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Chapter 8: Problem 29

Would you favor spending more federal tax money on the arts? This question was asked by a research group on behalf of The National Institute (Reference: Painting by Numbers, J. Wypijewski, University of California Press). Of a random sample of \(n_{1}=220\) women, \(r_{1}=59\) responded yes. Another random sample of \(n_{2}=175\) men showed that \(r_{2}=56\) responded yes. Does this information indicate a difference (either way) between the population proportion of women and the population proportion of men who favor spending more federal tax dollars on the arts? Use \(\alpha=0.05\)

Short Answer

Expert verified
There is no significant difference between the proportions of women and men favoring more federal tax money on the arts.

Step by step solution

01

Define Hypotheses

We need to set up our null and alternative hypotheses. The null hypothesis is that there is no difference in the population proportions: \( H_0: p_1 - p_2 = 0 \). The alternative hypothesis states there is a difference: \( H_a: p_1 - p_2 eq 0 \).
02

Calculate Sample Proportions

Calculate the sample proportions for women and men. For women, \( \hat{p}_1 = \frac{r_1}{n_1} = \frac{59}{220} \approx 0.268 \). For men, \( \hat{p}_2 = \frac{r_2}{n_2} = \frac{56}{175} \approx 0.32 \).
03

Calculate Pooled Proportion

The pooled proportion is calculated as \( \hat{p} = \frac{r_1 + r_2}{n_1 + n_2} = \frac{59 + 56}{220 + 175} = \frac{115}{395} \approx 0.291 \).
04

Calculate Standard Error

The standard error for the difference between two proportions is \( SE = \sqrt{\hat{p}(1 - \hat{p})(\frac{1}{n_1} + \frac{1}{n_2})} \). Substituting the values, we get \( SE = \sqrt{0.291(1 - 0.291)(\frac{1}{220} + \frac{1}{175})} \approx 0.058 \).
05

Calculate Test Statistic

The test statistic is calculated as \( z = \frac{\hat{p}_1 - \hat{p}_2}{SE} = \frac{0.268 - 0.32}{0.058} \approx -0.897 \).
06

Determine Critical Value and Decision

Using a significance level \( \alpha = 0.05 \) and a two-tailed test, the critical z-value is approximately \( \pm 1.96 \). Since \(-0.897\) is within the range \(-1.96\) and \(1.96\), 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.

Sample Proportions
When conducting hypothesis testing with proportions, the first step is to determine the sample proportions. These help in understanding the ratio of individuals in each sample responding positively to a question or condition. In the exercise, we look at two groups: women and men.
  • For women, the sample proportion is calculated as: \[\hat{p}_1 = \frac{r_1}{n_1} = \frac{59}{220} \approx 0.268\]This means approximately 26.8% of the sampled women favor spending more on the arts.
  • For men, the sample proportion is:\[\hat{p}_2 = \frac{r_2}{n_2} = \frac{56}{175} \approx 0.32\]Thus, about 32% of sampled men support the idea.
Sample proportions give a preliminary glimpse into the data but aren't yet conclusive by themselves in determining statistical significance.
Pooled Proportion
The pooled proportion combines the successes from both samples, providing a common proportion estimate. This is particularly useful when comparing two samples. For the given exercise, the pooled proportion is calculated as:\[\hat{p} = \frac{r_1 + r_2}{n_1 + n_2} = \frac{59 + 56}{220 + 175} = \frac{115}{395} \approx 0.291\]This represents the overall proportion of individuals favoring arts spending, combining both groups.The pooled proportion helps to unify the two datasets, giving as a crucial component for further calculations like the standard error and eventually in validating our hypothesis.
Standard Error
The standard error provides a measure of variability or dispersion within the sample proportion differences. Understanding this helps in assessing how much the observed difference in sample proportions might diverge from the true population difference. The formula for the standard error of the difference between the two proportions is:\[SE = \sqrt{\hat{p}(1 - \hat{p})\left(\frac{1}{n_1} + \frac{1}{n_2}\right)}\]Substituting the values from the exercise, we get:\[SE = \sqrt{0.291(1 - 0.291)\left(\frac{1}{220} + \frac{1}{175}\right)} \approx 0.058\]This value of standard error reflects how much variation we can expect if there was no real difference in the population proportions.
Z-value
The z-value is calculated in hypothesis testing to determine how far the sample proportion difference is from the hypothesized no-difference scenario, relative to the standard error.For this exercise, the z-value is given by:\[z = \frac{\hat{p}_1 - \hat{p}_2}{SE} = \frac{0.268 - 0.32}{0.058} \approx -0.897\]A z-value of this size indicates how unusual our observed result is under the assumption that the null hypothesis holds.In hypothesis testing, the z-value is crucial for deciding whether to accept or reject the null hypothesis, especially when compared to a critical value obtained from standard statistical tables.
Significance Level
The significance level, denoted by \( \alpha \), is a threshold set for deciding whether an observed effect is statistically significant in hypothesis testing. It reflects the probability of rejecting the null hypothesis when it is true (Type I error).
  • In this exercise, the significance level is set at \( \alpha = 0.05 \).
  • This means we allow a 5% chance of incorrectly stating a difference exists when it actually doesn鈥檛.
Using a two-tailed test with \( \alpha = 0.05 \), the standard critical z-values are \( \pm 1.96 \). In our case, since the calculated z-value \(-0.897\) falls within the range \(-1.96\) to \(1.96\), we don't have enough evidence to reject the null hypothesis.Thus, based on this significance level, we conclude that there is no statistically significant difference in the proportions of men and women favoring more arts spending.

