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)
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 = p_2 ), ( h_1: p_1
eq 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 = 6.72 ) (round to two decimal places as needed)
identify the p - value.
p - value = 0.000 (round to three decimal places as needed)
what is the conclusion based on the hypothesis test?
the p - value is less than the significance level of ( alpha = 0.05 ), so reject the null hypothesis. there is sufficient evidence to support the claim that the proportion of people over 55 who dream in black and white is greater than the proportion for those under 25.
b. test the claim by constructing an appropriate confidence interval.
the 90% confidence interval is ( square < (p_1 - p_2) < square ) (round to three decimal places as needed)

Explanation:

Part a: Hypothesis Testing

Step 1: 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 \( p_1 \leq p_2 \), and the alternative hypothesis (\( H_1 \)) is \( p_1 > p_2 \). Looking at the options, this matches option E: \( H_0: p_1 = p_2 \) (note: \( p_1 \leq p_2 \) is equivalent to \( p_1 = p_2 \) in the null for a right - tailed test setup here) and \( H_1: p_1 > p_2 \).

Step 2: Test Statistic and P - value

We are given the test statistic \( z = 6.72 \) and the P - value \( = 0.000 \).

Step 3: Conclusion

Since the significance level \( \alpha=0.05 \) and the P - value (\( 0.000<0.05 \)), we reject the null hypothesis. There is sufficient evidence to support the claim that the proportion of people over 55 who dream in black and white is greater than that for those under 25.

Part b: Confidence Interval for \( p_1 - p_2 \)

First, we calculate the sample proportions:

• For the over - 55 group: \( n_1 = 288 \), \( x_1 = 64 \), so \( \hat{p}_1=\frac{64}{288}\approx0.2222 \)
• For the under - 25 group: \( n_2 = 299 \), \( x_2 = 11 \), so \( \hat{p}_2=\frac{11}{299}\approx0.0368 \)

The formula for the confidence interval for \( p_1 - p_2 \) is:

$$ (\hat{p}_1-\hat{p}_2)\pm z_{\alpha/2}\sqrt{\frac{\hat{p}_1(1 - \hat{p}_1)}{n_1}+\frac{\hat{p}_2(1 - \hat{p}_2)}{n_2}} $$

For a 90% confidence interval, \( \alpha = 0.10 \), so \( \alpha/2=0.05 \) and \( z_{\alpha/2}=z_{0.05}=1.645 \)

Step 1: Calculate \( \hat{p}_1-\hat{p}_2 \)

\( \hat{p}_1-\hat{p}_2=0.2222 - 0.0368 = 0.1854 \)

Step 2: Calculate the standard error (SE)

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Step 3: Calculate the margin of error (ME)

\( ME = z_{\alpha/2}\times SE=1.645\times0.0268\approx0.0441 \)

Step 4: Calculate the confidence interval

Lower bound: \( 0.1854-0.0441 = 0.1413 \)
Upper bound: \( 0.1854 + 0.0441=0.2295 \)

Rounding to three decimal places, the 90% confidence interval for \( (p_1 - p_2) \) is \( 0.141

Final Answers

Part a

• Hypotheses: E. \( H_0: p_1 = p_2 \), \( H_1: p_1 > p_2 \)
• Test statistic: \( z = 6.72 \)
• P - value: \( 0.000 \)
• Conclusion: Reject \( H_0 \). There is sufficient evidence to support the claim.

Part b

The 90% confidence interval is \( 0.141<(p_1 - p_2)<0.230 \)

Answer:

Part a: Hypothesis Testing

Step 1: 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 \( p_1 \leq p_2 \), and the alternative hypothesis (\( H_1 \)) is \( p_1 > p_2 \). Looking at the options, this matches option E: \( H_0: p_1 = p_2 \) (note: \( p_1 \leq p_2 \) is equivalent to \( p_1 = p_2 \) in the null for a right - tailed test setup here) and \( H_1: p_1 > p_2 \).

Step 2: Test Statistic and P - value

We are given the test statistic \( z = 6.72 \) and the P - value \( = 0.000 \).

Step 3: Conclusion

Since the significance level \( \alpha=0.05 \) and the P - value (\( 0.000<0.05 \)), we reject the null hypothesis. There is sufficient evidence to support the claim that the proportion of people over 55 who dream in black and white is greater than that for those under 25.

Part b: Confidence Interval for \( p_1 - p_2 \)

First, we calculate the sample proportions:

• For the over - 55 group: \( n_1 = 288 \), \( x_1 = 64 \), so \( \hat{p}_1=\frac{64}{288}\approx0.2222 \)
• For the under - 25 group: \( n_2 = 299 \), \( x_2 = 11 \), so \( \hat{p}_2=\frac{11}{299}\approx0.0368 \)

The formula for the confidence interval for \( p_1 - p_2 \) is:

$$ (\hat{p}_1-\hat{p}_2)\pm z_{\alpha/2}\sqrt{\frac{\hat{p}_1(1 - \hat{p}_1)}{n_1}+\frac{\hat{p}_2(1 - \hat{p}_2)}{n_2}} $$

For a 90% confidence interval, \( \alpha = 0.10 \), so \( \alpha/2=0.05 \) and \( z_{\alpha/2}=z_{0.05}=1.645 \)

Step 1: Calculate \( \hat{p}_1-\hat{p}_2 \)

\( \hat{p}_1-\hat{p}_2=0.2222 - 0.0368 = 0.1854 \)

Step 2: Calculate the standard error (SE)

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Step 3: Calculate the margin of error (ME)

\( ME = z_{\alpha/2}\times SE=1.645\times0.0268\approx0.0441 \)

Step 4: Calculate the confidence interval

Lower bound: \( 0.1854-0.0441 = 0.1413 \)
Upper bound: \( 0.1854 + 0.0441=0.2295 \)

Rounding to three decimal places, the 90% confidence interval for \( (p_1 - p_2) \) is \( 0.141

Final Answers

Part a

• Hypotheses: E. \( H_0: p_1 = p_2 \), \( H_1: p_1 > p_2 \)
• Test statistic: \( z = 6.72 \)
• P - value: \( 0.000 \)
• Conclusion: Reject \( H_0 \). There is sufficient evidence to support the claim.

Part b

The 90% confidence interval is \( 0.141<(p_1 - p_2)<0.230 \)