Question

Are They Tracking Me? A survey in 2019 asked a random sample of U.S. adults if they believed the government was tracking all or most of their activities online or on their cell phones. Of the 671 adults aged 18–29, 396 said “yes.” Of the 977 adults aged 65 or older, 293 said “yes.”

Answer 1

a. The sample sizes for both age groups (671 for adults aged 18–29 and 977 for adults aged 65 or older) are large enough to construct a confidence interval.

b. 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 is approximately 0.243 to 0.337.

a. The guidelines for large-sample confidence intervals typically require sufficiently large sample sizes, which is the case here.

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, you can use the following formula:

`[ \text{Confidence Interval} = (\hat{p}_1 - \hat{p}_2) \pm Z \sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2}} ]`

Where:

`( \hat{p}_1 ) and ( \hat{p}_2 )` = the sample proportions for adults aged 18–29 and 65 or older, respectively.

`( n_1 ) and ( n_2 )` = the sample sizes for adults aged 18–29 and 65 or older, respectively.

`( Z )` = the critical value for a 95% confidence interval, approximately 1.96.

The number of adults aged 18–29, 396 who said “yes" that they believed the government was tracking all or most of their activities online or on their cell phones = 671

The number of adults aged 65 or older who said "yes" that they believed the government was tracking all or most of their activities online or on their cell phones = 293

The total number of adults surveyed = 977

First, we calculate the sample proportions:

`[ \hat{p}_1 = (396)/(671) \approx 0.590 ] [ \hat{p}_2 = (293)/(977) \approx 0.300 ]`

Then, we calculate the standard error:

`[ SE = \sqrt{(0.590(1-0.590))/(671) + (0.300(1-0.300))/(977)} \approx 0.024 ]`

Finally, we calculate the confidence interval:

`[ (0.590 - 0.300) \pm 1.96(0.024) ] [ 0.290 \pm 0.047 ] [ 0.243 \text{ to } 0.337 ]`

Thus, we can conclude that 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 is approximately 0.243 to 0.337.

Complete Question:

A survey in 2019 asked a random sample of U.S. adults if they believed the government was tracking all or most of their activities online or on their cell phones. Of the 671 adults aged 18–29, 396 said “yes.” Of the 977 adults aged 65 or older, 293 said “yes.” a. Do these samples satisfy the guidelines for the large-sample confidence interval? b. Give a 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 all or most of their activities online or on their cell phones.