males	675	$1884.52	$1368.37
Females	621	$1360.39	$1037.46

Part (a) Step 2: Explanation

Since the earnings cannot be negative and the standard deviation is nearly identical to the mean. Because the negative values appear within two standard deviations of the mean, these values lie inside the range of regular values for asymmetric distribution, hence the distribution must be substantially skewed.
The sample sizes $(675 / 621)$ are at least $30$ and the distribution is yet considered to be almost normal.
The standard deviation is nearly identical to the mean.
At least $30$ in sample size is required.

Part (b) Step 3: Given information

A $90 %$ confidence interval for the difference between the mean summer earnings of male and female students at this university.

Part (b) Step 4: Explanation

Using a $T i - 83$ calculator, the $90 %$ confidence interval for the difference between the mean wages of female and male university students can be determined as follows:

The confidence interval thus becomes $(413.62, 634.64)$ .
Therefore, the mean difference in earnings has a $90 %$ of probability is between $413.62$ and $634.64$ .

Part (c) Step 5: Given information

To interpret the $90 %$ confidence level in the context of the study.

Part (c) Step 6: Explanation

The $90 %$ confidence interval means that $90 %$ of all potential data will have a confidence interval that includes the genuine population mean difference in earnings.

Hence, the confidence interval reveals that on average $90 %$ of the samples had confidence intervals reflecting the mean difference in men and women's earnings.