Question

A group of 59 randomly selected students have a mean score of 29.5 with a standard deviation of 5.2 on a placement test. What is the 95% confidence interval for the mean score, , of all students taking the test

Answer 1

The 95% confidence interval for the mean score, , of all students taking the test is

`28.37< L\ 30.63`

Explanation:

From the question we are told that

The sample size is
`n = 59`

The mean score is
`\= x = 29.5`

The standard deviation
`\sigma = 5.2`

Generally the standard deviation of mean is mathematically represented as

`\sigma _(\= x) = (\sigma )/(√(n) )`

substituting values

`\sigma _(\= x) = (5.2 )/(√(59) )`

`\sigma _(\= x) = 0.677`

The degree of freedom is mathematically represented as

`df = n - 1`

substituting values

`df = 59 -1`

`df = 58`

Given that the confidence interval is 95% then the level of significance is mathematically represented as

`\alpha = 100 -95`

`\alpha =`5%

`\alpha = 0.05`

Now the critical value at this significance level and degree of freedom is

`t_(df , \alpha ) = t_(58, 0.05 ) = 1.672`

Obtained from the critical value table

So the the 95% confidence interval for the mean score, , of all students taking the test is mathematically represented as

`\= x - t*(\sigma_(\= x)) < L\ \= x + t*(\sigma_(\= x))`

substituting value

`(29.5 - 1.672* 0.677) < L\ (29.5 + 1.672* 0.677)`

`28.37< L\ 30.63`