Question

a study was conducted to determine the proportion of people who dream in black and white instead of color. among 288 people over the age of 55, 64 dream in black and white, and among 299 people under the age of 25, 11 dream in black and white. use a 0.05 significance level to test the claim that the proportion of people over 55 who dream in black and white is greater than the proportion for those under 25. complete parts (a) through (c) below.

a. test the claim using a hypothesis test
consider the first sample to be the sample of people over the age of 55 and the second sample to be the sample of people under the age of 25. what are the null and alternative hypotheses for the hypothesis test?
a. ( h_0: p_1
eq p_2 )
( h_1: p_1 = p_2 )
b. ( h_0: p_1 = p_2 )
( h_1: p_1
eq p_2 )
c. ( h_0: p_1 leq p_2 )
( h_1: p_1
eq p_2 )
d. ( h_0: p_1 geq p_2 )
( h_1: p_1
eq p_2 )
e. ( h_0: p_1 = p_2 )
( h_1: p_1 > p_2 )
f. ( h_0: p_1 = p_2 )
( h_1: p_1 < p_2 )
identify the test statistic.
( z = square )
(round to two decimal places as needed.)

Explanation:

Step1: Identify Hypotheses

The claim is that the proportion of people over 55 (\(p_1\)) who dream in black and white is greater than that for those under 25 (\(p_2\)). So, the null hypothesis \(H_0\) is \(p_1 = p_2\) (equality), and the alternative hypothesis \(H_1\) is \(p_1 > p_2\). Looking at the options, option E matches: \(H_0: p_1 = p_2\), \(H_1: p_1 > p_2\).

Step2: Calculate Sample Proportions

For over 55: \(n_1 = 288\), \(x_1 = 64\), so \(\hat{p}_1=\frac{64}{288}\approx0.2222\). For under 25: Wait, the problem says "among 299 people under the age of 25, 11..." (assuming 11 dream in B&W, maybe typo? Wait, original: "among 288 people over 55, 64 dream... and among 299 people under 25, 11..." Let's confirm. So \(n_2 = 299\), \(x_2 = 11\), \(\hat{p}_2=\frac{11}{299}\approx0.0368\).

Step3: Calculate Pooled Proportion

\(\bar{p}=\frac{x_1 + x_2}{n_1 + n_2}=\frac{64 + 11}{288 + 299}=\frac{75}{587}\approx0.1278\), \(\bar{q}=1 - \bar{p}\approx0.8722\).

Step4: Calculate Test Statistic

The formula for the z - statistic for two proportions is \(z=\frac{\hat{p}_1 - \hat{p}_2}{\sqrt{\bar{p}\bar{q}(\frac{1}{n_1}+\frac{1}{n_2})}}\).

Substitute values: \(\hat{p}_1 - \hat{p}_2 = 0.2222 - 0.0368 = 0.1854\).

\(\sqrt{\bar{p}\bar{q}(\frac{1}{n_1}+\frac{1}{n_2})}=\sqrt{0.1278\times0.8722\times(\frac{1}{288}+\frac{1}{299})}\).

Calculate \(\frac{1}{288}+\frac{1}{299}\approx0.00347 + 0.00334 = 0.00681\).

Then \(0.1278\times0.8722\times0.00681\approx0.1278\times0.00594\approx0.00076\).

\(\sqrt{0.00076}\approx0.0276\).

Now, \(z=\frac{0.1854}{0.0276}\approx6.72\)? Wait, no, wait, maybe I misread \(x_2\). Wait, the original says "among 299 people under 25, 11" – no, maybe "11" is a typo? Wait, no, let's check again. Wait, the user's image: "Among 288 people over the age of 55, 64 dream in black and white, and among 299 people under the age of 25, 11..." So \(x_2 = 11\). Then recalculate:

\(\bar{p}=\frac{64 + 11}{288 + 299}=\frac{75}{587}\approx0.1278\) (correct).

\(\hat{p}_1 = 64/288 ≈ 0.2222\), \(\hat{p}_2 = 11/299 ≈ 0.0368\) (correct).

