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Perform these steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. In a random sample of 200 men, 130 said they used seat belts. In a random sample of 300 women, 63 said they used seat belts. Test the claim that men are more safety-conscious than women, at \(\alpha=0.01\). Use the \(P\) -value method.

Short Answer

Expert verified
Men are statistically more safety-conscious than women.

Step by step solution

01

State the Hypotheses and Identify the Claim

First, define the populations and claim. We are comparing two proportions: the proportion of men using seat belts (\( p_1 \)) and the proportion of women using seat belts (\( p_2 \)). The null hypothesis (\( H_0 \)) states that \( p_1 = p_2 \). The alternative hypothesis (\( H_a \)) states that \( p_1 > p_2 \), which aligns with the claim that men are more safety-conscious than women.
02

Calculate Sample Proportions

Calculate the sample proportions for each group. For men: \( \hat{p}_1 = \frac{130}{200} = 0.65 \). For women: \( \hat{p}_2 = \frac{63}{300} = 0.21 \).
03

Find the Critical Values

For \( \alpha = 0.01 \) and a one-tailed test, find the critical value using the standard normal distribution (z-distribution). The critical value \( z_c \) for \( \alpha = 0.01 \) is approximately 2.33.
04

Compute the Test Statistic

First, find the pooled proportion \( \hat{p} \): \[\hat{p} = \frac{x_1 + x_2}{n_1 + n_2} = \frac{130 + 63}{200 + 300} = 0.386.\]Then calculate the standard error:\[SE = \sqrt{\hat{p}(1 - \hat{p})\left(\frac{1}{n_1} + \frac{1}{n_2}\right)} = \sqrt{0.386 \times 0.614 \times \left(\frac{1}{200} + \frac{1}{300}\right)} = 0.043.\]Compute the z-test statistic:\[z = \frac{\hat{p}_1 - \hat{p}_2}{SE} = \frac{0.65 - 0.21}{0.043} \approx 10.0.\]
05

Make the Decision

Compare the computed z-test value to the critical z-value. Since the computed z (\( 10.0 \)) is greater than the critical value (\( 2.33 \)), we reject the null hypothesis.
06

Summarize the Results

Since the null hypothesis is rejected, there is enough statistical evidence to support the claim that men are more safety-conscious than women, at the \( \alpha = 0.01 \) level.

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Key Concepts

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

Null Hypothesis
The null hypothesis is a fundamental part of hypothesis testing, often denoted as \( H_0 \). It represents the default position or statement that there is no effect or no difference. In this scenario, the null hypothesis asserts that the proportion of men using seat belts is equal to the proportion of women, or mathematically, \( p_1 = p_2 \).

This acts as the starting point for statistical testing. To challenge or reject this hypothesis, we gather evidence through data analysis. Only significant statistical results can lead us to reject the null hypothesis. In our exercise, rejecting \( H_0 \) would imply there's enough evidence to say men (\( p_1 \)) are more safety-conscious than women (\( p_2 \)).
Alternative Hypothesis
Contrary to the null hypothesis, the alternative hypothesis represents what we aim to support with our data. This is often denoted as \( H_a \). For this problem, the alternative hypothesis is that the proportion of men using seat belts is greater than that of women, or \( p_1 > p_2 \).

The alternative hypothesis is what researchers want to prove. It suggests a real effect or difference, supporting the initial claim. To accept \( H_a \), statistical methods compare the observed data to what we'd expect if \( H_0 \) were true. Here, it leads to testing whether there is substantial evidence that men are more safety-conscious.
P-value Method
The \( P \)-value method provides a way to make statistical decisions by determining the probability of observing results as extreme as the ones obtained if \( H_0 \) were true. In hypothesis testing, the \( P \)-value helps decide whether to reject \( H_0 \). A small \( P \)-value, typically less than the chosen significance level \( \alpha \), suggests that the observed effect is statistically significant.

In our exercise, if the \( P \)-value is less than \( \alpha = 0.01 \), the null hypothesis will be rejected. This method gives a probabilistic measure that helps statisticians and researchers confidently state the chances of their observations occurring under the null hypothesis.
Critical Value
A critical value is a point on the test distribution that is compared to the test statistic to determine whether to reject \( H_0 \). Derived from the significance level \( \alpha \), it acts as a threshold. For a \( 0.01 \) significance level in a one-tailed test, the critical z-value is approximately 2.33.

