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91Ó°ÊÓ

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 a \(1 \%\) 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 \(\mathrm{B}\) ?

Short Answer

Expert verified
a. The 98\% confidence interval for the difference between the proportions of all sales with at least one item returned at the two stores is calculated using the given sample proportions and standard error.\nb. Using 1\% significance level, if the p-value is less than 0.01, then we can conclude that the proportion of all sales with at least one item returned is higher for Store A than for Store B. The p-value is determined based on the test statistic z-score.

Step by step solution

01

Calculate Sample Proportions

For store A, the sample proportion (p_A) of sales with at least one returned item is 280/800 = 0.35. For store B, the sample proportion (p_B) of sales with at least one returned item is 279/900 = 0.31.
02

Construct the Confidence Interval

To construct a 98\% confidence interval for the difference between the proportions (p_A - p_B), we first find the standard error (SE) of the difference using the formula: SE = sqrt[(p_A*(1-p_A)/n_A) + (p_B*(1-p_B)/n_B)]. Here, n_A=800 and n_B=900. The critical value (z) for a 98\% confidence interval can be found from the standard normal table, which is approximately 2.33. The confidence interval is thus (p_A - p_B) ± z*SE.
03

Perform the Hypothesis Test

The null hypothesis is 'There is no difference between the proportions of all sales where at least one item is returned between store A and store B'. The alternative hypothesis is 'The proportion of all sales where at least one item is returned is higher for store A than for store B'. We find the test statistic using the formula: z = (p_A - p_B - 0)/SE. The p-value is then the area under the normal curve to the right of this z-score. If p-value < 0.01 (significance level), we reject the null hypothesis.

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Key Concepts

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

Sample Proportions
Sample proportions give us a way to understand a part of a whole by looking at just a sample of it. In this case, for Store A, 280 out of 800 sales had items returned, giving us a sample proportion for Store A as \( p_A = \frac{280}{800} = 0.35 \). Similarly, for Store B, 279 out of 900 sales had items returned, providing Store B's sample proportion as \( p_B = \frac{279}{900} = 0.31 \). This helps us summarize significant data about the two stores' sales.
  • Sample proportions are often expressed in decimal form.
  • They provide insight into the behavior of larger populations.
Understanding sample proportions allows us to make statistical inferences about the whole group from which the sample was drawn.
Hypothesis Testing
Hypothesis testing is a method to determine if there's enough evidence to support a particular belief about a parameter. In our scenario, we want to test if Store A has a higher proportion of returns than Store B. The null hypothesis \( H_0 \) states that there is no difference in return proportions between the two stores. In contrast, the alternative hypothesis \( H_a \) suggests that the proportion is higher for Store A. The test involves computing a test statistic, which helps in making a decision.
  • If the test statistic is significant, we might reject the null hypothesis.
  • Decisions are made based on a comparison of the calculated p-value with a pre-determined significance level.
Hypothesis testing helps in making data-driven decisions with a defined level of confidence.
Significance Level
The significance level, often denoted by \( \alpha \), is a threshold used to determine whether a hypothesis test result is statistically significant. In our case, a 1% significance level \( (\alpha = 0.01) \) is used.Using a lower significance level means we're looking for stronger evidence to reject the null hypothesis. For instance, with \( \alpha = 0.01 \), we only reject the null hypothesis if there's less than a 1% chance that the observed data could occur if the null hypothesis were true.
  • Common significance levels are 0.05 (5%) and 0.01 (1%).
  • A lower significance level reduces the risk of a Type I error (incorrectly rejecting a true null hypothesis).
The significance level helps control the confidence with which we make statistical inferences.
Standard Error
Standard error provides a measure of variability or deviation from the mean in a sample distribution. For proportions, it helps in constructing confidence intervals and hypothesis tests. In our example, the standard error for the difference in proportions is calculated as:\[SE = \sqrt{\left( \frac{p_A(1-p_A)}{n_A} \right) + \left( \frac{p_B(1-p_B)}{n_B} \right)}\]Here, \( n_A = 800 \) and \( n_B = 900 \) are the sample sizes for stores A and B respectively.
  • Standard error decreases with larger sample sizes.
  • It helps determine how much the sample mean is expected to vary from the actual population mean.
Understanding standard error allows us to estimate the precision of the sample proportions.

