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Chapter 9: Problem 18

A manufacturer of walk-behind push mowers receives refurbished small engines from two new suppliers, \(A\) and \(B\). It is not uncommon that some of the refurbished engines need to be lightly serviced before they can be fitted into mowers. The mower manufacturer recently received 100 engines from each supplier. In the shipment from \(A, 13\) needed further service. In the shipment from \(B, 10\) needed further service. Test, at the \(10 \%\) level of significance, whether the data provide sufficient evidence to conclude that there exists a difference in the proportions of engines from the two suppliers needing service.

Short Answer

Expert verified
No significant evidence of a difference between suppliers.

Step by step solution

01

Define Hypotheses

First, identify the null and alternative hypotheses for the problem. The null hypothesis (H_0) is that there is no difference in the proportion of engines needing service between suppliers A and B. The alternative hypothesis (H_a) is that there is a difference.\[ \begin{align*} H_0 &: p_A = p_B \ H_a &: p_A eq p_B \end{align*} \]Where \( p_A \)and \( p_B \)are the proportions of engines needing further service from suppliers A and B respectively.
02

Calculate Sample Proportions

The sample proportions of engines needing service from each supplier can be calculated as follows:\[ \hat{p}_A = \frac{13}{100} = 0.13 \]\[ \hat{p}_B = \frac{10}{100} = 0.10 \]
03

Calculate Pooled Sample Proportion

The pooled sample proportion is calculated by combining the successes (engines needing service) and the total shipments:\[ \hat{p} = \frac{13 + 10}{100 + 100} = \frac{23}{200} = 0.115 \]
04

Compute Standard Error

The standard error of the difference in sample proportions is computed with the formula:\[ SE = \sqrt{\hat{p}(1-\hat{p}) \left( \frac{1}{n_A} + \frac{1}{n_B} \right)} \]where \( n_A = n_B = 100 \).Substitute the values:\[ SE = \sqrt{0.115(1-0.115) \left( \frac{1}{100} + \frac{1}{100} \right)} \approx 0.047 \]
05

Calculate Test Statistic

The test statistic for the difference in proportions is given by:\[ z = \frac{\hat{p}_A - \hat{p}_B}{SE} = \frac{0.13 - 0.10}{0.047} \approx 0.638 \]
06

Determine Critical Value and Make Decision

For a two-tailed test at the \(10\%\) significance level, the critical z-values are approximately \(\pm 1.645\). Since the calculated z-value \(0.638\) is within the range \([-1.645, 1.645]\), we fail to reject the null hypothesis.

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Key Concepts

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

Proportions
When discussing hypothesis testing, particularly with proportions, we are comparing parts of a whole in different groups. Here, let's take a closer look at what proportions mean in the context of our exercise.
A proportion is simply a fraction or a percentage that signifies a part of a whole group. For instance, in this scenario:
  • Supplier A delivered 100 engines, and 13 of these needed more service. This gives us a proportion for Supplier A of 0.13 or 13%.
  • Supplier B also provided 100 engines, with 10 needing service, resulting in a proportion of 0.10 or 10% for Supplier B.
These proportions highlight the engines requiring further service over the total delivered engines from each supplier. We use these numbers to understand if there's a significant difference in the quality or requirements of the engines between the two suppliers.
Significance Level
The significance level is a crucial component in hypothesis testing. It helps in determining the threshold for deciding whether to support or reject the null hypothesis. In this exercise, the significance level is set at 10%, often represented as \( \alpha = 0.10 \).
This means we are allowing a 10% risk of concluding a difference when there is none.
In hypothesis testing:
  • The significance level indicates how strictly we are judging the results of our test.
  • A lower significance level means a stricter criterion to reject the null hypothesis, making it less likely to claim a difference exists unless there is strong evidence.
  • The 10% level tells us to reject the null hypothesis if our test statistic exceeds the critical values within this level, which are around 1.645 in both directions for a two-tailed test.
If our test statistic falls outside these bounds, then the evidence is considered significant enough to suggest a difference between proportions.
Null Hypothesis
The null hypothesis is a starting assumption for any hypothesis testing. It proposes that no relationship or difference exists between the groups being compared.
In our exercise, the null hypothesis ( \( H_0 \) ) asserts that there is no difference in the proportions of engines needing service from suppliers A and B. It's expressed as:
\[ H_0: p_A = p_B \] Here are the basics to remember about the null hypothesis:
  • The null hypothesis is what we initially assume to be true.
  • It is typically tested with the idea of trying to find evidence against it.
  • In statistical terms, we "fail to reject" the null hypothesis; we never "accept" it. This wording reflects our conservative stance in statistics.
In this case, we are testing to see if our collected engine data provide sufficient evidence to reject this assumption.
Alternative Hypothesis
The alternative hypothesis represents the scenario we are trying to find evidence for in hypothesis testing. When our assumptions under the null hypothesis do not hold, we look to the alternative.
Here for the given exercise, the alternative hypothesis ( \( H_a \) ) argues that there is a difference between the two suppliers in terms of the proportion of engines needing service. Expressed mathematically as:
\[ H_a: p_A eq p_B \] Points to recall about the alternative hypothesis:
  • It contains the change or difference we are interested in proving with statistical evidence.
  • The goal is to gather sufficient evidence to demonstrate that the alternative is more likely than the null.
  • Reaching this hypothesis can lead to changes or decisions based on the newfound evidence.
In our context, it would suggest a significant difference in the quality control between suppliers.