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Most popular questions from this chapter

Pyramid Lake is on the Paiute Indian Reservation in Nevada. The lake is famous for cutthroat trout. Suppose a friend tells you that the average length of trout caught in Pyramid Lake is \(\mu=19\) inches. However, the Creel Survey (published by the Pyramid Lake Paiute Tribe Fisheries Association) reported that of a random sample of 51 fish caught, the mean length was \(\bar{x}=18.5\) inches, with estimated standard deviation \(s=3.2\) inches. Do these data indicate that the average length of a trout caught in Pyramid Lake is less than \(\mu=19\) inches? Use \(\alpha=0.05\).

The U.S. Department of Transportation, National Highway Traffic Safety Administration, reported that \(77 \%\) of all fatally injured automobile drivers were intoxicated. A random sample of 27 records of automobile driver fatalities in Kit Carson County, Colorado, showed that 15 involved an intoxicated driver. Do these data indicate that the population proportion of driver fatalities related to alcohol is less than \(77 \%\) in Kit Carson County? Use \(\alpha=0.01\).

Bill Alther is a zoologist who studies Anna's hummingbird (Calypte anna) (Reference: Hummingbirds by K. Long and W. Alther). Suppose that in a remote part of the Grand Canyon, a random sample of six of these birds was caught, weighed, and released. The weights (in grams) were \(\begin{array}{llllll}3.7 & 2.9 & 3.8 & 4.2 & 4.8 & 3.1\end{array}\) The sample mean is \(\bar{x}=3.75\) grams. Let \(x\) be a random variable representing weights of Anna's hummingbirds in this part of the Grand Canyon. We assume that \(x\) has a normal distribution and \(\sigma=0.70\) gram. It is known that for the population of all Anna's hummingbirds, the mean weight is \(\mu=4.55\) grams. Do the data indicate that the mean weight of these birds in this part of the Grand Canyon is less than \(4.55\) grams? Use \(\alpha=0.01\).

REM (rapid eye movement) sleep is sleep during which most dreams occur. Each night a person has both REM and non-REM sleep. However, it is thought that children have more REM sleep than adults (Reference: Secrets of Sleep by Dr. A. Borbely). Assume that REM sleep time is normally distributed for both children and adults. A random sample of \(n_{1}=10\) children (9 years old) showed that they had an average REM sleep time of \(\bar{x}_{1}=2.8\) hours per night. From previous studies, it is known that \(\sigma_{1}=0.5\) hour. Another random sample of \(n_{2}=10\) adults showed that they had an average REM sleep time of \(\bar{x}_{2}=2.1\) hours per night. Previous studies show that \(\sigma_{2}=0.7\) hour. Do these data indicate that, on average, children tend to have more REM sleep than adults? Use a \(1 \%\) level of significance.

A random sample of size 16 from a normal distribution with \(\sigma=3\) produced a sample mean of \(4.5\). (a) Cbeck Requirements Is the \(\bar{x}\) distribution normal? Explain. (b) Compute the sample test statistic \(z\) under the null hypothesis \(H_{0}: \mu=6.3\). (c) For \(H_{1}: \mu<6.3\), estimate the \(P\) -value of the test statistic. (d) For a level of significance of \(0.01\) and the hypotheses of parts (b) and (c), do you reject or fail to reject the null hypothesis? Explain.
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