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Testing Claims About Proportions. In Exercises 7–22, test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value or critical value(s), then state the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim.

Dreaming in Black and White A study was conducted to determine the proportion of people who dream in black and white instead of color. Among 306 people over the age of 55, 68 dream in black and white, and among 298 people under the age of 25, 13 dream in black and white (based on data from “Do We Dream in Color?” by Eva Murzyn, Consciousness and Cognition, Vol. 17, No. 4). We want to use a 0.01 significance level to test the claim that the proportion of people over 55 who dream in black and white is greater than the proportion of those under 25.

a. Test the claim using a hypothesis test.

b. Test the claim by constructing an appropriate confidence interval.

c. An explanation given for the results is that those over the age of 55 grew up exposed to media that was mostly displayed in black and white. Can the results from parts (a) and (b) be used to verify that explanation?

Short Answer

Expert verified

a.There is sufficient evidence to support the claimthat the proportion of people over 55 who dream in black and white is greater than the proportion of people under 25 who dream in black and white.

b.The confidence interval is equal to (0.1167, 0.2405), and it suggeststhat the proportion of people over 55 who dream in black and white is greater than the proportion of people under 25 who dream in black and white.

c.The sample results can not be used to verify the cause of the difference in the two proportions.

Step by step solution

01

Given information

For a sample of 306 people over the age of 55, 68 dream in black and white, while for another sample of 298 people under the age of 25,13 dream in black and white.

02

Describe the hypotheses to be tested

Null hypothesis:The proportion of people over 55 who dream in black and white is equal to the proportion of people under 25 who dream in black and white.

\({H_0}:{p_1} = {p_2}\)

Alternate hypothesis: The proportion of people over 55 who dream in black and white is greater than the proportion of people under 25 who dream in black and white.

\({\rm{ }}{H_1}:{p_1} > {p_2}\)

03

Calculate the sample statistics

Let\({n_1}\)denote the sample size of people over the age of 55.

\({n_1} = 306\)

Let\({n_2}\)denote the sample size of people under the age of 25.

\({n_2} = 298\)

Assume that\({x_1}\)and\({x_2}\)are the number ofpeople over 55 who dream in black and whiteand the number of peopleunder 25 who dream in black and whiterespectively.

Let \({\hat p_1}\)be the sample proportion ofpeople over 55 who dream in black and white.

\(\begin{array}{c}{{\hat p}_1} = \frac{{{x_1}}}{{{n_1}}}{\rm{ }}\\ = \frac{{68}}{{306}}\\ = 0.2222\end{array}\)

Thus,

\(\begin{array}{c}{{\hat q}_1} = 1 - {{\hat p}_1}\\ = 0.7778\end{array}\)

Let \({\hat p_2}\)be the sample proportion ofpeople under 25 who dream in black and white.\(\begin{array}{c}{\rm{ }}{{\hat p}_2} = \frac{{{x_2}}}{{{n_2}}}\\ = \frac{{13}}{{298}}\\ = 0.0436\end{array}\)

Thus,

\(\begin{array}{c}{{\hat q}_2} = 1 - {{\hat p}_2}\\ = 0.9564\end{array}\)

The value of the pooled sample proportion is equal to:

\(\begin{array}{c}\bar p = \frac{{{x_1} + {x_2}}}{{{n_1} + {n_2}}}\\ = \frac{{68 + 13}}{{306 + 298}}\\ = 0.1341\end{array}\)

Hence,

\(\begin{array}{c}\bar q = 1 - \bar p\\ = 0.8659\end{array}\)

04

Compute the value of test statistic

The test statistic is equal to:

\(\begin{array}{c}z = \frac{{\left( {{{\hat p}_1} - {{\hat p}_2}} \right) - \left( {{p_1} - {p_2}} \right)}}{{\sqrt {\frac{{\bar p\bar q}}{{{n_1}}} + \frac{{\bar p\bar q}}{{{n_2}}}} }}\\ = \frac{{\left( {0.2222 - 0.0436} \right) - 0}}{{\sqrt {\frac{{\left( {0.1341} \right)\left( {0.8659} \right)}}{{306}} + \frac{{\left( {0.1341} \right)\left( {0.8659} \right)}}{{298}}} }}\\ = 6.440\end{array}\)

Referring to the standard normal distribution table, the critical value of z corresponding to\(\alpha = 0.01\)for a right-tailed test is equal to 2.33.

Referring to the standard normal distribution table, the corresponding p-value is equal to 0.000.

Here, the value of the test statistic is greater than the critical value, and the p-value is less than 0.01.

Therefore, the null hypothesis is rejected.

05

Conclusion of the test

a.

There is sufficient evidence to support the claimthat the proportion of people over 55 who dream in black and white is greater than the proportion of people under25 who dream in black and white.

06

Describe the confidence interval

b.

If the level of significance for a one-tailed test is equal to 0.01, then the corresponding confidence level to construct the confidence interval is equal to 98%. Thus, the level of significance\(\left( \alpha \right)\)for the confidence interval method is 0.02.

