/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 75 A company that has many departme... [FREE SOLUTION] | 91Ó°ÊÓ

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A company that has many department stores in the southern states wanted to find at two such stores the percentage of sales for which at least one of the items was returned. A sample of 800 sales randomly selected from Store A showed that for 280 of them at least one item was returned. Another sample of 900 sales randomly selected from Store B showed that for 279 of them at least one item was returned. a. Construct a \(98 \%\) confidence interval for the difference between the proportions of all sales at the two stores for which at least one item is returned. b. Using the \(1 \%\) significance level, can you conclude that the proportions of all sales for which at least one item is returned is higher for Store A than for Store \(B\) ?

Short Answer

Expert verified
a. The 98% confidence interval for the difference between the proportions of all sales at the two stores for which at least one item is returned is \(-0.0317, 0.1117\).\nb. Using the 1% significance level, we cannot conclude that the proportion of all sales for which at least one item is returned is higher for Store A than for Store B.

Step by step solution

01

Identify the Sample Proportions for Each Store

Calculate the proportion of sales which had at least one item returned for each store. For Store A, this is \(\frac{280}{800} = 0.35\) and for Store B, this is \(\frac{279}{900} = 0.31\).
02

Calculate the Standard Error

The standard error \(SE\) for the difference in proportions is calculated using the formula: \(SE = \sqrt{\frac{{p1(1-p1)}}{{n1}} + \frac{{p2(1-p2)}}{{n2}}}\) where \(p1\) and \(p2\) are the proportions in store A and B, and \(n1\) and \(n2\) are the number of sales in store A and B. Substituting the given values, \(SE = \sqrt{\frac{{0.35(1-0.35)}}{{800}} + \frac{{0.31(1-0.31)}}{{900}}} = 0.03075 \)
03

Construct the Confidence Interval

The confidence interval is given by \((p1 - p2) \pm Z_{\alpha/2} * SE\). Here, \(\alpha = 0.02\) since the level of confidence is 98%, so \(Z_{\alpha/2} = Z_{0.01} = 2.33\). Substituting the values, we find the confidence interval to be \((0.35 - 0.31) \pm 2.33 * 0.03075 = 0.04 \pm 0.0717\), which is equivalent to \(-0.0317, 0.1117\)
04

Perform the Hypothesis Test

We perform a one-sided z-test. The null hypothesis \(H0\) is \(p1 <= p2\) and the alternative hypothesis \(H1\) is \(p1 > p2\). The test statistic is \(z = \frac{p1 - p2}{SE} = \frac{0.35 - 0.31}{0.03075} = 1.30\). Using a 1% significance level, the critical value from the z-table is 2.33. Since the test statistic of 1.30 is less than 2.33, we fail to reject the null hypothesis.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Confidence Interval
A confidence interval is a range of values that estimates a population parameter based on sample data. It provides an upper and a lower limit within which we can say, with a certain level of confidence, that the true value of the parameter lies.
In hypothesis testing, especially when comparing proportions as in our exercise, constructing a confidence interval helps us understand the difference in proportions between two groups or samples.
For instance, when we say we have a 98% confidence interval for the difference in store return rates, it means we are 98% sure that the true difference between the two stores' return rates falls within this calculated range.
The confidence interval for the difference in proportions is calculated using the formula: \[(p1 - p2) \pm Z_{\alpha/2} \times SE\] Where:
  • \(p1, p2\) are the sample proportions
  • \(Z_{\alpha/2}\) is the z-value corresponding to the desired confidence level
  • \(SE\) is the standard error calculated for these proportions
This ensures we incorporate both sampling variation and desired confidence level when estimating the parameter range.
Proportions
Proportions represent the fraction of the total sample that holds a particular characteristic. In the context of our exercise, we are looking at the proportion of sales with returned items at each department store.
Calculating these proportions involves dividing the number of favorable outcomes by the total number of events. So for Store A, the proportion is: - \(\text{Proportion A} = \frac{280}{800} = 0.35\)
Similarly, for Store B, the calculation is: - \(\text{Proportion B} = \frac{279}{900} = 0.31\)
Using proportions provides a more standardized way to compare different groups since it accounts for varied sample sizes. Whether using smaller or larger samples, proportions allow us to see the ratio of occurrences in a consistent manner.
Standard Error
The standard error (SE) is a measure of the variability or dispersion of sample statistics from the true population parameter. In simpler terms, it shows how much the sample proportion could vary from the actual population proportion due to sampling errors.
For proportions, the formula for standard error is: \[SE = \sqrt{\frac{p1(1-p1)}{n1} + \frac{p2(1-p2)}{n2}}\]Where:
  • \(p1, p2\) are the sample proportions of Store A and Store B
  • \(n1, n2\) are the respective sample sizes of Store A and Store B
The calculated SE helps us understand the precision of our estimate when comparing two proportions.
In our solution, the SE was found to be approximately 0.03075, indicating the likely fluctuation in the sample difference of return rates due to sampling randomness.
Significance Level
The significance level, often denoted as \(\alpha\), represents the probability of rejecting the null hypothesis when it is actually true. It is the benchmark for deciding whether observed outcomes are statistically significant or just due to chance.
In our problem, a 1% significance level was used, meaning there is a 1% risk of concluding that the proportion of returned sales in Store A is higher than Store B when it is not.
This critical value derived from the significance level is crucial in hypothesis testing as it dictates the threshold for our test statistic. If the test statistic exceeds the critical value, we reject the null hypothesis.
However, in this exercise, the test statistic was 1.30, which is less than the critical value of 2.33 at the 1% significance level, leading us to fail to reject the null hypothesis.

