/*! 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 51 A state that requires periodic e... [FREE SOLUTION] | 91Ó°ÊÓ

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A state that requires periodic emission tests of cars operates two emission test stations, \(\mathrm{A}\) and \(\mathrm{B}\), in one of its towns. Car owners have complained of lack of uniformity of procedures at the two stations, resulting in different failure rates. A sample of 400 cars at Station A showed that 53 of those failed the test; a sample of 470 cars at Station B found that 51 of those failed the test. a. What is the point estimate of the difference between the two population proportions? b. Construct a \(95 \%\) confidence interval for the difference between the two population proportions. c. Testing at a \(5 \%\) significance level, can you conclude that the two population proportions are different? Use both the critical-value and the \(p\) -value approaches.

Short Answer

Expert verified
a. The point estimate for the difference between the two sample proportions is 0.024. b. The 95% confidence interval for the difference in proportions is (-0.02, 0.068). c. Based on the p-value of 0.276, there is insufficient evidence at a 5% significance level to conclude that the proportions are different at the two stations.

Step by step solution

01

Compute proportions and their difference

First, calculate the proportions of cars that failed the test at each station. For station A, the failure rate is \( \frac{53}{400} = 0.1325 \) (or 13.25%). For station B, the failure rate is \( \frac{51}{470} = 0.1085 \) (or 10.85%). The point estimate of the difference between the two population proportions (Station A - Station B) is \(0.1325 - 0.1085 = 0.024\).
02

Calculate the standard error

Next, calculate the standard error (SE), which reflects the variability in the estimated difference in failure rates: The formula for SE when comparing two proportions is \[SE = \sqrt{\frac{pA(1-pA)}{nA} + \frac{pB(1-pB)}{nB}}\]Substituting given values\[SE = \sqrt{\frac{0.1325(1-0.1325)}{400} + \frac{0.1085(1-0.1085)}{470}} = 0.022\]
03

Construct 95% Confidence Interval

A 95% confidence interval (CI) calculates the range within which the true difference in failure proportions between the two stations likely lies. Using the formula \[(\hat{p}_1-\hat{p}_2)\pm z*\sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2}}\]Substitute the values into the formula\[(0.024)\pm 1.96*0.022\]The 95% CI for the difference in proportions is (-0.02, 0.068). This means that with 95% confidence, the difference in failure proportions between the two stations can be said to range from -0.02 to 0.068.
04

Conduct Hypothesis Test

A hypothesis test is performed to check whether the difference in proportions is statistically significant or not.The null hypothesis is that there's no difference in the proportions (\(p_A = p_B\) or, equivalently, \(p_A - p_B = 0\)), while the alternative hypothesis is that there's a difference (\(p_A \neq p_B\) or, equivalently, \(p_A - p_B \neq 0\)).The test statistic \(Z\) is given by \(\frac{(\hat{p}_A - \hat{p}_B) - 0}{SE}\).Substituting the determined values, the test statistic \(Z\) becomes \( \frac{0.024 - 0}{0.022} = 1.09\).From a z-table or using a calculator, the p-value for \(Z = 1.09\) is approximately 0.276.Since the p-value is more than the significance level (0.05), the null hypothesis cannot be rejected. This means that the difference in proportions is not statistically significant.

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Key Concepts

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

Understanding Population Proportions
Population proportions help in understanding the portion of a population that displays certain characteristics. In our emissions test exercise, population proportions represent the fraction of cars that failed at the test stations. To compute the proportion for a station, divide the number of failures by the total cars inspected.

For example:
  • Station A: Failures = 53, Total = 400, Proportion = \( \frac{53}{400} = 0.1325 \).
  • Station B: Failures = 51, Total = 470, Proportion = \( \frac{51}{470} = 0.1085 \).
This gives a clear picture of how each station is performing in terms of adherence to pollution controls compared to the total tests conducted. Knowing population proportions helps in making comparisons between different groups or stations in statistical evaluations.
What is a Confidence Interval?
A confidence interval (CI) provides a range of values within which the true parameter, such as a population proportion, is expected to fall with a certain level of confidence. In the emissions test example, a 95% confidence interval indicates that we are 95% certain that the true difference in proportions lies within this calculated range.

