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46. Happy customers As the Hispanic population in the United States has grown, businesses have tried to understand what Hispanics like. One study inter-viewed a random sample of customers leaving a bank. Customers were classified as Hispanic if they preferred to be interviewed in Spanish or as Anglo if they preferred English. Each customer rated the importance
of several aspects of bank service on a 10-point scale. Here are summary results for the importance of 鈥渞eliability鈥 (the accuracy of account records and so on):

(a) The distribution of reliability ratings in each group is not Normal. A graph of the data reveals no outliers. The use of two-sample t procedures is
justified. Why?
(b) Construct and interpret a 95% confidence interval for the difference between the mean ratings of the importance of reliability for Anglo and
Hispanic bank customers.
(c) Interpret the 95% confidence level in the context of this study.

Short Answer

Expert verified

(a) The distribution is about normal because the sample sizes (92/86)are at least 30.

(b) The mean difference in ratings has a 95%of probability is between 0.2264and 0.6936.

(c) The confidence interval level shows that on an average 95%of the samples have confidence interval that are containing the mean difference of the ratings of the reliability for Anglo and Hispanic bank customers

Step by step solution

01

Part (a) Step 1: Given information

The summary results for the importance of 鈥渞eliability鈥 is:

Group
n
x
sx
Anglo
92
6.37
0.60
Hispanic
86
5.91
0.93
02

Part (a) Step 2: Explanation

Let, assume the sample sizes are 92and 86, which are both greater than the required sample size of 30, implying that the distribution is normally distributed..

Therefore, the distribution is about normal because the sample sizes(92/86) are at least 30.
At least 30 in sample size is required.

03

Part (b) Step 3: Given information

A 95%confidence interval for the difference between the mean ratings of the importance of reliability for Anglo and Hispanic bank customers.

04

Part (b) Step 4: Explanation

Using a Ti-83 calculator, the 95%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(0.2264,0.6936)

Therefore, the mean difference in ratings has a 95%of probability is between 0.2264and 0.6936.

05

Part (c) Step 5: Given information

To interpret the 95%confidence level in the context of this study.

06

Part (c) Step 6: Explanation

Since, the 95%confidence interval means that 95% of all available samples will have a confidence interval that includes the genuine population mean difference between the mean ratings of the significance of reliability for Anglo and Hispanic bank customers.

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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 =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

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- 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-2.83)=1-0.0023=0.9977.

Conclude: The p-value, 0.9977, is greater than =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.

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