Final answer:
a. Yes, the sample sizes satisfy the guidelines for the large-sample confidence interval. b. The 95% confidence interval for the difference between the proportions is (0.441, 0.491).
Step-by-step explanation:
a. To determine if these samples satisfy the guidelines for the large-sample confidence interval, we need to check if the sample sizes are sufficiently large.
The guideline is that both sample sizes, n1 and n2, should be greater than or equal to 5 and both sample sizes multiplied by their respective proportions should be greater than or equal to 5.
In this case, both sample sizes are greater than 5 (671 and 977) and the product of the sample size and the proportion for both groups are also greater than 5 (396*671/671 > 5 and 293*977/977 > 5).
Therefore, these samples satisfy the guidelines for the large-sample confidence interval.
b. To calculate the 95% confidence interval for the difference between the proportions of adults aged 18–29 and adults aged 65 or older who believed the government was tracking their activities online or on their cell phones, we can use the formula:
CI = (p1 - p2) ± Z * sqrt((p1 * (1 - p1) / n1) + (p2 * (1 - p2) / n2))
Where:
p1 and p2 are the proportions for the two groups (396/671 and 293/977)
Z is the Z-score corresponding to the desired confidence level (1.96 for a 95% confidence level)
n1 and n2 are the sample sizes for the two groups (671 and 977)
Plugging in the values, the 95% confidence interval for the difference between the proportions is:
(396/671 - 293/977) ± 1.96 * sqrt(((396/671) * (1 - 396/671) / 671) + ((293/977) * (1 - 293/977) / 977))