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Most popular questions from this chapter

Assuming that the two populations have unequal and unknown population standard deviations, construct a \(99 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) for the following. $$ \begin{array}{lll} n_{1}=48 & \bar{x}_{1}=.863 & s_{1}=.176 \\ n_{2}=46 & \bar{x}_{2}=.796 & s_{2}=.068 \end{array} $$

The following information was obtained from two independent samples selected from two normally distributed populations with unknown but equal standard deviations. $$ \begin{array}{lllllllllllll} \text { Sample 1: } & 2.18 & 2.23 & 1.96 & 2.24 & 2.72 & 1.87 & 2.68 & 2.15 & 2.49 & 2.05 & & \\ \text { Sample 2: } & 1.82 & 1.26 & 2.00 & 1.89 & 1.73 & 2.03 & 1.43 & 2.05 & 1.54 & 2.50 & 1.99 & 2.13 \end{array} $$ a. Let \(\mu_{1}\) be the mean of population 1 and \(\mu_{2}\) be the mean of population \(2 .\) What is the point estimate of \(\mu_{1}-\mu_{2} ?\) b. Construct a \(99 \%\) confidence interval for \(\mu_{1}-\mu_{2}\). c. Test at the \(2.5 \%\) significance level if \(\mu_{1}\) is lower than \(\mu_{2}\).

A sample of 500 observations taken from the first population gave \(x_{1}=305\). Another sample of 600 observations taken from the second population gave \(x_{2}=348\). a. Find the point estimate of \(p_{1}-p_{2}\). b. Make a \(97 \%\) confidence interval for \(p_{1}-p_{2}\). c. Show the rejection and nonrejection regions on the sampling distribution of \(\hat{p}_{1}-\hat{p}_{2}\) for \(H_{0}: p_{1}=p_{2}\) versus \(H_{1}: p_{1}>p_{2} .\) Use a significance level of \(2.5 \% .\) d. Find the value of the test statistic \(z\) for the test of part \(\mathrm{c}\). e. Will you reject the null hypothesis mentioned in part \(\mathrm{c}\) at a significance level of \(2.5 \%\) ?

The following information is obtained from two independent samples selected from two normally distributed populations. $$ \begin{array}{lll} n_{1}=18 & \bar{x}_{1}=7.82 & \sigma_{1}=2.35 \\ n_{2}=15 & \bar{x}_{2}=5.99 & \sigma_{2}=3.17 \end{array} $$ a. What is the point estimate of \(\mu_{1}-\mu_{2}\) ? b. Construct a \(99 \%\) confidence interval for \(\mu_{1}-\mu_{2}\). Find the margin of error for this estimate.

Construct a \(95 \%\) confidence interval for \(p_{1}-p_{2}\) for the following. $$ n_{1}=100 \quad \hat{p}_{1}=.81 \quad n_{2}=150 \quad \hat{p}_{2}=.77 $$

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