To compute a 95% confidence interval for the difference in failure rates between two stations, use the formula:\[(\hat{p}_1-\hat{p}_2)\pm z*SE\]Here, \(z\) is the standard normal deviate, which is 1.96 for a 95% CI, and SE is the standard error. Applying the values, we derive a CI of (-0.02, 0.068). This means the true difference might be negative or positive, but mostly, it lies in this range. It gives us insights without a need for definitive proof, helping to make informed decisions based on statistical evidence.
Exploring Significance Level
Significance level, often denoted as \(\alpha\), is the threshold used in hypothesis testing to determine the evidence required to reject the null hypothesis. It represents the probability of a Type I error, which is rejecting a true null hypothesis. Common significance levels are 0.05, 0.01, and 0.10.

In this emissions test case, a 5% significance level (\(\alpha = 0.05\)) is used. This means that there is a 5% risk of concluding that there is a difference in the failure rates when there isn't one. If the p-value obtained from the hypothesis test is less than 0.05, we reject the null hypothesis, suggesting that the proportions are indeed different. Conversely, if it's greater, as in the calculated p-value of 0.276 here, we do not reject the null hypothesis. In simpler terms, the evidence is not strong enough to declare a statistically significant difference in the emission test results between the two stations.
Clarifying Point Estimate
A point estimate is a single value estimate of a population parameter. It provides the "best guess" of the parameter's true value based on sample data. In the context of two population proportions, the point estimate is the difference between the sample proportions.

From our emissions test example, the point estimate is calculated as the difference between the failure proportions at the two stations. It is expressed as:\[0.1325 - 0.1085 = 0.024\]This point estimate suggests that Station A has a higher failure rate by 2.4% points than Station B. While this provides useful information, it doesn’t account for variability, making confidence intervals critical for better reliability. Thus, point estimates offer immediate insights but are best used in conjunction with other statistical measures like confidence intervals for broader decision-making.

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Most popular questions from this chapter

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.

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.

Using data from the U.S. Census Bureau and other sources, www.nerdwallet.com estimated that considering only the households with credit card debts, the average credit card debt for U.S. households was \(\$ 15,523\) in 2014 and \(\$ 15,242\) in 2013 . Suppose that these estimates were based on random samples of 600 households with credit card debts in 2014 and 700 households with credit card debts in 2013 . Suppose that the sample standard deviations for these two samples were \(\$ 3870\) and \(\$ 3764\), respectively. Assume that the standard deviations for the two populations are unknown and unequal. a. Let \(\mu_{1}\) and \(\mu_{2}\) be the average credit card debts for all such households for the years 2014 and 2013 , respectively. What is the point estimate of \(\mu_{1}-\mu_{2}\) ? b. Construct a \(98 \%\) confidence interval for \(\mu_{1}-\mu_{2}\). c. Using a \(1 \%\) significance level, can you conclude that the average credit card debt for such households was higher in 2014 than in 2013 ? Use both the \(p\) -value and the critical-value approaches to make this test.

The management of a supermarket chain wanted to investigate if the percentages of men and women who prefer to buy national brand products over the store brand products are different. A sample of 600 men shoppers at the company's supermarkets showed that 246 of them prefer to buy national brand products over the store brand products. Another sample of 700 women shoppers at the company's supermarkets showed that 266 of them prefer to buy national brand products over the store brand products. a. What is the point estimate of the difference between the two population proportions? b. Construct a \(98 \%\) confidence interval for the difference between the proportions of all men and all women shoppers at these supermarkets who prefer to buy national brand products over the store brand products. c. Testing at a \(1 \%\) significance level, can you conclude that the proportions of all men and all women shoppers at these supermarkets who prefer to buy national brand products over the store brand products are different?

The management at New Century Bank claims that the mean waiting time for all customers at its branches is less than that at the Public Bank, which is its main competitor. A business consulting firm took a sample of 200 customers from the New Century Bank and found that they waited an average of \(4.5\) minutes before being served. Another sample of 300 customers taken from the Public Bank showed that these customers waited an average of \(4.75\) minutes before being served. Assume that the standard deviations for the two populations are \(1.2\) and \(1.5\) minutes, respectively. a. Make a \(97 \%\) confidence interval for the difference between the two population means. b. Test at a \(2.5 \%\) significance level whether the claim of the management of the New Century Bank is true. c. Calculate the \(p\) -value for the test of part b. Based on this \(p\) -value, would you reject the null hypothesis if \(\alpha=.01\) ? What if \(\alpha=.05\) ?

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