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?

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.

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.

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

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,

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

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.

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.

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:

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

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

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

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.

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.