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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 1296students 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 30in sample size is required.

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

01

Part (a) Step 1: Given information

The data in summary form:

Group
n
x炉
sx
Males
675
$1884.52
$1368.37
Females
621
$1360.39
$1037.46
02

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.

03

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.

04

Part (b) Step 4: Explanation

Using a Ti-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 413.62and 634.64.

05

Part (c) Step 5: Given information

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

06

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.

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Most popular questions from this chapter

A large university is considering the establishment of a schoolwide recycling program. To gauge interest in the program by means of a questionnaire, the university takes separate random samples of undergraduate students, graduate students, faculty, and staff. This is an example of what type of sampling design?

(a) Simple random sample

(b) Stratified random sample

(c) Convenience sample

(d) Cluster sample

(e) Systematic sample

Men versus women The National Assessment of Educational Progress (NAEP) Young Adult Literacy Assessment Survey interviewed a random sample of 1917people 21to 25years old. The sample contained 840men and 1077women. 49The mean and standard deviation of scores on the NAEP's test of quantitative skills were x炉1=272.40and s1=59.2for the men in the sample. For the women, the results were x炉2=274.73and s2=57.5. Is the difference between the mean scores for men and women significant at the 1% level? Give appropriate statistical evidence to justify your answer.

A driving school wants to find out which of its two instructors is more effective at preparing students to pass the state鈥檚 driver鈥檚 license exam. An incoming class of 100students is randomly assigned to two groups, each of size 50. One group is taught by Instructor A; the other is taught by Instructor B. At the end of the course, 30of Instructor A鈥檚 students and 22of Instructor B鈥檚 students pass the state exam. Do these results give convincing evidence that Instructor A is more effective?

Min Jae carried out the significance test shown below to answer this question. Unfortunately, he made some mistakes along the way. Identify as many mistakes as you can, and tell how to correct each one.

State: I want to perform a test of

H0:p1-p2=0

Ha:p1-p2>0

where p1=the proportion of Instructor A's students that passed the state exam and p2=the proportion of Instructor B's students that passed the state exam. Since no significance level was stated, I'll use 蟽=0.05

Plan: If conditions are met, I鈥檒l do a two-sample ztest for comparing two proportions.

Random The data came from two random samples of 50students.

- Normal The counts of successes and failures in the two groups -30,20,22, and 28-are all at least 10.

- Independent There are at least 1000 students who take this driving school's class.

Do: From the data, p^1=2050=0.40and p^2=3050=0.60. So the pooled proportion of successes is

p^C=22+3050+50=0.52

- Test statistic

localid="1650450621864" z=(0.40-0.60)-00.52(0.48)100+0.52(0.48)100=-2.83

- p-value From Table A, localid="1650450641188" P(z鈮�-2.83)=1-0.0023=0.9977.

Conclude: The p-value, 0.9977, is greater than 伪=0.05, so we fail to reject the null hypothesis. There is no convincing evidence that Instructor A's pass rate is higher than Instructor B's.

Expensive ads Consumers who think a product鈥檚 advertising is expensive often also think the product must be of high quality. Can other information undermine this effect? To find out, marketing researchers did an experiment. The subjects were 90 women from the clerical and administrative staff of a large organization. All subjects read an ad that described a fictional line of food products called 鈥淔ive Chiefs.鈥� The ad also described the major TV commercials that would soon be shown, an unusual expense for this type of product. The 45women who were randomly assigned to the control group read nothing else. The 45in the 鈥渦ndermine group鈥� also read a news story headlined 鈥淣o Link between Advertising Spending and New Product Quality.鈥� All the subjects then rated the quality of Five Chefs products on a seven-point scale. The study report said, 鈥淭he mean quality ratings were significantly lower in the undermine treatment (xA  4.56) than in the control treatment x炉C=5.05;t=2.64,P<0.01.

(a). Explain why the Random and Independent conditions are met in this case.

(b) The distribution of individual responses is not Normal, because there is only a seven-point scale. Why is it still proper to use a two-sample t-test?

(c) Interpret the P-value in context.

A random sample of 100 of a certain popular car model last year found that 20 had a certain minor defect in the brakes. The car company made an adjustment in the production process to try to reduce the proportion of cars with the brake problem. A random sample of 350 of this year's model found that 50 had the minor brake defect.

(a) Was the company's adjustment successful? Carry out an appropriate test to support your answer.

(b) Describe a Type I error and a Type Il error in this setting, and give al possible consequence of each.
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