Question

A random sample of 64 students at a university showed an average age of 20 years and a sample standard deviation of 4 years. The 90% confidence interval for the true average age of all students in the university is

Answer 1

The 90% confidence level is
`19.15< L < 20.85`

Explanation:

From the question we are told that

The sample size is
`n = 64`

The mean age is
`\= x = 20 \ years`

The standard deviation is
`\sigma = 4 \ years`

Generally the degree of freedom for this data set is mathematically represented as

`df = n - 1`

substituting values

`df = 64 - 1`

`df = 63`

Given that the level of confidence is 90% the significance level is mathematically evaluated as

`\alpha = 100 - 90`

`\alpha =`10 %

`\alpha = 0.10`

Now
`(\alpha )/(2) = (0.10)/(2) = 0.05`

Since we are considering a on tail experiment

The critical value for half of this significance level at the calculated degree of freedom is obtained from the critical value table as

`t_{df, ( \alpha)/(2) } = t_(63, 0.05 ) = 1.669`

The margin for error is mathematically represented as

`MOE = t_{df , (\alpha )/(2) } * (\sigma)/(√(n) )`

substituting values

`MOE = 1.699 * (4 )/(√(64) )`

`MOE = 0.85`

he 90% confidence interval for the true average age of all students in the university is evaluated as follows

`\= x - MOE < L < \= x + E`

substituting values

`20 - 0. 85 < L < 20 + 0.85`

`19.15< L < 20.85`