Question 16, 9.1.11-T
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A study was conducted to determine the proportion of people who dream in black and white instead of color. Among 304 people over the age of 55,79 dream in black and white, and among 299 people under the age of 25, 12 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.

P-value $=0$
(Round to three decimal places as needed.)
What is the conclusion based on the hypothesis test?
The P-value is less than the signifibance 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 $\square< \left(p_{1}-p_{2}\right)< \square$.
(Round to three decimal places as needed.)
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Step 1 ：Define the null hypothesis (H0) as the proportion of people over 55 who dream in black and white is equal to the proportion of those under 25. The alternative hypothesis (H1) is that the proportion of people over 55 who dream in black and white is greater than the proportion of those under 25.

Step 2 ：Given the P-value is 0, which is less than the significance level of 0.05, we reject the null hypothesis. This suggests that 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.

Step 3 ：Calculate the sample proportions and their difference: For people over 55: \(p1 = \frac{79}{304} = 0.260\), For people under 25: \(p2 = \frac{12}{299} = 0.040\), Difference: \(p1 - p2 = 0.260 - 0.040 = 0.220\)

Step 4 ：Calculate the standard error (SE) of the difference: \(SE = \sqrt{\left(\frac{p1*(1-p1)}{304}\right) + \left(\frac{p2*(1-p2)}{299}\right)} = \sqrt{\left(\frac{0.260*0.740}{304}\right) + \left(\frac{0.040*0.960}{299}\right)} = 0.027\)

Step 5 ：Construct a 90% confidence interval for the difference in proportions: \(p1 - p2 \pm Z*SE\), where Z is the Z-score for a 90% confidence interval, which is 1.645. Lower limit: \(0.220 - 1.645*0.027 = 0.176\), Upper limit: \(0.220 + 1.645*0.027 = 0.264\)

Step 6 ：\(\boxed{\text{Therefore, the 90% confidence interval for the difference in proportions is } 0.176 < (p1 - p2) < 0.264. \text{ This interval does not contain 0, which further supports the conclusion that the proportion of people over 55 who dream in black and white is greater than the proportion for those under 25.}}\)