This means that if our test statistic is greater than 2.33, we have enough evidence to reject the null hypothesis under the assumption of statistical significance. Critical values form a barrier which, if crossed by the test statistic, leads to rejecting \( H_0 \).
Test Statistic
The test statistic is a standardized value that is calculated from sample data during a hypothesis test. In many tests, this statistic follows a known distribution and is used to compare against the critical value. For our exercise, a z-test statistic is used to analyze the differences in proportions.

Calculating this involves several steps:
  • Find the sample proportions for men and women.
  • Calculate the pooled proportion.
  • Determine the standard error based on the pooled proportion.
  • Compute the z-test statistic using these values.
Finally, the calculated z-value (approximately 10.0) significantly exceeded the critical z-value (2.33), reinforcing the decision to reject the null hypothesis.

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Most popular questions from this chapter

Perform each of the following steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. A researcher wishes to see if there is a difference in the manual dexterity of athletes and that of band members. Two random samples of 30 are selected from each group and are given a manual dexterity test. The mean of the athletes' test was \(87,\) and the mean of the band members' test was \(92 .\) The population standard deviation for the test is \(7.2 .\) At \(\alpha=0.01,\) is there a significant difference in the mean scores?

Perform the following steps. Assume that all variables are normally distributed. a. State the hypotheses and identify the claim. b. Find the critical value. c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. The numbers of calories contained in \(\frac{1}{2}\) -cup servings of randomly selected flavors of ice cream from two national brands are listed. At the 0.05 level of significance, is there sufficient evidence to conclude that the variance in the number of calories differs between the two brands? $$ \begin{array}{cc|cc} &{\text { Brand A }} &{\text { Brand B }} \\ \hline 330 & 300 & 280 & 310 \\ 310 & 350 & 300 & 370 \\ 270 & 380 & 250 & 300 \\ 310 & 300 & 290 & 310 \end{array} $$

Perform the following steps. Assume that all variables are normally distributed. a. State the hypotheses and identify the claim. b. Find the critical value. c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. The weights in ounces of a random sample of running shoes for men and women are shown. Calculate the variances for each sample, and test the claim that the variances are equal at \(\alpha=0.05\). Use the \(P\) -value method. $$ \begin{array}{rrr|rrr} && {\text { Men }} & {\text { Women }} \\ \hline 11.9 & 10.4 & 12.6 & 10.6 & 10.2 & 8.8 \\ 12.3 & 11.1 & 14.7 & 9.6 & 9.5 & 9.5 \\ 9.2 & 10.8 & 12.9 & 10.1 & 11.2 & 9.3 \\ 11.2 & 11.7 & 13.3 & 9.4 & 10.3 & 9.5 \\ 13.8 & 12.8 & 14.5 & 9.8 & 10.3 & 11.0 \end{array} $$

Perform each of these steps. Assume that all variables are normally or approximately normally distributed a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Retention Test Scores A random sample of nonEnglish majors at a selected college was used in a study to see if the student retained more from reading a 19 th-century novel or by watching it in DVD form. Each student was assigned one novel to read and a different one to watch, and then they were given a 100 -point written quiz on each novel. The test results are shown. At \(\alpha=0.05,\) can it be concluded that the book scores are higher than the DVD scores? $$ \begin{array}{l|lllllll} \text { Book } & 90 & 80 & 90 & 75 & 80 & 90 & 84 \\ \hline \text { DVD } & 85 & 72 & 80 & 80 & 70 & 75 & 80 \end{array} $$

Perform the following steps. Assume that all variables are normally distributed. a. State the hypotheses and identify the claim. b. Find the critical value. c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. A random sample of daily high temperatures in January and February is listed. At \(\alpha=0.05,\) can it be concluded that there is a difference in variances in high temperature between the two months? $$ \begin{array}{l|cccccccccc} \text { Jan. } & 31 & 31 & 38 & 24 & 24 & 42 & 22 & 43 & 35 & 42 \\ \hline \text { Feb. } & 31 & 29 & 24 & 30 & 28 & 24 & 27 & 34 & 27 & \end{array} $$

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