Question

A college admissions director wishes to estimate the mean age of all students currently enrolled. In a random sample of 19 students, the mean age is found to be 22.4 years. From past studies, the ages of enrolled students are normally distributed with a standard deviation of 9.5 years. Construct a 90% confidence interval for the mean age of all students currently enrolled.

1. What is the critical value?

2. What is the standard deviation of the sample mean?

3. What is the margin of error?

4. What is the lower limit of the interval?

5. What is the upper limit of the interval?

Answer 1

Solution:-

1. The critical value = 1.734

2. The standard deviation of the sample mean = 2.1794
=> sd = s/sqrt(n) = 9.5/sqrt(19) = 2.1794

3. The margin of error = 3.779

4. The lower limit = 18.621

5. The upper limit = 26.179
  

Explanation:-

The provided sample mean is X = 22.4 and the sample standard deviation is s = 9.5. The size of the sample is n = 19 and the required confidence level is 90%. The number of degrees of freedom are df = 19 -1 = 18, and the significance level is a = 0.1. Based on the provided information, the critical t-value for a = 0.1 and df = 18 degrees of freedom is te = 1.734. The 90% confidence for the population mean u is computed using the following expression CI= (1 - tovo + tema) Therefore, based on the information provided, the 90 % confidence for the population mean is C1= (22. - .758 95, 924 + 1.7581705 V19 = (22.4 - 3.779, 22.4 +3.779) = (18.621, 26.179)