The confidence interval has the following expression:

\(\left( {{{\hat p}_1} - {{\hat p}_2}} \right) - E < {p_1} - {p_2} < \left( {{{\hat p}_1} - {{\hat p}_2}} \right) + E\)

07

Calculate the margin of error

E is the margin of error and has the following formula:

\(\begin{array}{c}E = {z_{\frac{\alpha }{2}}}\sqrt {\frac{{{{\hat p}_1}{{\hat q}_1}}}{{{n_1}}} + \frac{{{{\hat p}_2}{{\hat q}_2}}}{{{n_2}}}} \\ = 2.33 \times \sqrt {\frac{{\left( {0.2222} \right)\left( {0.7778} \right)}}{{306}} + \frac{{\left( {0.0436} \right)\left( {0.9564} \right)}}{{298}}} \\ = 0.0619\end{array}\)

08

Construct the confidence interval.

b.

Substituting the required values, the following confidence interval is obtained:

\(\begin{array}{c}\left( {{{\hat p}_1} - {{\hat p}_2}} \right) - E < {p_1} - {p_2} < \left( {{{\hat p}_1} - {{\hat p}_2}} \right) + E\\(0.2222 - 0.0436) - 0.0619 < {p_1} - {p_2} < (0.2222 - 0.0436) + 0.0619\\0.1167 < {p_1} - {p_2} < 0.2405\end{array}\)

Thus, the 99% confidence interval is equal to (0.1167, 0.2405).

This confidence interval does not contain zero and consists of only positive values. This implies that there is a significant difference between the two proportions of people who dream in black in white.

Therefore, there is sufficient evidence to support the claimthat the proportion of people over 55 who dream in black and white is greater than the proportion of people under 25 who dream in black and white.

09

Reason behind the difference in the proportions

c.

The sample proportion of people over 55 who dream in black and white is equal to 22.22%.

The sample proportion of people under 25 who dream in black and white is equal to 4.36%.

It is believed that most people over the age of 55 dream in black and white as they have witnessed black and white media to a great extent.

Although the sample results support the belief, the actual reason behind the results cannot be verified.

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Most popular questions from this chapter

In Exercises 5–20, assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. (Note: Answers in Appendix D include technology answers based on Formula 9-1 along with “Table” answers based on Table A-3 with df equal to the smaller of\({n_1} - 1\)and\({n_2} - 1\).) Bad Stuff in Children’s Movies Data Set 11 “Alcohol and Tobacco in Movies” in Appendix B includes lengths of times (seconds) of tobacco use shown in animated children’s movies. For the Disney movies, n = 33,\(\bar x\)= 61.6 sec, s = 118.8 sec. For the other movies, n = 17,\(\bar x\)= 49.3 sec, s = 69.3 sec. The sorted times for the non-Disney movies are listed below.

a. Use a 0.05 significance level to test the claim that Disney animated children’s movies and other animated children’s movies have the same mean time showing tobacco use.

b. Construct a confidence interval appropriate for the hypothesis test in part (a).

c. Conduct a quick visual inspection of the listed times for the non-Disney movies and comment on the normality requirement. How does the normality of the 17 non-Disney times affect the results?

0 0 0 0 0 0 1 5 6 17 24 55 91 117 155 162 205

Testing Claims About Proportions. In Exercises 7–22, test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value or critical value(s), then state the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim.

Overlap of Confidence Intervals In the article “On Judging the Significance of Differences by Examining the Overlap Between Confidence Intervals,” by Schenker and Gentleman (American Statistician, Vol. 55, No. 3), the authors consider sample data in this statement: “Independent simple random samples, each of size 200, have been drawn, and 112 people in the first sample have the attribute, whereas 88 people in the second sample have the attribute.”

a. Use the methods of this section to construct a 95% confidence interval estimate of the difference p1-p2. What does the result suggest about the equality of p1andp2?

b. Use the methods of Section 7-1 to construct individual 95% confidence interval estimates for each of the two population proportions. After comparing the overlap between the two confidence intervals, what do you conclude about the equality ofp1andp2?

c. Use a 0.05 significance level to test the claim that the two population proportions are equal. What do you conclude?

d. Based on the preceding results, what should you conclude about the equality ofp1andp2? Which of the three preceding methods is least effective in testing for the equality ofp1andp2?

Testing Claims About Proportions. In Exercises 7–22, test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value or critical value(s), then state the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim.

Are Seat Belts Effective? A simple random sample of front-seat occupants involved in car crashes is obtained. Among 2823 occupants not wearing seat belts, 31 were killed. Among 7765 occupants wearing seat belts, 16 were killed (based on data from “Who Wants Airbags?” by Meyer and Finney, Chance, Vol. 18, No. 2). We want to use a 0.05 significance level to test the claim that seat belts are effective in reducing fatalities.

a. Test the claim using a hypothesis test.

b. Test the claim by constructing an appropriate confidence interval.

c. What does the result suggest about the effectiveness of seat belts?

Critical Thinking: Did the NFL Rule Change Have the Desired Effect? Among 460 overtime National Football League (NFL) games between 1974 and 2011, 252 of the teams that won the overtime coin toss went on to win the game. During those years, a team could win the coin toss and march down the field to win the game with a field goal, and the other team would never get possession of the ball. That just didn’t seem fair. Starting in 2012, the overtime rules were changed. In the first three years with the new overtime rules, 47 games were decided in overtime and the team that won the coin toss won 24 of those games. Analyzing the Results

First explore the two proportions of overtime wins. Does there appear to be a difference? If so, how?

Braking Reaction Times: Normal? The accompanying normal quantile plot is obtained by using the braking reaction times of females listed in Exercise 6. Interpret this graph.

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