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]
### 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]