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?

Answer 1

The critical value for $\alpha = 0.05$ is $z_c = z_{1-\alpha/2} = 1.96.$ The corresponding confidence interval is computed as shown below:

$$CI=(\bar{X}-z_c\times\dfrac{\sigma}{\sqrt{n}}, \bar{X}+z_c\times\dfrac{\sigma}{\sqrt{n}})$$

$$=(10798-1.96\times\dfrac{1560}{\sqrt{150}}, 10798+1.96\times\dfrac{1560}{\sqrt{150}})$$

$$=(10548.35, 11047.65)$$

Therefore, based on the data provided, the 95% confidence interval for the population mean is $10548.35 < \mu < 11047.65,$ which indicates that we are 95% confident that the true population mean $\mu$ is contained by the interval $(10548.35, 11047.65).$

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 $|z| = 3.619 > 1.96=z_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is $p=2P(Z>3.6193)=0.000295,$ and since $p=0.000295<0.05=\alpha,$ 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.