Standard error: \(\sqrt{\bar{p}(1 - \bar{p})(\frac{1}{n_1}+\frac{1}{n_2})}=\sqrt{0.1278\times0.8722\times(\frac{1}{288}+\frac{1}{299})}\).

Calculate \(\frac{1}{288}≈0.00347\), \(\frac{1}{299}≈0.00334\), sum ≈ 0.00681.

0.1278×0.8722≈0.1115, 0.1115×0.00681≈0.00076.

\(\sqrt{0.00076}≈0.0276\).

Then \(z=\frac{0.2222 - 0.0368}{0.0276}=\frac{0.1854}{0.0276}≈6.72\)? Wait, that seems high. Wait, maybe \(x_2\) is 111? Maybe a typo. Wait, original image: "among 299 people under the age of 25, 11" – maybe 11 is wrong. Wait, let's check the problem again. The user's image: "A study was conducted to determine the proportion of people who dream in black and white instead of color. Among 288 people over the age of 55, 64 dream in black and white, and among 299 people under the age of 25, 11..." Maybe it's 111? Let's assume maybe a typo, but proceed with given numbers. Wait, maybe I misread \(x_2\). Wait, the problem says "11" – so \(x_2 = 11\). Then:

\(\hat{p}_1 = 64/288 ≈ 0.2222\), \(\hat{p}_2 = 11/299 ≈ 0.0368\), \(\bar{p} = 75/587 ≈ 0.1278\).

Standard error: \(\sqrt{0.1278×0.8722×(1/288 + 1/299)} ≈ \sqrt{0.1115×0.00681} ≈ \sqrt{0.00076} ≈ 0.0276\).

Then \(z = (0.2222 - 0.0368)/0.0276 ≈ 0.1854/0.0276 ≈ 6.72\). But that seems very high. Wait, maybe \(x_2\) is 110? Let's check: if \(x_2 = 110\), then \(\hat{p}_2 = 110/299 ≈ 0.368\), \(\bar{p} = (64 + 110)/587 = 174/587 ≈ 0.2964\), standard error: \(\sqrt{0.2964×0.7036×(1/288 + 1/299)} ≈ \sqrt{0.2086×0.00681} ≈ \sqrt{0.00142} ≈ 0.0377\), \(z = (0.2222 - 0.368)/0.0377 ≈ -3.87\), which is different. But the original problem says "11" – maybe that's correct. Wait, maybe I made a mistake. Let's re - calculate:

\(n_1 = 288\), \(x_1 = 64\), \(\hat{p}_1 = 64/288 = 8/36 = 2/9 ≈ 0.2222\).

\(n_2 = 299\), \(x_2 = 11\), \(\hat{p}_2 = 11/299 ≈ 0.0368\).

Pooled proportion \(\bar{p} = (64 + 11)/(288 + 299) = 75/587 ≈ 0.1278\).

Standard error \(SE=\sqrt{\bar{p}(1 - \bar{p})(\frac{1}{n_1}+\frac{1}{n_2})}=\sqrt{0.1278×0.8722×(\frac{1}{288}+\frac{1}{299})}\).

Calculate \(\frac{1}{288}+\frac{1}{299}=\frac{299 + 288}{288×299}=\frac{587}{86112}≈0.00681\).

Then \(0.1278×0.8722 = 0.1278×0.8722≈0.1115\).

\(0.1115×0.00681≈0.00076\).

\(SE=\sqrt{0.00076}≈0.0276\).

\(z=\frac{\hat{p}_1 - \hat{p}_2}{SE}=\frac{0.2222 - 0.0368}{0.0276}=\frac{0.1854}{0.0276}≈6.72\).

Answer:

Hypotheses: E. \(H_0: p_1 = p_2\), \(H_1: p_1 > p_2\)

Test Statistic: \(z\approx6.72\) (assuming \(x_2 = 11\); if \(x_2\) is different, recalculate)

Wait, but the problem's first part is choosing the hypotheses. The correct option for hypotheses is E: \(H_0: p_1 = p_2\), \(H_1: p_1 > p_2\) because the claim is \(p_1 > p_2\). For the test statistic, with \(x_2 = 11\), we get \(z\approx6.72\).