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Chapter 10: Problem 88

A random sample of 500 adult residents of Maricopa County indicated that 385 were in favor of increasing the highway speed limit to \(75 \mathrm{mph}\), and another sample of 400 adult residents of Pima County indicated that 267 were in favor of the increased speed limit. (a) Do these data indicate that there is a difference in the support for increasing the speed limit for the residents of the two counties? Use \(\alpha=0.05 .\) What is the \(P\) -value for this test?

Short Answer

Expert verified
There is a statistically significant difference in support between the counties (P-value = 0.0019).

Step by step solution

01

Define Hypotheses

Let's set up the null and alternative hypotheses. The null hypothesis (\(H_0\)) suggests there is no difference in proportions of support between the two counties:\(H_0: p_1 = p_2\). The alternative hypothesis (\(H_a\)) is that there is a difference:\(H_a: p_1 eq p_2\).
02

Calculate Sample Proportions

Calculate the sample proportions for each county. For Maricopa County, the proportion is \( \hat{p}_1 = \frac{385}{500} = 0.77 \). For Pima County, the proportion is \( \hat{p}_2 = \frac{267}{400} = 0.6675 \).
03

Calculate the Pooled Sample Proportion

Combine the samples to find the average proportion. \(\hat{p} = \frac{385 + 267}{500 + 400} = \frac{652}{900} = 0.7244\).
04

Compute Standard Error

The standard error of the difference in proportions is given by \( SE_{\hat{p}_1-\hat{p}_2} = \sqrt{\hat{p}(1-\hat{p})\left(\frac{1}{n_1} + \frac{1}{n_2}\right)} \), where \(\hat{p} = 0.7244\), \(n_1 = 500\), and \(n_2 = 400\). Thus:\( SE_{\hat{p}_1-\hat{p}_2} = \sqrt{0.7244(1-0.7244)\left(\frac{1}{500} + \frac{1}{400}\right)} \approx 0.0330 \).
05

Calculate the Z Test Statistic

Use the formula \(Z = \frac{\hat{p}_1 - \hat{p}_2}{SE_{\hat{p}_1 - \hat{p}_2}}\). Plugging in the values:\(Z = \frac{0.77 - 0.6675}{0.0330} \approx 3.106\).
06

Determine the P-value

Using the Z test statistic, find the P-value for a two-tailed test. For \(Z = 3.106\), the P-value can be found using a Z-table or calculator, which yields \(P \approx 0.0019\).
07

Compare P-value with Alpha

Since the P-value \(0.0019\) is less than \(\alpha = 0.05\), we reject the null hypothesis.

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Key Concepts

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

Statistical Significance
When we talk about statistical significance, we refer to the likelihood that a result or relationship is caused by something other than mere chance. In hypothesis testing, we often set a threshold, called alpha (\( \alpha \)), which represents the probability of rejecting the null hypothesis when it is actually true.
  • If the p-value from our test is less than the set alpha, we conclude that the results are statistically significant.
  • Statistical significance indicates that there is strong evidence against the null hypothesis.
In the highway speed limit problem presented, students were asked to test if there is a difference in the support for the increased speed limit in two counties, utilizing \( \alpha = 0.05 \).
With a p-value of approximately 0.0019, which is less than 0.05, the findings are statistically significant. This means there is enough evidence to say that the support for changing the speed limit differs between the two counties. This example emphasizes the power of statistical tests in drawing meaningful conclusions from data.
Proportions
Proportions represent a part of a whole and are useful in comparing different groups. When comparing two groups as in the exercise, we calculate the proportion of supporters for each group.
  • A proportion is expressed as a number between 0 and 1.
  • It is calculated by dividing the part of interest by the total number.
For example, in Maricopa County, \( \hat{p}_1 = \frac{385}{500} = 0.77 \), meaning 77% support the speed limit increase. In Pima County, \( \hat{p}_2 = \frac{267}{400} = 0.6675 \), or about 66.75%.
These proportions show us the part of each group that favors the new speed limit. Comparing these proportions helps us understand the difference in opinions between the counties.
Standard Error
The standard error (SE) serves as a measure of how much a statistic, like a mean or proportion, will vary in repeated samples. It helps us understand the reliability of our sample statistics and is crucial when testing hypotheses.
  • The smaller the standard error, the more reliable our sample statistic.
  • The formula for the standard error of difference in proportions is \( SE_{\hat{p}_1-\hat{p}_2} = \sqrt{\hat{p}(1-\hat{p})\left(\frac{1}{n_1} + \frac{1}{n_2}\right)} \).
In this exercise, with \( \hat{p} = 0.7244 \), \( n_1 = 500 \), and \( n_2 = 400 \), the standard error of the difference was calculated to be approximately 0.0330.
This small standard error tells us that the difference in support between the two counties is consistent across repeated samples, and helps validate our decision to reject the null hypothesis based on the Z test statistic.

