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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.

H0:p1=p2

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.

H1:p1>p2

03

Calculate the sample statistics

Let n1denote the sample size of people over the age of 55.

n1=306

Let n2denote the sample size of people under the age of 25.

n2=298

Assume that x1and x2are the number ofpeople over 55 who dream in black and whiteand the number of peopleunder 25 who dream in black and whiterespectively.

Letp^1be the sample proportion ofpeople over 55 who dream in black and white.

p^1=x1n1=68306=0.2222

Thus,

q^1=1-p^1=0.7778

Let p^2be the sample proportion of people under 25 who dream in black and white.

p^2=x2n2=13298=0.0436

Thus,

q^2=1-p^2=0.9564

The value of the pooled sample proportion is equal to:

p¯=x1+x2n1+n2=68+13306+298=0.1341

Hence,

q¯=1-p¯=0.8659

04

Compute the value of test statistic

The test statistic is equal to:

z=p^1-p^2-p1-p2p¯q¯n1+p¯q¯n2=0.2222-0.0436-00.13410.8659306+0.13410.8659298=6.440

Referring to the standard normal distribution table, the critical value of z corresponding to α=0.01for 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 significanceα for the confidence interval method is 0.02.

The confidence interval has the following expression:

p^1-p^2-E<p1-p2<p^1-p^2+E

07

Calculate the margin of error

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

E=zα2p^1q^1n1+p^2q^2n2=2.33×0.22220.7778306+0.04360.9564298=0.0619

08

Construct the confidence interval.

b.

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

p^1-p^2-E<p1-p2<p^1-p^2+E(0.2222-0.0436)-0.0619<p1-p2<(0.2222-0.0436)+0.06190.1167<p1-p2<0.2405

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

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.

Is Echinacea Effective for Colds? Rhinoviruses typically cause common colds. In a test of the effectiveness of Echinacea, 40 of the 45 subjects treated with Echinacea developed rhinovirus infections. In a placebo group, 88 of the 103 subjects developed rhinovirus infections (based on data from “An Evaluation of Echinacea Angustifolia in Experimental Rhinovirus Infections,” by Turner et al., New England Journal of Medicine, Vol. 353, No. 4). We want to use a 0.05 significance level to test the claim that Echinacea has an effect on rhinovirus infections.

a. Test the claim using a hypothesis test.

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

c. Based on the results, does Echinacea appear to have any effect on the infection rate?

Confidence Interval for Haemoglobin

Large samples of women and men are obtained, and the haemoglobin level is measured in each subject. Here is the 95% confidence interval for the difference between the two population means, where the measures from women correspond to population 1 and the measures from men correspond to population 2:\(\)\( - 1.76g/dL < {\mu _1} - {\mu _2} < - 1.62g/dL\).

a. What does the confidence interval suggest about equality of the mean hemoglobin level in women and the mean hemoglobin level in men?

b. Write a brief statement that interprets that confidence interval.

c. Express the confidence interval with measures from men being population 1 and measures from women being population 2.

Find and interpret 95 % confidence interval for the proportion of all US adults who never clothes-shop online.

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.

Cardiac Arrest at Day and Night A study investigated survival rates for in hospital patients who suffered cardiac arrest. Among 58,593 patients who had cardiac arrest during the day, 11,604 survived and were discharged. Among 28,155 patients who suffered cardiac arrest at night, 4139 survived and were discharged (based on data from “Survival from In-Hospital Cardiac Arrest During Nights and Weekends,” by Puberty et al., Journal of the American Medical Association, Vol. 299, No. 7). We want to use a 0.01 significance level to test the claim that the survival rates are the same for day and night.

a. Test the claim using a hypothesis test.

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

c. Based on the results, does it appear that for in-hospital patients who suffer cardiac arrest, the survival rate is the same for day and night?

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 n1−1 and n2−1.)

Color and Creativity Researchers from the University of British Columbia conducted trials to investigate the effects of color on creativity. Subjects with a red background were asked to think of creative uses for a brick; other subjects with a blue background were given the same task. Responses were scored by a panel of judges and results from scores of creativity are given below. Higher scores correspond to more creativity. The researchers make the claim that “blue enhances performance on a creative task.”

a. Use a 0.01 significance level to test the claim that blue enhances performance on a creative task.

b. Construct the confidence interval appropriate for the hypothesis test in part (a). What is it about the confidence interval that causes us to reach the same conclusion from part (a)?

Red Background: n = 35, x = 3.39, s = 0.97

Blue Background: n = 36, x = 3.97, s = 0.63

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