Answer to Question #212659 in Statistics and Probability for Nurin

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. Do these results indicate that the average expenditure has changed? Choose your own level of significance.

Answer 1

Sample size is greater than 30 and the population standard deviation is given. Thus, in order to test

H₀: µ =$10,337

H₁: µ≠$10,337

The appropriate test is one sample z test.

The formula for one sample z test:

$Z= \frac{x- \mu}{\sigma / \sqrt{n}} \\ x=10798 \\ \mu=10337 \\ \sigma=1560 \\ n=150 \\ Z = \frac{10798-10337}{1560/ \sqrt{150}}=3.62$

Computation of p-value:

The EXCEL formula to find the p-value for a two-tailed z-test is

“=2*(1-NORM.S.DIST(3.62, TRUE))”

Thus, the p-value obtained is 0.00029

Conclusion using α = 0.05:

Rejection rule using p-value:

If p-value ≤ α, then reject the null hypothesis.

Conclusion:

Since, 0.00029 < 0.05, there is sufficient evidence to reject the null hypothesis. Therefore, the true mean expenditure per student is different from $10,337.