Question

A study done at a polling organization Randomly surveyed 1031 adults by phone. The poll found that 42% of the respondents said they spend less than three hours a day watching TV. Based on this finding calculate to 2 decimals places the ((% confidence interval for the proportion of US adults who spend less than three hours watching TV on an average day.

Answer 1

Answer: To calculate the % confidence interval for the proportion of US adults who spend less than three hours watching TV on an average day, we can use the following formula:

p +/- z*(sqrt(p(1-p)/n))

Where:

p = proportion of US adults who spend less than three hours watching TV on an average day (0.42)

z = standard normal score for a given level of confidence (e.g. for a 95% confidence level, z = 1.96)

n = sample size (1031)

Plugging in the given values, we get:

0.42 +/- 1.96*(sqrt(0.42*(1-0.42)/1031))

Simplifying, we get:

0.42 +/- 0.0156

So the 95% confidence interval for the proportion of US adults who spend less than three hours watching TV on an average day is approximately (0.40,0.44)

So the answer is (0.40,0.44)

Explanation: