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A sample of 300 urban adult residents of a particular state revealed 63 who favored increasing the highway speed limit from 55 to \(65 \mathrm{mph}\), whereas a sample of 180 rural residents yielded 75 who favored the increase. Does this data indicate that the sentiment for increasing the speed limit is different for the two groups of residents? a. Test \(H_{0}: p_{1}=p_{2}\) versus \(H_{a}: p_{1} \neq p_{2}\) using \(\alpha=.05\), where \(p_{1}\) refers to the urban population. a b. If the true proportions favoring the increase are actually \(p_{1}=.20\) (urban) and \(p_{2}=.40\) (rural), what is the probability that \(H_{0}\) will be rejected using a level \(.05\) test with \(m=300, n=180 ?\)

Short Answer

Expert verified
Yes, the sentiment is different. Power is approximately 0.9854, indicating high probability of correctly rejecting \( H_0 \).

Step by step solution

01

Define the null and alternative hypotheses

The null hypothesis \( H_0 \) states that the proportion of urban residents favoring the increase \( p_1 \) is equal to the proportion of rural residents favoring the increase \( p_2 \). The alternative hypothesis \( H_a \) states that these proportions are not equal. Therefore, \( H_0: p_1 = p_2 \) and \( H_a: p_1 eq p_2 \).
02

Calculate sample proportions

Compute the sample proportions for both groups. For urban residents: \( \hat{p_1} = \frac{63}{300} = 0.21 \). For rural residents: \( \hat{p_2} = \frac{75}{180} = 0.4167 \).
03

Compute the pooled proportion

The pooled proportion (\( \hat{p} \)) is calculated as \( \hat{p} = \frac{x_1 + x_2}{n_1 + n_2} = \frac{63 + 75}{300 + 180} = \frac{138}{480} = 0.2875 \).
04

Calculate the standard error

Calculate the standard error using the formula \( SE = \sqrt{\hat{p}(1 - \hat{p})(\frac{1}{n_1} + \frac{1}{n_2})} = \sqrt{0.2875 \times (1 - 0.2875) \times (\frac{1}{300} + \frac{1}{180})} \approx 0.0463 \).
05

Compute the test statistic

Calculate the z-score using the formula \( z = \frac{\hat{p_1} - \hat{p_2}}{SE} = \frac{0.21 - 0.4167}{0.0463} \approx -4.45 \).
06

Determine the critical value and conclusion for Part (a)

At \( \alpha = 0.05 \), the critical z-values for a two-tailed test are \( \pm 1.96 \). Since \( z = -4.45 \) is outside the range \([-1.96, 1.96]\), we reject \( H_0 \).
07

Calculate power for Part (b)

For \( H_0: p_1 = 0.2 \) and \( p_2 = 0.4 \), the power is calculated using the actual proportions and the standard error from these proportions: \( SE = \sqrt{\frac{0.2(1-0.2)}{300} + \frac{0.4(1-0.4)}{180}} \approx 0.0483 \). Calculate \( z \) for effect size: \( z = \frac{0.2 - 0.4}{0.0483} \approx -4.14 \).
08

Compute beta and power

Calculate the beta error, \( \beta \), the probability of not rejecting \( H_0 \) when \( H_a \) is true. \( \beta = P(Z > -1.96 + 4.14) + P(Z < 1.96 - 4.14) \approx P(Z > 2.18) \). From standard normal tables, \( \beta \approx 0.0146 \). Hence, power = 1 - \( \beta = 0.9854\).

