Answer to Question #284595 in Statistics and Probability for mercy

Question

Student Expenditures 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

The following null and alternative hypotheses need to be tested:

$H_0:\mu=10337$

$H_1:\mu\not=10337$

This corresponds to a two-tailed test, for which a z-test for one mean, with known population standard deviation will be used.

Based on the information provided, the significance level is $\alpha = 0.05,$ and the critical value for a two-tailed test is $z_c = 1.96.$

The rejection region for this two-tailed test is $R = \{z: |z| > 1.96\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{10798−10337}{1560/\sqrt{150}}$$

$$\approx3.6193$$

Since it is observed that $|z| = 3.6193 >1.96= z_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is $p=2P(Z>3.6193)\approx 0.000295,$ and since $p= 0.000295<0.05=\alpha,$ 0.05 it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$ is different than $10337,$ at the $\alpha = 0.05$ significance level.

Therefore, there is enough evidence to claim that the average expenditure has changed, at the $\alpha = 0.05$ significance level.