/*! 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 614 Refer to the data in the previou... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Refer to the data in the previous problem concerning the Selective Service data to test the hypothesis that the proportions are equal against the alternative that \(\pi_{1}>\pi_{2}\). Use \(\alpha=.05\).

Short Answer

Expert verified
Based on the data from the Selective Service, we have tested the hypothesis that the proportions are equal against the alternative that \(\pi_{1}>\pi_{2}\) using a significance level of \(\alpha=0.05\). By calculating the pooled proportion, test statistic, and comparing it to the critical value, we can either reject or fail to reject the null hypothesis. If the test statistic \(Z > 1.645\), we reject the null hypothesis, concluding that \(\pi_{1} > \pi_{2}\). Otherwise, we do not have enough evidence to support the claim that the proportions are unequal.

Step by step solution

01

1. State the null and alternative hypotheses

The null hypothesis, \(H_{0}\), is that the proportions are equal: \( \pi_{1} = \pi_{2} \). The alternative hypothesis, \(H_{1}\), is that proportion 1 is greater than proportion 2: \( \pi_{1} > \pi_{2} \).
02

2. Calculate the pooled proportion

To calculate the pooled proportion, we first need data from the previous problem about the number of successes and the sample sizes in each group. Let's assume the data is as follows: - Group 1: \(x_{1}\) successes and \(n_{1}\) total observations - Group 2: \(x_{2}\) successes and \(n_{2}\) total observations. Calculate the pooled proportion, \(\hat{p}\), using the formula: \[\hat{p}=\frac{x_{1}+x_{2}}{n_{1}+n_{2}}\]
03

3. Calculate the test statistic

Next, calculate the test statistic \(Z\) using the formula: \[Z=\frac{(\hat{p}_{1}-\hat{p}_{2})-0}{\sqrt{\hat{p}(1-\hat{p})(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}\] where \(\hat{p}_{1}=\frac{x_{1}}{n_{1}}\) and \(\hat{p}_{2}=\frac{x_{2}}{n_{2}}\).
04

4. Find the critical value and rejection region

Now, we need to determine the critical value and the rejection region. Since this is a one-tailed test with a significance level of \(\alpha = 0.05\), we look for the critical value in the standard normal distribution table or use a calculator to find the value \(Z_{\alpha}\) such that \(P(Z > Z_{\alpha})=\alpha\). Thus, we find that \(Z_{0.05} \approx 1.645\). The rejection region is defined as all values of the test statistic greater than the critical value: \(Z > Z_{0.05}\) or \(Z > 1.645\).
05

5. Make a decision and interpret the results

We now compare our calculated test statistic \(Z\) to the critical value and the rejection region. If \(Z > 1.645\), we reject the null hypothesis in favor of the alternative hypothesis, concluding that there is enough evidence to support the claim that the proportions are unequal and \(\pi_{1} > \pi_{2}\). If \(Z \le 1.645\), we do not reject the null hypothesis, and we do not have enough evidence to conclude that \(\pi_{1} > \pi_{2}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Null Hypothesis
When conducting hypothesis testing, the null hypothesis often represents a statement of no effect or no difference. In the context of testing proportion equality, the null hypothesis, denoted as \(H_0\), suggests that there is no difference between the proportions in the groups being compared. This is mathematically expressed as \( \pi_1 = \pi_2 \). By carrying out hypothesis testing, we aim to determine whether there is statistically significant evidence to reject this assumed equality.
This process involves comparing calculated statistics to predetermined thresholds to decide whether observed differences could occur by random chance.
Alternative Hypothesis
In contrast to the null hypothesis, the alternative hypothesis proposes a specific difference or effect. For this exercise, the alternative hypothesis is that the first proportion is greater than the second: \( \pi_1 > \pi_2 \). This is a directional hypothesis because it specifies the direction of the expected difference rather than just stating that there is a difference.
The formulation of the alternative hypothesis is crucial because it directly impacts the selection of the statistical test and the determination of the rejection region. If evidence supports the alternative hypothesis, we reject the null hypothesis, concluding that the proportions are not equal.
Test Statistic
The test statistic is a crucial component in hypothesis testing, acting as a standardized value that is calculated from sample data. It allows us to assess how far the observed sample statistic deviates from the null hypothesis. In this scenario, the test statistic \(Z\) is used, derived from the formula: \[ Z = \frac{(\hat{p}_1 - \hat{p}_2) - 0}{\sqrt{\hat{p}(1-\hat{p})\left(\frac{1}{n_1} + \frac{1}{n_2}\right)}} \]
  • \(\hat{p}_1\) and \(\hat{p}_2\) represent the sample proportions of success for each group.
  • The pooled proportion \(\hat{p}\) is the combined success rate from both groups.

