Question

A college admissions director wishes to estimate the mean age of all students currently enrolled. In a random sample of 16 students, the mean age is found to be 21.8 years. From past studies, the ages of enrolled students are normally distributed with a standard deviation of 10.1 years. Construct a 90% confidence interval for the mean age of all students currently enrolled.

1. The critical value:

2. The standard deviation of the sample mean:

3. The margin of error:

4. The lower limit of the interval:

5. The upper limit of the interval:

Answer 1

Given that,

= 21.8

The standard deviation of the sample mean =s =10.1

n =16

Degrees of freedom = df = n - 1 =16 - 1 = 15

critical value = At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

t _/2,df = t_0.05,15=   1.753

critical value = =1.753    ( using student t table)

Margin of error = E = t_/2,df * (s /n)

= 1.753* ( 10.1/ 16) = 4.4263

The 90% confidence interval estimate of the population mean is,

- E < < + E

21.8 -4.4263 < <21.8 + 4.4263

17.3737< < 26.2263

( The lower limit of the interval =17.3737,  The upper limit of the interval =26.2263)