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Chapter 11: Problem 40

In a recent poll, American adults were asked, if they have a choice, would they prefer to live in a city, suburb, or countryside. The following table shows the frequencies for the three choices. $$ \begin{array}{l|ccc} \hline \text { Response } & \text { City } & \text { Suburb } & \text { Countryside } \\ \hline \text { Frequency } & 640 & 790 & 570 \\ \hline \end{array} $$ Test at a \(1 \%\) significance level if these three places are equally preferred by American adults.

Short Answer

Expert verified
Based on Chi-Square goodness of fit test, there is no significant difference in preferences between living in a city, suburb, or countryside among American adults at the 1% significance level.

Step by step solution

01

State the hypotheses

The null hypothesis \(H_0\): The three places (city, suburb, and countryside) are equally preferred by American adults. \nThe alternative hypothesis \(H_a\): At least one place is preferred more than the others.
02

Calculate the Expected Frequencies

Under the null hypothesis, each place should be equally preferred. Find the total number of responses (Total = City + Suburb + Countryside = 640 + 790 + 570 = 2000). The expected frequency for each category is Total/3 = 2000/3 = 666.67.
03

Compute the Chi-Square Test Statistic

The test statistic is computed using the formula \(\chi^2 = \sum \frac{(O-E)^2}{E}\), where O refers to the observed frequency and E refers to the expected frequency. \(\chi^2 = \frac{(640-666.67)^2}{666.67} + \frac{(790-666.67)^2}{666.67} + \frac{(570-666.67)^2}{666.67} = 7.34\)
04

Determine the P-value

The P-value is the probability of getting a \(\chi^2\) value of 7.34 or more extreme, if the null hypothesis is true. For df=2 (three categories -1), and \(\alpha\) = 0.01, the critical value for \(\chi^2\) is 9.21. We observe that calculated \(\chi^2\) (7.34) < critical \(\chi^2\) (9.21), hence we cannot reject the null hypothesis.
05

Conclusion

Because the P-value is not less than the significance level (0.01), we fail to reject the null hypothesis. Hence, at the 1% level of significance, there is insufficient evidence to say that American adults don't have equal preferences for living in the city, suburb, and countryside.

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Key Concepts

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

Hypothesis Testing
Hypothesis testing is a fundamental statistical procedure used to determine whether the data supports a particular hypothesis or not. When engaging in hypothesis testing, you'll typically start with two hypotheses: the null hypothesis and the alternative hypothesis.
  • Null Hypothesis ( H_0 ): This hypothesis posits that there is no significant effect or difference. In our context, it suggests that American adults equally prefer living in a city, suburb, or countryside.
  • Alternative Hypothesis ( H_a ): This hypothesis claims that there is a significant effect or difference. Here, it implies that at least one of these living places is preferred more than the others.
The goal of hypothesis testing is to assess if the observed results reflect true preferences or if they occurred by random chance. Through statistical analysis, we can make informed decisions about the validity of the null hypothesis.
Significance Level
The significance level, denoted as \( \alpha \) , is a threshold set by the researcher to decide whether to reject the null hypothesis. It quantifies the risk of making a Type I error, which occurs when reality favors the null hypothesis, but based on the sample data, we incorrectly reject \(H_0\) .
  • For this problem, the significance level is set at 1% ( \( \alpha = 0.01 \) ), meaning we are willing to accept a 1% chance of mistakenly rejecting the null hypothesis.
  • A lower \( \alpha \) means more stringent criteria for demonstrating an effect or difference. This is important for ensuring the results are statistically valid.
The chosen significance level balances the need for robust evidence with the risk of identifying random variation as meaningful.
Expected Frequency
Expected frequencies are calculated under the assumption that the null hypothesis is true. They reflect the frequency count we would expect in each category if there were no true preference difference.
Let's see how expected frequencies are determined in action.
  • With a total of 2000 responses across three categories (city, suburb, countryside), the expected frequency per category is computed by dividing the total responses by the number of categories:
  • Expected frequency per category = \( \frac{2000}{3} \approx 666.67 \)
Comparing expected frequencies with observed data helps in measuring the extent of deviation, revealing whether the preferences are equal or not.
P-value
The P-value is an essential concept when interpreting the results of hypothesis testing. It is the probability of observing the test statistic, or something more extreme, under the null hypothesis.
  • For this exercise, the Chi-square statistic was computed to be 7.34, and with 2 degrees of freedom, the corresponding P-value is checked against the critical Chi-square value (9.21) for \( \alpha = 0.01 \).
  • If the P-value is less than \( \alpha \) , we reject the null hypothesis. If it is larger, we fail to reject \(H_0\) .
  • Here, since 7.34 is less than 9.21, the P-value does not lead us to reject the null hypothesis.
This indicates insufficient evidence to say preferences vary, maintaining that American adults may equally prefer all three living options.

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Most popular questions from this chapter

Explain how the expected frequencies for cells of a contingency table are calculated in a test of independence or homogeneity.

Of all students enrolled at a large undergraduate university, \(19 \%\) are seniors, \(23 \%\) are juniors, \(27 \%\) are sophomores, and \(31 \%\) are freshmen. A sample of 200 students taken from this university by the student senate to conduct a survey includes 50 seniors, 46 juniors, 55 sophomores, and 49 freshmen. Using a \(2.5 \%\) significance level, test the null hypothesis that this sample is a random sample.

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The manufacturer of a certain brand of lightbulbs claims that the variance of the lives of these bulbs is 4200 square hours. A consumer agency took a random sample of 25 such bulbs and tested them. The variance of the lives of these bulbs was found to be 5200 square hours. Assume that the lives of all such bulbs are (approximately) normally distributed. a. Make the \(99 \%\) confidence intervals for the variance and standard deviation of the lives of all such bulbs. b. Test at a \(5 \%\) significance level whether the variance of such bulbs is different from 4200 square hours.

Sandpaper is rated by the coarseness of the grit on the paper. Sandpaper that is more coarse will remove material faster. Jobs that involve the final sanding of bare wood prior to painting or sanding in between coats of paint require sandpaper that is much finer. A manufacturer of sandpaper rated 220, which is used for the final preparation of bare wood, wants to make sure that the variance of the diameter of the particles in their 220 sandpaper does not exceed \(2.0\) micrometers. Fifty-one randomly selected particles are measured. The variance of the particle diameters is \(2.13\) micrometers. Assume that the distribution of particle diameter is approximately normal. a. Construct the \(95 \%\) confidence intervals for the population variance and standard deviation. b. Test at a \(2.5 \%\) significance level whether the variance of the particle diameters of all particles in 220 -rated sandpaper is greater than \(2.0\) micrometers.
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