A study was conducted to determine the proportion of people who dream in black and white instead of color. Among 295 people over the age of 55, 69 dream in black and white, and among 290 people under the age of 25, 11 dream in black and white. 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 for those under 25. Complete parts (a) through (c) below.

a. State the hypotheses:
A. [tex]\( H_0: p_1 \leq p_2 \)[/tex] [tex]\( H_1: p_1 \neq p_2 \)[/tex]
B.
[tex]\[
\begin{array}{l}
H_0: p_1 = p_2 \\
H_1: p_1 \ \textless \ p_2
\end{array}
\][/tex]
C. [tex]\( H_0: p_1 \geq p_2 \)[/tex]
D. [tex]\( H_0: p_1 = p_2 \)[/tex] [tex]\( H_1: p_1 \neq p_2 \)[/tex]

b. Identify the test statistic:
[tex]\[
z = 6.90
\][/tex]
(Round to two decimal places as needed.)

c. Identify the [tex]\( P \)[/tex]-value:
[tex]\[
P \text{-value} = 0.000
\][/tex]
(Round to three decimal places as needed.)

d. What is the conclusion based on the hypothesis test?
The [tex]\( P \)[/tex]-value is less than the significance level of [tex]\( \alpha = 0.01 \)[/tex], so 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 the proportion for those under 25.

e. Test the claim by constructing an appropriate confidence interval:
The [tex]\( 98\% \)[/tex] confidence interval is [tex]\( \square \, \square \, \square \ \textless \ \left( p_1 - p_2 \right) \ \textless \ \square \, \square \, \square \)[/tex].
(Round to three decimal places as needed.)



Answer :

Let's go through the problem step-by-step, including parts (a) through (c).

### Step-by-Step Solution

#### Part (a): Formulating Hypotheses

To test the claim that the proportion of people over 55 ([tex]\( p_1 \)[/tex]) who dream in black and white is greater than the proportion for those under 25 ([tex]\( p_2 \)[/tex]), we need to set up our hypotheses:
- [tex]\( H_0 \)[/tex]: [tex]\( p_1 \leq p_2 \)[/tex]
- [tex]\( H_1 \)[/tex]: [tex]\( p_1 > p_2 \)[/tex]

Given the options provided:
- None of the hypotheses pair fits exactly this scenario, but the closest is option A:
- [tex]\( H_0: p_1 \leq p_2 \)[/tex]
- [tex]\( H_1: p_1 \neq p_2 \)[/tex]

However, it should be noted that the correct null and alternative hypotheses for the given claim would be:
- [tex]\( H_0: p_1 \leq p_2 \)[/tex]
- [tex]\( H_1: p_1 > p_2 \)[/tex]


#### Part (b): Identifying the Test Statistic

The test statistic for comparing two proportions is calculated using the formula for the z-score. Given the results:
- The test statistic [tex]\( z \)[/tex] is 6.90 (rounded to two decimal places).

So, the test statistic is:
[tex]\[ z = 6.90 \][/tex]

#### Part (c): Identifying the P-value

The P-value corresponds to the probability of observing a test statistic as extreme as the one observed, under the null hypothesis. Given the calculated z-score:

- The P-value is extremely small: 0.000 (rounded to three decimal places).

So, the P-value is:
[tex]\[ \text{P-value} = 0.000 \][/tex]

#### Conclusion Based on the Hypothesis Test

Since the P-value is less than the significance level of [tex]\( \alpha = 0.01 \)[/tex]:
- 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 the proportion for those under 25.

#### Part (d): Constructing a Confidence Interval

We are asked to construct a 98% confidence interval for the difference in proportions ([tex]\( p_1 - p_2 \)[/tex]). The confidence interval formula for the difference in proportions is:
[tex]\[ \left( p_1 - p_2 \right) \pm \left( z^* \times \text{SE} \right) \][/tex]

Given the results:
- The 98% confidence interval is: [tex]\((0.129870932, 0.262063609)\)[/tex] (rounded to three decimal places).

So, the 98% confidence interval for the difference in proportions ([tex]\( p_1 - p_2 \)[/tex]) is:
[tex]\[ 0.130 < (p_1 - p_2) < 0.262 \][/tex]

### Final Answers Summary
1. Hypotheses: [tex]\( H_0: p_1 \leq p_2 \)[/tex], [tex]\( H_1: p_1 > p_2 \)[/tex]
2. Test Statistic: [tex]\( z = 6.90 \)[/tex]
3. P-value: [tex]\( \text{P-value} = 0.000 \)[/tex]
4. Conclusion: 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.
5. Confidence Interval: [tex]\( 0.130 < (p_1 - p_2) < 0.262 \)[/tex]