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Most popular questions from this chapter

According to a Bureau of Labor Statistics report released on March 25,2015 , statisticians earn an average of \(\$ 84,010\) a year and accountants and auditors earn an average of \(\$ 73,670\) a year (www.bls. gov). Suppose that these estimates are based on random samples of 2000 statisticians and 1800 accountants and auditors. Further, assume that the sample standard deviations of the annual earnings of these two groups are \(\$ 15,200\) and \(\$ 14,500\), respectively, and the population standard deviations are unknown but equal for the two groups. a. Construct a \(98 \%\) confidence interval for the difference in the mean annual earnings of the two groups, statisticians and accountants and auditors. b. Using a \(1 \%\) significance level, can you conclude that the average annual earnings of statisticians is higher than that of accountants and auditors?

The standard recommendation for automobile oil changes is once every 5000 miles. A local mechanic is interested in determining whether people who drive more expensive cars are more likely to follow the recommendation. Independent random samples of 45 customers who drive luxury cars and 40 customers who drive compact lowerprice cars were selected. The average distance driven between oil changes was 5187 miles for the luxury car owners and 5214 miles for the compact lower-price cars. The sample standard deviations were 424 and 507 miles for the luxury and compact groups, respectively. Assume that the two population distributions of the distances between oil changes have the same standard deviation. a. Construct a \(95 \%\) confidence interval for the difference in the mean distances between oil changes for all luxury cars and all compact lower-price cars. b. Using a \(1 \%\) significance level, can you conclude that the mean distance between oil changes is less for all luxury cars than that for all compact lower-price cars?

A local college cafeteria has a self-service soft ice cream machine. The cafeteria provides bowls that can hold up to 16 ounces of ice cream. The food service manager is interested in comparing the average amount of ice cream dispensed by male students to the average amount dispensed by female students. A measurement device was placed on the ice cream machine to determine the amounts dispensed. Random samples of 85 male and 78 female students who got ice cream were selected. The sample averages were \(7.23\) and \(6.49\) ounces for the male and female students, respectively. Assume that the population standard deviations are \(1.22\) and \(1.17\) ounces, respectively. a. Let \(\mu_{1}\) and \(\mu_{2}\) be the population means of ice cream amounts dispensed by all male and all female students at this college, respectively. What is the point estimate of \(\mu_{1}-\mu_{2} ?\) b. Construct a \(95 \%\) confidence interval for \(\mu_{1}-\mu_{2}\). c. Using a \(1 \%\) significance level, can you conclude that the average amount of ice cream dispensed by all male college students is larger than the average amount dispensed by all female college students? Use both approaches to make this test.

Explain when would you use the paired-samples procedure to make confidence intervals and test hypotheses.

The Pew Research Center conducted a poll in January 2014 of online adults who use social networking sites. According to this poll, \(89 \%\) of the \(18-29\) year olds and \(82 \%\) of the \(30-49\) year olds who are online use social networking sites (www.pewinternet.org). Suppose that this survey included 562 online adults in the \(18-29\) age group and 624 in the \(30-49\) age group. a. Let \(p_{1}\) and \(p_{2}\) be the proportion of all online adults in the age groups \(18-29\) and \(30-49\), respectively, who use social networking sites. Construct a \(95 \%\) confidence interval for \(p_{1}-p_{2}\) b. Using a \(1 \%\) significance level, can you conclude that \(p_{1}\) is different from \(p_{2} ?\) Use both the critical-value and the \(p\) -value approaches.

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