Answer :
To solve this problem, we need to find a 95% confidence interval for the difference in proportions between two groups.
Given:
- Group 1 (with gift card): 65 out of 500 responded.
- Group 2 (without gift card): 45 out of 500 responded.
Step 1: Calculate the sample proportions for each group.
The sample proportion [tex]\hat{p}_1[/tex] for Group 1 is calculated as:
[tex]\hat{p}_1 = \frac{65}{500} = 0.13[/tex]
The sample proportion [tex]\hat{p}_2[/tex] for Group 2 is calculated as:
[tex]\hat{p}_2 = \frac{45}{500} = 0.09[/tex]
Step 2: Calculate the standard error of the difference between the two proportions.
The standard error (SE) of the difference can be calculated using:
[tex]SE = \sqrt{\frac{\hat{p}_1(1 - \hat{p}_1)}{n_1} + \frac{\hat{p}_2(1 - \hat{p}_2)}{n_2}}[/tex]
where [tex]n_1[/tex] and [tex]n_2[/tex] are the sample sizes, both equal to 500.
[tex]SE = \sqrt{\frac{0.13(1 - 0.13)}{500} + \frac{0.09(1 - 0.09)}{500}}[/tex]
[tex]SE = \sqrt{\frac{0.13 \times 0.87}{500} + \frac{0.09 \times 0.91}{500}}[/tex]
[tex]SE = \sqrt{\frac{0.1131}{500} + \frac{0.0819}{500}}[/tex]
[tex]SE \approx \sqrt{0.0002262 + 0.0001638} \approx \sqrt{0.00039} \approx 0.0197[/tex]
Step 3: Calculate the 95% confidence interval for the difference in proportions.
The formula for the confidence interval is:
[tex](\hat{p}_1 - \hat{p}_2) \pm Z \times SE[/tex]
where [tex]Z[/tex] is the Z-value for a 95% confidence level, which is 1.96.
Calculate the difference in sample proportions:
[tex]\hat{p}_1 - \hat{p}_2 = 0.13 - 0.09 = 0.04[/tex]
Determine the margin of error:
[tex]1.96 \times 0.0197 \approx 0.0386[/tex]
Thus, the confidence interval is:
[tex]0.04 \pm 0.0386[/tex]
So, the interval is:
[tex](0.0014, 0.0786)[/tex]
Conclusion:
(a-1) The 95% confidence interval for the difference in proportions is approximately [tex](0.0014, 0.0786)[/tex].
(a-2) Does the interval include zero? The interval does not include zero, which indicates there is a statistically significant difference between the two proportions.
Final Answer: No
