The owner of Limp Pines Resort wanted to know the average age of its clients. A random sample of 25 tourists is taken. It shows a mean age of 46 years with a standard deviation of 5 years. The width of a 95 percent CI for the true mean client age is approximately:

Given Information:

Sample size = n = 25

Mean age of client = 46 years

Standard deviation of age of client = 5 years

Confidence level = 95%

Required Information:

Width of the confidence interval = ?

Step-by-step explanation:

The width for the true mean client age is given by

Where n is the sample size, s is the standard deviation of age of client, and is the t-score corresponding to 95% confidence level.

The t-score corresponding to 95% confidence level is

Significance level = 1 - 0.95 = 0.05/2 = 0.025

Degree of freedom = n - 1 = 25 - 1 = 24

From the t-table at α = 0.025 and DF = 24

t-score = 2.064

Therefore, width of the 95% confidence Interval for the true mean client age is approximately ±2.064.

Bonus:

The corresponding 95% confidence interval is given by

Where x_bar is the mean client age

Which means that we are 95% sure that the true mean client age is within the range of (44, 48) years.

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