By comparing the calculated \(Z\)-value to a critical value obtained from a standard normal distribution, we can determine whether to reject the null hypothesis.
Pooled Proportion
The pooled proportion is key in hypothesis testing involving proportions because it provides a combined estimate of the success probability, assuming the null hypothesis is true. By aggregating the data from both groups, the pooled proportion \(\hat{p}\) is computed using the formula: \[ \hat{p} = \frac{x_1 + x_2}{n_1 + n_2} \]
  • \(x_1\) and \(x_2\) are the numbers of successes in each group.
  • \(n_1\) and \(n_2\) are the total number of observations in each group.

This weighted average of the sample proportions informs the computation of the test statistic, enabling the assessment of variance in observed differences given the null hypothesis.
One-Tailed Test
A one-tailed test evaluates the possibility of a parameter exceeding (or being less than) a certain value. In this exercise, we are interested in testing whether one proportion is greater than the other, making it a right-tailed test.
This type of test focuses all significance into one end of the distribution, increasing the test's ability to detect an effect from the hypothesized direction, compared to a two-tailed test.
For significance level \( \alpha = 0.05 \), the critical value marks the boundary of the rejection region. If the test statistic falls into this region, we reject the null hypothesis, concluding evidence exists in favor of the alternative hypothesis that \( \pi_1 > \pi_2 \).

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The standard deviation of a particular dimension of a metal component is small enough so that it is satisfactory in subsequent assembly. A new supplier of metal plate is under consideration and will be preferred if the standard deviation of his product is not larger, because the cost of his product is lower than that of the present supplier. A sample of 100 items from each supplier is obtained. The results are as follows: New supplier: \(\mathrm{S}_{1}^{2}=.0058\) Old supplier: \(\mathrm{S}_{2}{ }^{2}=.0041\) Should the new supplier's metal plates be purchased? Test at the \(5 \%\) level of significance.

An experimenter tested for differences in attitudes toward smoking before and after a film on lung cancer was shown. He found a difference which was significant between the \(.05\) and .02 levels. a. What is the assumed hypothesis in words? b. Which level of significance indicates the greater degree of significance, \(.05\) or \(.02 ?\) c. If his \(\alpha\) level is \(.05\), will he reject \(\mathrm{H}_{1} ?\) Will he reject it If he employs the \(.02\) level? In choosing \(\alpha=.02\) instead of \(\alpha=.05\), he increases the risk of making one of the two types of error. Which type?

A manufacturer of transistors makes two brands, \(\mathrm{A}\) and \(\mathrm{B}\). Brand \(\mathrm{A}\) is supposed to have an average life of 60 hours more than brand \(\mathrm{B}\). To verify whether this is true, each month 9 transistors of each brand are tested and a t- value corresponding to the difference of the sample means is computed. The manufacturer is satisfied if the computed \(\mathrm{t}\) -value falls between \(-\mathrm{t}_{05}\) and \(\mathrm{t}_{.05}\). During one month, the sample of nine transistors from brand \(\mathrm{A}\) had a mean life-span of 1000 hours and a standard deviation of 60 hours, while those of brand \(\mathrm{B}\) had a mean life-span of 925 hours with a standard deviation of 50 hours. Assuming that the life-span of both brand \(\mathrm{A}\) and brand \(\mathrm{B}\) is normal, should the manufacturer be satisfied?

If \(99 \%\) of a certain publisher's books are ordinarily bound perfectly, would the finding of 3 imperfectly bound books in an order of 60 books be significant of a lowered quality, or might this occurrence come about as a result of mere chance?

Suppose that you want to decide which of two equally-priced brands of light bulbs lasts longer. You choose a random sample of 100 bulbs of each brand and find that brand \(\mathrm{A}\) has sample mean of 1180 hours and sample standard deviation of 120 hours, and that brand \(\mathrm{B}\) has sample mean of 1160 hours and sample standard deviation of 40 hours. What decision should you make at the \(5 \%\) significance level?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.