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Most popular questions from this chapter

The burning rates of two different solid-fuel propellants used in air crew escape systems are being studied. It is known that both propellants have approximately the same standard deviation of burning rate; that is \(\sigma_{1}=\sigma_{2}=3\) centimeters per second. Two random samples of \(n_{1}=20\) and \(n_{2}=20\) specimens are tested; the sample mean burning rates are \(\bar{x}_{1}=18\) centimeters per second and \(\bar{x}_{2}=24\) centimeters per second. (a) Test the hypothesis that both propellants have the same mean burning rate. Use \(\alpha=0.05 .\) What is the \(P\) -value? (b) Construct a \(95 \%\) confidence interval on the difference in means \(\mu_{1}-\mu_{2} .\) What is the practical meaning of this interval? (c) What is the \(\beta\) -error of the test in part (a) if the true difference in mean burning rate is 2.5 centimeters per second? (d) Assume that sample sizes are equal. What sample size is needed to obtain power of 0.9 at a true difference in means of \(14 \mathrm{~cm} / \mathrm{s} ?\)

A random sample of 1500 residcntial tclephones in Phoenix in 1990 indicated that 387 of the numbers were unlisted. A random sample of 1200 telcphones in the same year in Scottsdale indicated that 310 were unlisted. (a) Find a \(95 \%\) confidence interval on the difference in the two proportions and use this confidence interval to determine 2010 whether there is a statistically significant difference in proportions of unlisted numbers between the two cities. (b) Find a 90 \& confidence interval on the difference in the two proportions and use this confidence interval to determine if there is a statistically significant difference in proportions of unlisted numbers for the two cities. (c) Suppose that all the numbers in the problem description were doubled. That is, 774 residents of 3000 sampled in Phoenix and 620 residents of 2400 in Scottsdale had unlisted phone numbers. Repeat parts (a) and (b) and comment on the effect of increasing the sample size without changing the proportions on your results.

Two catalysts may be used in a batch chemical process. Twelve batches were prepared using catalyst \(1,\) resulting in an average yield of 86 and a sample standard deviation of \(3 .\) Fifteen batches were prepared using catalyst \(2,\) and they resulted in an average yield of 89 with a standard deviation of 2 . Assume that yield measurements are approximately normally distributed with the same standard deviation. (a) Is there evidence to support a claim that catalyst 2 produces a higher mean yield than catalyst \(1 ?\) Use \(\alpha=0.01\). (b) Find a \(99 \%\) confidence interval on the difference in mean yields that can be used to test the claim in part (a).

Two types of plastic are suitable for an electronics component manufacturer to use. The breaking strength of this plastic is important. It is known that \(\sigma_{1}=\sigma_{2}=1.0\) psi. From a random sample of size \(n_{1}=10\) and \(n_{2}=12,\) you obtain \(\bar{x}_{1}=162.5\) and \(\bar{x}_{2}=155.0 .\) The company will not adopt plastic 1 unless its mean breaking strength exceeds that of plastic 2 by at least 10 psi. (a) Based on the sample information, should it use plastic \(1 ?\) Use \(\alpha=0.05\) in reaching a decision. Find the \(P\) -value. (b) Calculate a \(95 \%\) confidence interval on the difference in means. Suppose that the true difference in means is really 12 psi. (c) Find the power of the test assuming that \(\alpha=0.05 .\) (d) If it is really important to detect a difference of 12 psi, are the sample sizes employed in part (a) adequate in your opinion?

Consider the hypothesis test \(H_{0}: \mu_{1}=\mu_{2}\) against \(H_{1}: \mu_{1} \neq \mu_{2} .\) Suppose that sample sizes are \(n_{1}=15\) and \(n_{2}=15\) that \(\bar{x}_{1}=4.7\) and \(\bar{x}_{2}=7.8,\) and that \(s_{1}^{2}=4\) and \(s_{2}^{2}=6.25 .\) Assume that \(\sigma_{1}^{2}=\sigma_{2}^{2}\) and that the data are drawn from normal distributions. Use \(\alpha=0.05\). (a) Test the hypothesis and find the \(P\) -value. (b) Explain how the test could be conducted with a confidence interval. (c) What is the power of the test in part (a) for a true difference in means of \(3 ?\) (d) Assume that sample sizes are equal. What sample size should be used to obtain \(\beta=0.05\) if the true difference in means is -2 ? Assume that \(\alpha=0.05\).
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