Question

In a sample of 41 students, the mean scholarship amount received was $1138, with a standard deviation of $18.82. Contruct a 90% confidence interval for the mean scholarship award for all students. Round your answer to the nearest cent.

________ < μ< _________

Answer 1

Solution :

Given that,

Point estimate = sample mean = = $ 1138

sample standard deviation = s = $ 18.82

sample size = n = 41

Degrees of freedom = df = n - 1 = 41 - 1 = 40

At 90% confidence level

= 1 - 90%

=1 - 0.90 =0.10

/2 = 0.05

t/2,df = t_0.05,40 = 1.684

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

= 1.684 * (18.82 / 41)

Margin of error = E = 4.95

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

    - E < < + E

1138 - 4.95 < < 1138 + 4.95

( 1133.05 < < 1142.95 )