Question

The paper referenced in the previous exercise also gave the following sample statistics for the percentage of study time that occurred in the 24 hours prior to the final exam: $$ n=411 \quad \bar{x}=43.18 \quad s=21.46 $$ Construct and interpret a \(90 \%\) confidence interval for the mean percentage of study time that occurs in the 24 hours prior to the final exam.

Short answer

The 90% confidence interval for the mean percentage of study time that occurs in the 24 hours prior to the final exam is (41.415%, 44.945%).

Step-by-step solution

Identify sample statistics

The sample mean (\(\bar{x}\)) is 43.18, the sample standard deviation (s) is 21.46, and the sample size (n) is 411.

Find z-score for 90% confidence interval

For a 90% confidence interval, the z-score is 1.645. This value corresponds to the number of standard deviations we must go from the mean to capture 90% of the values in a population, assuming a normal distribution.

Calculate margin of error

The margin of error (E) for a confidence interval is calculated using the formula: E = z * (s / sqrt(n)). Substituting the given values into the formula gives: E = 1.645 * (21.46 / sqrt(411)) = 1.765.

Construct the confidence interval

The 90% confidence interval is calculated as: \(\bar{x}\) ± E, or 43.18 ± 1.765. So, the interval is (41.415, 44.945).

Interpret the confidence interval

The interpretation of this 90% confidence interval is that we are 90% confident that the true mean percentage of study time that occurs in the 24 hours prior to the final exam is between 41.415% and 44.945%.