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Most popular questions from this chapter

A neighborhood home owners association suspects that the recent appraisal values of the houses in the neighborhood conducted by the county government for taxation purposes is too high. It hired a private company to appraise the values of ten houses in the neighborhood. The results, in thousands of dollars, are \(\begin{array}{|l|c|c|} \hline \text { House } & \text { County Government } & \text { Private Company } \\ \hline 1 & 217 & 219 \\ \hline 2 & 350 & 338 \\ \hline 3 & 296 & 291 \\ \hline 4 & 237 & 237 \\ \hline \end{array}\) \(\begin{array}{|l|c|c|} \hline \text { House } & \text { County Government } & \text { Private Company } \\ \hline 5 & 237 & 235 \\ \hline 6 & 272 & 269 \\ \hline 7 & 257 & 239 \\ \hline 8 & 277 & 275 \\ \hline 9 & 312 & 320 \\ \hline 10 & 335 & 335 \\ \hline \end{array}\) a. Give a point estimate for the difference between the mean private appraisal of all such homes and the government appraisal of all such homes. b. Construct the \(99 \%\) confidence interval based on these data for the difference. c. Test, at the \(1 \%\) level of significance, the hypothesis that appraised values by the county government of all such houses is greater than the appraised values by the private appraisal company.

An educational researcher wishes to estimate the difference in average scores of elementary school children on two versions of a 100-point standardized test, at \(99 \%\) confidence and to within two points. Estimate the minimum equal sample sizes necessary if it is known that the standard deviation of scores on different versions of such tests is 4.9

To assess to the relative happiness of men and women in their marriages, a marriage counselor plans to administer a test measuring happiness in marriage ton randomly selected married couples, record the their test scores, find the differences, and then draw inferences on the possible difference. Let \(\mu 1\) and \(\mu_{2}\) be the true average levels of happiness in marriage for men and women respectively as measured by this test. Suppose it is desired to find a \(90 \%\) confidence interval for estimating \(\mu d=\mu 1-\mu 2\) to within two test points. Suppose further that, from prior studies, it is known that the standard deviation of the differences in test scores is \(\sigma d \approx 10 .\) What is the minimum number of married couples that must be included in this study?

A journalist plans to interview an equal number of members of two political parties to compare the proportions in each party who favor a proposal to allow citizens with a proper license to carry a concealed handgun in public parks. Let \(p_{1}\) and \(p_{2}\) be the true proportions of members of the two parties who are in favor of the proposal. Suppose it is desired to find a \(95 \%\) confidence interval for estimating \(p_{1-p 2}\) to within 0.05. Estimate the minimum equal number of members of each party that must be sampled to meet these criteria.

Estimate the minimum equal sample sizes \(n 1=n 2\) necessary in order to estimate \(p 1-p_{2}\) as specified. a. \(80 \%\) confidence, to within 0.02 (two percentage points) a. when no prior knowledge of \(p_{1}\) or \(p_{2}\) is available b. when prior studies indicate that \(p 1 \approx 0.78\) and \(p 2 \approx 0.65\) b. \(90 \%\) confidence, to within 0.05 (two percentage points) a. when no prior knowledge of \(p_{1}\) or \(p_{2}\) is available b. when prior studies indicate that \(p 1 \approx 0.12\) and \(p 2 \approx 0.24\) c. \(95 \%\) confidence, to within 0.10 (ten percentage points) a. when no prior knowledge of \(p_{1}\) or \(p_{2}\) is available b. When prior studies indicate that \(p_{1} \approx 0.14\) and \(p_{2} \approx 0.21\)
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