A poll conducted in 1972 asked 1003 people, " During the past year, about how many books did you read either all or part of the way through?" Results of the survey indicated that x-bar=18.9 books and s=1

9.9 books.

A)Construct a 95% confidence interval for the mean number of books read either all or part of during the preceding year. Interpret the interval.

____,_____

See previous post.

To construct a confidence interval for the mean number of books read, we can use the following formula:

CI = x̄ ± Z * (s / √n)

Where:
- CI is the confidence interval
- x̄ is the sample mean (18.9 books in this case)
- Z is the Z-score corresponding to the desired confidence level (in this case, 95% confidence level)
- s is the sample standard deviation (19.9 books in this case)
- n is the sample size (1003 in this case)

First, find the Z-score corresponding to a 95% confidence level. This can be looked up in a Z-table or calculated using a calculator. For a 95% confidence level, the Z-score is approximately 1.96.

Next, substitute the values into the formula:

CI = 18.9 ± 1.96 * (19.9 / √1003)

Calculating this equation will give you the lower and upper limits of the confidence interval.

Interpretation of the confidence interval:
A 95% confidence interval implies that if we were to conduct the same survey multiple times and calculate the confidence interval each time, approximately 95% of the intervals would contain the true mean number of books read in the population. In other words, we can be reasonably confident that the true mean number of books read in the population falls within this interval.