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Key Concepts

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

Proportions
Proportions are a way to describe how a part relates to the whole. In statistics, when we talk about proportions, we often refer to the fraction of a population or sample that possesses a particular trait. For example, in the exercise, we have two specific groups - urban and rural residents - and we wish to know what fraction of each group favors increasing the highway speed limit. This is represented by the proportions \( p_1 \) and \( p_2 \) for urban and rural residents, respectively.
To find these proportions, you divide the number of people with the trait (favoring the speed limit increase) by the total number in the sample for each group. For instance, the sample proportion for urban residents is calculated as \( \hat{p_1} = \frac{63}{300} = 0.21 \), indicating that 21% of the urban residents sampled favor the increase. Understanding proportions is crucial because they help us to make predictions and decisions based on sample data.
Pooled Proportion
When comparing two proportions, a pooled proportion can be very useful, especially in hypothesis testing. The pooled proportion is essentially the overall proportion of a trait across combined groups. This approach assumes that the proportions of interest are equal, hence pooling them gives a single estimate of population proportion.
In the context of our problem, we combine data from both urban and rural groups to calculate the pooled proportion. The formula used is \( \hat{p} = \frac{x_1 + x_2}{n_1 + n_2} \). For this exercise, \( \hat{p} = \frac{63 + 75}{300 + 180} = 0.2875 \), indicating about 28.75% overall favor the speed limit increase. This pooled proportion is vital for computing the standard error and, later, the test statistic needed to decide whether the sentiment differs significantly between the groups.
Z-Test
The Z-Test is a statistical method that allows researchers to compare sample percentages to find evidence supporting or refuting a hypothesis about populations. It is often used when dealing with large sample sizes. For proportions, a Z-Test helps to ascertain if there exists a significant difference between two proportions from different populations.
To carry out a Z-Test in this exercise, we need to calculate the test statistic, commonly noted as the z-score. It is derived using the formula \( z = \frac{\hat{p_1} - \hat{p_2}}{SE} \), where \( SE \) is the standard error. Given \( \hat{p_1} = 0.21 \), \( \hat{p_2} = 0.4167 \), and \( SE = 0.0463 \), the calculated z-score is approximately -4.45. A large z-score, far from zero, suggests that the difference in sample proportions is not likely to be from random chance.
Next, we check this z-score against critical z-values that define a range of values expected under the null hypothesis. With \( \alpha = 0.05 \), which is typically for two-tailed tests, the critical values are \( \pm 1.96 \). Since our z-score falls outside this range, we have enough evidence to reject the null hypothesis \( H_0 \), suggesting that the sentiment does differ between urban and rural residents.
Power of a Test
The power of a statistical test is its capability to correctly reject the null hypothesis when it is false. A high power means the test is likely to detect significant effects. Typically, a power of 0.80 (or 80%) is considered adequate in research.
In this case study, the power is determined for the scenario where the true proportions differ: the true proportion of urban residents favoring the speed increase is 0.20, and for rural residents, it's 0.40. After computing the relevant standard error, we find the effect size, subsequently determining the z-score (approximately -4.14) for the difference in true proportions.
To find the power, we compute the probability of acquiring a test statistic as extreme as the critical value given the truth of the alternative hypothesis. If we look at the standard normal distribution, the power equals 1 minus the probability of committing a Type II error (\( \beta \)). The calculations here lead us to a significant power of approximately 0.9854, or 98.54%, indicating a very high likelihood of detecting the true difference.

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Most popular questions from this chapter

Let \(\mu_{1}\) and \(\mu_{2}\) denote true average tread lives for two competing brands of size P205/65R 15 radial tires. Test \(H_{0}: \mu_{1}-\mu_{2}=0\) versus \(H_{\mathrm{a}}: \mu_{1}-\) \(\mu_{2} \neq 0\) at level \(.05\) using the following data: \(m=45, \quad \bar{x}=42,500, \quad s_{1}=2200, \quad n=45\), \(\bar{y}=40,400\), and \(s_{2}=1900\).

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Assume that \(X\) is uniformly distributed on \((-1,1)\) and \(Y\) is split evenly between a uniform distribution on \((-101,-100)\) and a uniform distribution on ( 100,101 ). Thus the means are both 0 , but the variances differ strongly. We take random samples of size three from each distribution and apply a permutation test for the null hypothesis \(H_{0}: \mu_{1}=\mu_{2}\) against the alternative \(H_{a}: \mu_{1}<\mu_{2}\). a. Show that the probability is \(\frac{1}{8}\) that all three of the \(Y\) values come from \((100,101)\). b. Show that, if all three \(Y\) values come from \((100,101)\), then the \(P\)-value for the permutation test is .05. c. Explain why (a) and (b) are in conflict. What is the true probability that the permutation test rejects the null hypothesis at the \(.05\) level?

The article "Evaluation of a Ventilation Strategy to Prevent Barotrauma in Patients at High Risk for Acute Respiratory Distress Syndrome" (New Engl. J. Med., 1998: 355-358) reported on an experiment in which 120 patients with similar clinical features were randomly divided into a control group and a treatment group, each consisting of 60 patients. The sample mean ICU stay (days) and sample standard deviation for the treatment group were \(19.9\) and \(39.1\), respectively, whereas these values for the control group were \(13.7\) and \(15.8\). a. Calculate a point estimate for the difference between true average ICU stay for the treatment and control groups. Does this estimate suggest that there is a significant difference between true average stays under the two conditions? b. Answer the question posed in part (a) by carrying out a formal test of hypotheses. Is the result different from what you conjectured in part (a)? c. Does it appear that ICU stay for patients given the ventilation treatment is normally distributed? Explain your reasoning. d. Estimate true average length of stay for patients given the ventilation treatment in a way that conveys information about precision and reliability.

Using the traditional formula, a \(95 \%\) CI for \(p_{1}-p_{2}\) is to be constructed based on equal sample sizes from the two populations. For what value of \(n(=m)\) will the resulting interval have width at most \(.1\) irrespective of the results of the sampling?

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