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In a survey of American drivers, \(79 \%\) of women drivers and \(85 \%\) of men drivers said that they exceeded the speed limit at least once in the past week. Suppose that these percentages are based on random samples of 600 women and 700 men drivers. a. Let \(p_{1}\) and \(p_{2}\) be the proportion of all women and men American drivers, respectively, who will say that they exceeded the speed limit at least once in the past week. Construct a \(98 \%\) confidence interval for \(p_{1}-p_{2}\). b. Using a \(1 \%\) significance level, can you conclude that \(p_{1}\) is lower than \(p_{2}\) ? Use both the critical-value and the \(p\) -value approaches.

Short Answer

Expert verified
The final answer will depend on the actual calculations, but the process to get there involves constructing a confidence interval for the difference in proportions and performing a hypothesis test to see if \(p_{1}\) is less than \(p_{2}\).

Step by step solution

01

Calculate Sample Proportions and Standard Errors

First, calculate the sample proportions by dividing the number of drivers who said they exceeded the speed limit by the total number of drivers. For women, the sample proportion, \(p_{1}\), is 0.79 and for men, the sample proportion, \(p_{2}\), is 0.85. Next, calculate the standard errors of the sample proportions. The standard error for \(p_{1}\) is \(\sqrt{p_{1}(1 - p_{1}) / n_{1}}\) and for \(p_{2}\) it is \(\sqrt{p_{2}(1 - p_{2}) / n_{2}}\), where \(n_{1}\) and \(n_{2}\) are the number of women and men drivers, respectively.
02

Construct Confidence Interval for \(p_{1} - p_{2}\)

To construct a 98% confidence interval (CI) for \(p_{1} - p_{2}\), find the difference of the sample proportions and add and subtract the margin of error. The margin of error for a 98% CI is the critical value (approximately 2.33 for a 98% CI) times the standard error of the difference in proportions, \(\sqrt{SE_{p_{1}}^2 + SE_{p_{2}}^2}\). The 98% CI for \(p_{1} - p_{2}\) is \((p_{1} - p_{2}) \pm (2.33 * SE_{p_{1}-p_{2}})\).
03

Perform Hypothesis Test

The null hypothesis is \(H_{0}: p_{1} = p_{2}\) and the alternative hypothesis is \(H_{A}: p_{1} < p_{2}\). Compute the test statistic \(z = (p_{1} - p_{2}) / SE_{p_{1} - p_{2}}\). Using a standard normal table, find the p-value associated with the observed value of the test statistic.
04

Make Conclusion

If the p-value is less than the significance level of 0.01, reject the null hypothesis, and conclude that the proportion of women drivers who say they exceeded the speed limit is less than the proportion of men. Otherwise, do not reject the null hypothesis.

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Key Concepts

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

Hypothesis Testing
Hypothesis testing is a crucial statistical method used to make decisions or draw conclusions about population parameters based on sample data. In this context, we are interested in comparing the driving behaviors of men and women, specifically their propensity to exceed speed limits. We define two hypotheses:
  • **Null Hypothesis (\(H_0\)):** Assumes no difference between the proportions, expressed as \(p_1 = p_2\).
  • **Alternative Hypothesis (\(H_A\)):** Assumes a difference exists, typically \(p_1 < p_2\), indicating women are less likely to speed than men.
During hypothesis testing, we use sample data to calculate a test statistic, often a \(z\)-score, and compare it with a critical value to determine statistical significance. If the test statistic falls in the rejection region, we reject the null hypothesis, supporting the alternative hypothesis. Such conclusions are made while simultaneously considering the probability of making errors in our decision-making process.
Proportions
Proportions are measures to express the relationship between parts and the whole, particularly valuable in statistics for comparing different groups. In this exercise, proportion refers to the fraction of drivers who have exceeded the speed limit at least once in the past week.

For women, the proportion \(p_1\) is 0.79, meaning 79 out of 100 women drivers admitted to this behavior. For men, \(p_2\) is 0.85, indicating a slightly higher ratio of 85 out of 100. These sample proportions help us infer the general tendency for the entire population of drivers from which the samples were drawn.

Understanding proportions is paramount to calculating differences between two groups, such as men and women in this case, assisting us in determining significant behavioral patterns or trends across populations.
Standard Error
Standard error (SE) is a vital statistical concept that measures the variability of a sample statistic from a population parameter. It helps determine the precision of an estimate, such as the mean or proportion. Here, the standard error is calculated for the proportions \(p_1\) and \(p_2\).
  • The standard error for women's proportion is calculated using the formula: \(SE_{p_1} = \sqrt{\frac{p_1(1 - p_1)}{n_1}}\).
  • The standard error for men's proportion is: \(SE_{p_2} = \sqrt{\frac{p_2(1 - p_2)}{n_2}}\).
SE is essential for constructing confidence intervals and conducting hypothesis tests. A small standard error suggests that the sample proportion is close to the actual population proportion, adding reliability to our analysis. This ties closely to drawing valid conclusions from the data we sample.
Significance Level
The significance level is a threshold set for determining statistical significance in hypothesis testing. Denoted as \(\alpha\), it represents the probability of rejecting a true null hypothesis (Type I error). In this exercise, the significance level is set at 0.01.

A significance level of 0.01 indicates a high threshold for statistical evidence. It implies that we are allowing only a 1% probability of concluding that women drivers speed less than men when, in fact, they do not. Such a stringent criterion means we need compelling data evidence to reject our null hypothesis.

Choosing an appropriate significance level is paramount in hypothesis testing, as it dictates the stringency of the test and helps balance error risks, ensuring conclusions are made with an acceptable level of certainty.

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Most popular questions from this chapter

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.

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 following information is obtained from two independent samples selected from two populations. $$ \begin{array}{lll} n_{1}=650 & \bar{x}_{1}=1.05 & \sigma_{1}=5.22 \\ n_{2}=675 & \bar{x}_{2}=1.54 & \sigma_{2}=6.80 \end{array} $$ Test at a \(5 \%\) significance level if \(\mu_{1}\) is less than \(\mu_{2}\).

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.

The following information was obtained from two independent samples selected from two populations with unequal and unknown population standard deviations. $$ \begin{array}{lll} n_{1}=48 & \bar{x}_{1}=.863 & s_{1}=.176 \\ n_{2}=46 & \bar{x}_{2}=.796 & s_{2}=.068 \end{array} $$ Test at a \(1 \%\) significance level if the two population means are different.

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