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 90% confidence interval for the mean

score, ", of all students taking the test?

Answer 1

`29.5-1.671(5.2)/(√(59))=28.37`

`29.5+1.671(5.2)/(√(59))=30.63`

And we can conclude that the true mean for the scores are given by this interval (28.37; 30.63) at 90% of confidence

Explanation:

Information given

`\bar X = 29.5` represent the sample mean

`\mu` population mean

s= 5.2 represent the sample standard deviation

n=59 represent the sample size

Confidence interval

The confidence interval for the true mean is given by the following formula:

`\bar X \pm t_(\alpha/2)(s)/(√(n))` (1)

The degrees of freedom are given by:

`df=n-1=59-1=58`

The Confidence interval is 0.90 or 90%, the significance is
`\alpha=0.1` and
`\alpha/2 =0.05`, and the critical value is
`t_(\alpha/2)=1.671`

Now we have everything in order to replace into formula (1):

`29.5-1.671(5.2)/(√(59))=28.37`

`29.5+1.671(5.2)/(√(59))=30.63`

And we can conclude that the true mean for the scores are given by this interval (28.37; 30.63) at 90% of confidence