A school administrator is interested in a schools performance in math. He administers a math test to...

A school administrator is interested in a schools performance in math. He administers a math test to a sample of 30 students and obtains a mean of 84.25. The standard deviation of this sample was 8.4. (T Test)

26. Construct a 95% confidence interval for this data.

27. How would you interpret this confidence interval?

Solutions

Expert Solution

olution :

Given that,

= 84.25

s =8.4

n =30

Degrees of freedom = df = n - 1 = 30- 1 = 29

a ) At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t_/2,df = t_0.025,29 = 2.045

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

= 2.045 * (8.4 / $\sqrt$ 30)

= 3.1363

The 95% confidence interval estimate of the sample mean is,

- E < < + E

84.25- 3.1363< <84.25 + 3.1363

81.1137< < 87.3863

( 81.1137 , 87.3863)

The 95% confidence interval estimate of the sample mean is,fall between the interval 81.1137 to 87.3863

milcah answered 3 days ago

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