Question

According to a ​poll, 22​% of adults from a certain region were very likely to watch some coverage of a certain sporting event on television. The survey polled 1400 adults from the region and had a margin of error of plus or minus 4 percentage points with a 90 ​% level of confidence. Complete parts​ (a) through​ (c) below. Question content area bottom Part 1 a. State the survey results in confidence interval form and interpret the interval. The confidence interval of the survey results is ​( enter your response here​, ____ enter your response here​). ​(Round to two decimal places as​ needed.)

Answer 1

the confidence interval of the survey results is (18%, 26%). We can interpret this interval as follows: We are 90% confident that the true percentage of adults from the region who are very likely to watch some coverage of the sporting event on television is between 18% and 26%.

Explanation:

The survey polled 1400 adults from the region and found that 22% of them were very likely to watch some coverage of a certain sporting event on television. The margin of error was plus or minus 4 percentage points with a 90% level of confidence.

(a) To state the survey results in confidence interval form, we can use the formula:

Confidence interval = sample statistic +/- margin of error

where the sample statistic is the percentage of adults who were very likely to watch the sporting event on television, and the margin of error is 4 percentage points.

Substituting the given values, we get:

Confidence interval = 22% +/- 4%

To calculate the endpoints of the interval, we need to add and subtract the margin of error from the sample statistic:

Lower endpoint = 22% - 4% = 18%

Upper endpoint = 22% + 4% = 26%

Therefore, the confidence interval of the survey results is (18%, 26%). We can interpret this interval as follows: We are 90% confident that the true percentage of adults from the region who are very likely to watch some coverage of the sporting event on television is between 18% and 26%.