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The null and alternate hypotheses are: $$ \begin{array}{l} H_{0}: \pi_{1} \leq \pi_{2} \\ H_{1}: \pi_{1}>\pi_{2} \end{array} $$ A sample of 100 observations from the first population indicated that \(X_{1}\) is \(70 .\) A sample of 150 observations from the second population revealed \(X_{2}\) to be \(90 .\) Use the .05 significance level to test the hypothesis. a. State the decision rule. b. Compute the pooled proportion. c. Compute the value of the test statistic. d. What is your decision regarding the null hypothesis? e. The null and alternate hypotheses are: $$ \begin{array}{l} H_{0}: \pi_{1}=\pi_{2} \\ H_{1}: \pi_{1} \neq \pi_{2} \end{array} $$

Short Answer

Expert verified
Fail to reject the first null hypothesis; test statistic (1.6) is less than critical value (1.645).

Step by step solution

01

Define Decision Rule for Initial Hypothesis

We have two populations and their sample sizes are \( n_1 = 100 \) and \( n_2 = 150 \). The null hypothesis is that \( \pi_1 \leq \pi_2 \) and the alternative hypothesis \( \pi_1 > \pi_2 \) makes this a right-tailed test. Using the 0.05 significance level, we find the critical value z from the standard normal distribution table that corresponds to 0.05 which is approximately 1.645. The decision rule is: Reject \( H_0 \) if \( Z > 1.645 \).
02

Compute the Pooled Proportion

The pooled proportion \( \hat{p} \) is calculated using the formula:\[\hat{p} = \frac {X_1 + X_2}{n_1 + n_2}\]Substitute the given values \( X_1 = 70 \), \( X_2 = 90 \), \( n_1 = 100 \), \( n_2 = 150 \) into the formula:\[\hat{p} = \frac {70 + 90}{100 + 150} = \frac {160}{250} = 0.64\]
03

Compute the Test Statistic

Calculate the test statistic using the formula for the difference of two proportions:\[Z = \frac{(\hat{p}_1 - \hat{p}_2) - D_0}{\sqrt{\hat{p}(1-\hat{p})(\frac{1}{n_1} + \frac{1}{n_2})}}\]where \( \hat{p}_1 = \frac{70}{100} = 0.7 \), \( \hat{p}_2 = \frac{90}{150} = 0.6 \) and \( D_0 = 0 \) since we are testing \( \pi_1 = \pi_2 \). Substituting in the values:\[Z = \frac{(0.7 - 0.6) - 0}{\sqrt{0.64(1-0.64)(\frac{1}{100} + \frac{1}{150})}}\]\[Z = \frac{0.1}{\sqrt{0.64 \times 0.36 \times 0.01667}} \approx \frac{0.1}{0.0625} \approx 1.6\]
04

Decision Regarding Null Hypothesis for Initial Hypothesis

Since the calculated test statistic \( Z = 1.6 \) is less than the critical value 1.645, we fail to reject the null hypothesis \( H_0: \pi_1 \leq \pi_2 \) at the 0.05 significance level.
05

Reminder of New Hypotheses

The new null hypothesis is \( H_0: \pi_1 = \pi_2 \) and the alternative hypothesis is \( H_1: \pi_1 eq \pi_2 \), which requires a two-tailed test using the same critical value strategy as before.

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Key Concepts

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

Null Hypothesis
In hypothesis testing, the null hypothesis is a critical concept that serves as a starting point for statistical testing. It is often denoted as \( H_0 \) and represents a statement of no effect or no difference. In this exercise, the null hypothesis for the first scenario is \( \pi_1 \leq \pi_2 \). This implies that the proportion of success in the first population is less than or equal to that in the second population. The purpose of testing the null hypothesis is to assess whether there is enough evidence to reject it in favor of the alternative hypothesis, \( H_1 \), which posits that \( \pi_1 > \pi_2 \), implying that the first proportion is greater. Understanding the null hypothesis is crucial because it establishes a baseline or default position that the test seeks to falsify. If the data provide sufficient evidence against the null hypothesis, we conclude that the alternative hypothesis is more likely true. However, if the evidence is insufficient, we fail to reject the null hypothesis.
Test Statistic
The test statistic is a standardized value used to determine whether to reject the null hypothesis. It offers a way to compare the sample data against what is expected under the null hypothesis. For a difference in proportions, the test statistic is calculated using the formula:\[Z = \frac{(\hat{p}_1 - \hat{p}_2) - D_0}{\sqrt{\hat{p}(1-\hat{p})(\frac{1}{n_1} + \frac{1}{n_2})}}\] Where:- \( \hat{p}_1 \) and \( \hat{p}_2 \) are the sample proportions.- \( D_0 \) is the hypothesized difference in proportions; in cases like this, it's often 0.- \( \hat{p} \) is the pooled proportion.The test statistic allows us to make comparisons based on standard normal distribution. In our example, the calculated \( Z \)-value is 1.6. We compare this value with a critical \( Z \)-value corresponding to the level of significance used, which is 0.05 for this test.
Pooled Proportion
The pooled proportion \( \hat{p} \) is an essential element in hypothesis testing for two proportions. It combines the proportions from two samples into a single, unified estimate. This is necessary when testing the null hypothesis that assumes the proportions are equal. The pooled proportion is calculated using:\[ \hat{p} = \frac{X_1 + X_2}{n_1 + n_2} \]Where \( X_1 \) and \( X_2 \) are the number of successes in the first and second samples, respectively, and \( n_1 \) and \( n_2 \) are the sizes of the two samples. In the example, with \( X_1 = 70 \) and \( X_2 = 90 \), and sample sizes of 100 and 150:- \( \hat{p} = \frac{160}{250} = 0.64 \)This pooled estimate is used in calculating the standard error of the difference in proportions, thereby enabling the test statistic calculation.
Decision Rule
The decision rule provides a guideline to decide when to reject the null hypothesis. It is based on the critical value obtained from the standard normal distribution, given a certain level of significance, often denoted by \( \alpha \). In this problem, with a significance level of 0.05, the decision is based on a right-tailed test with a critical value of approximately 1.645. - **Decision Rule**: Reject the null hypothesis if the test statistic \( Z \) is greater than 1.645.If \( Z \) is less than or equal to this critical value, we fail to reject the null hypothesis. In our example, the calculated test statistic is 1.6. Since this value does not exceed the critical value, the decision for the initial hypothesis is to fail to reject \( H_0: \pi_1 \leq \pi_2 \). This decision rule helps maintain the objectivity of the hypothesis testing process by providing a clear, quantitative threshold for decision-making.

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Most popular questions from this chapter

The null and alternate hypotheses are: $$ \begin{array}{l} H_{0}: \mu_{d}=0 \\ H_{1}: \mu_{d} \neq 0 \end{array} $$ The following paired observations show the number of traffic citations given for speeding by Officer Dhondt and Officer Meredith of the South Carolina Highway Patrol for the last five months. At the . 05 significance level, is there a difference in the mean number of citations given by the two officers?

Fairfield Homes is developing two parcels near Pigeon Fork, Tennessee. In order to test different advertising approaches, it uses different media to reach potential buyers. The mean annual family income for 15 people making inquiries at the first development is \(\$ 150,000\) with a standard deviation of \(\$ 40,000\). A corresponding sample of 25 people at the second development had a mean of \(\$ 180,000,\) with a standard deviation of \(\$ 30,000\). Assume the population standard deviations are the same. At the .05 significance level, can Fairfield conclude that the population means are different?

(a) state the decision rule, (b) compute the pooled estimate of the population variance, (c) compute the test statistic, (d) state your decision about the null hypothesis, and (e) estimate the \(p\) -value. The null and alternate hypotheses are: $$ \begin{array}{l} H_{0}: \mu_{1}=\mu_{2} \\ H_{1}: \mu_{1} \neq \mu_{2} \end{array} $$ A random sample of 10 observations from one population revealed a sample mean of 23 and a sample deviation of \(4 .\) A random sample of 8 observations from another population revealed a sample mean of 26 and a sample standard deviation of \(5 .\) At the .05 significance level, is there a difference between the population means?

A nationwide sample of influential Republicans and Democrats was asked as a part of a comprehensive survey whether they favored lowering environmental standards so that high-sulfur coal could be burned in coal-fired power plants. The results were: $$ \begin{array}{|lcc|} \hline & \text { Republicans } & \text { Democrats } \\ \hline \text { Number sampled } & 1,000 & 800 \\ \text { Number in favor } & 200 & 168 \end{array} $$ At the .02 level of significance, can we conclude that there is a larger proportion of Democrats in favor of lowering the standards? Determine the \(p\) -value.

A study was conducted to determine if there was a difference in the humor content in British and American trade magazine advertisements. In an independent random sample of 270 American trade magazine advertisements, 56 were humorous. An independent random sample of 203 British trade magazines contained 52 humorous ads. Does this data provide evidence at the .05 significance level that there is a difference in the proportion of humorous ads in British versus American trade magazines?

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