Question

1. An education publication claims that the mean expenditure per student in public elementary and secondary schools is at least $10,200. You want to test this claim. You randomly select 16 school districts and find the average expenditure per student. The results are listed below. At α = 0.01, can you say the mean expenditure is different either way from the amount in the publication? Assume the population is normally distributed.

9,242	10,641	9,763	9,545
9,847	9,851	8,935	9,784
10,065	10,377	10,157	9,974
10,857	9,364	9,969	9,962

Answer 1

Here, we have to use one sample t test for the population mean.

The null and alternative hypotheses are given as below:

Null hypothesis: H0: the mean expenditure per student in public elementary and secondary schools is at least $10,200.

Alternative hypothesis: Ha: the mean expenditure per student in public elementary and secondary schools is less than $10,200.

H0: µ ≥ $10,200 versus Ha: µ < $10,200

This is a lower or left tailed test.

The test statistic formula is given as below:

t = (Xbar - µ)/[S/sqrt(n)]

From given data, we have

µ = 10,200

Xbar = 9,896

S = 490.8820929

n = 16

df = n – 1 = 15

α = 0.01

Critical value = -2.6025

(by using t-table or excel)

t = (Xbar - µ)/[S/sqrt(n)]

t = (9,896 - 10,200)/[ 490.8820929/sqrt(16)]

t = -2.4787

P-value = 0.0128

(by using t-table)

P-value > α = 0.01

So, we do not reject the null hypothesis

There is sufficient evidence to conclude that the mean expenditure per student in public elementary and secondary schools is at least $10,200.