Question

The average expenditure per student (based on average daily attendance) for a certain school year was $10,337 with a population standard deviation of $1560. A survey for the next school year of 150 randomly selected students resulted in a sample mean of $10, 798. Find the 95% confidence level? Should the null hypothesis be rejected? please write in the computer.

Answer 1

Solution :

Given Data :

µ = $10,337
σ = $1560
CI = 95% = 0.95 , α = 1-0.95 = 0.05
x̄ = $10,798
n = 150

for 95% CI

z = 1.96 ( from table )

Margin of error = E = z(σ/√n) = 1.96(1560/√150) = $249.652

Lower limit = x̄ - E = $10,798 - $249.652 = $10,548.3480 = $10,548.35

Upper limit = x̄ + E = $10,798 + $249.652 = $11,047.6520 = $11,047.65

The 95% confidence interval is

( $10,548.35 , $11,047.65 )

Population mean $10,337 is not fall in the confidence interval hence Null hypothesis should be rejected.