208k views
0 votes
a random sample of 49 shoppers showed that they spend an average of 23.45 per visits at the Saturday Morning Bookstore. The standard deviation of the population is 2.80, Find the 90 percent confidence interval of the true mean.

User Jaggy
by
7.6k points

1 Answer

7 votes

The 90% confidence interval for the true mean is (22.79, 24.11).

How to solve

To establish the 90% confidence interval for the true mean spending at the bookstore:

Calculate the standard error: This measures the probable deviation of the sample mean from the actual mean. For this scenario, the standard error, indicating this variance, is about 0.36.

Determine the z-scores: These scores signify how many standard deviations a point differs from the mean. To encompass 90% of the data, z-scores approximately at -1.645 and 1.645 are used.

Compute the confidence interval: By employing these z-scores and the standard error, the margin of error is found.

we have:

The Lower Bound = 23.45 - (-1.645) * 0.36 ≈ 22.79

The Upper Bound = 23.45 + (1.645) * 0.36 ≈ 24.11

Therefore, the 90% confidence interval for the true mean is (22.79, 24.11).

User Mattacular
by
7.8k points