91影视

A sample of evaluation of the courses of 93 students is considered with a sample mean value equal to 3.91. The population standard deviation is equal to 0.53.

It is claimed that the population mean student evaluation is equal to 4.00.

The null hypothesis is written as follows:

The population mean student course evaluation is equal to 4.00.

$H_0:渭=4.00$

The alternative hypothesis is written as follows:

The population mean student course evaluation is not equal to 4.00.

$H_1:渭鈮�4.00$

The test is two-tailed.

The sample size is equal to n=93.

The sample mean student course evaluation is equal to 3.91.

The population mean student course evaluation is equal to 4.00.

The population standard deviation is equal to 0.53.

The value of the test statistic is computed below:

$$\begin{gathered} z=\frac{\stackrel{炉}{x}-渭}{\frac{蟽}{\sqrt{n}}} \\ =\frac{3.91-4.00}{\frac{0.53}{\sqrt{93}}} \\ =-1.638 \end{gathered}$$

Thus, z=-1.638.

Referring to the standard normal table, the critical value of z at $伪=0.05$ for a two-tailed test is equal to 1.96.

Referring to the standard normal table, the p-value for the test statistic value of -1.638 is equal to 0.1014.

Since the p-value is greater than 0.05, the null hypothesis is failed to reject.

There is not enough evidence to reject the claim that the population mean student evaluation course is equal to 4.00.

Refer to Exercise 13, when $蟽$ is unknown, the test statistic value is equal to -1.638, and the corresponding p-value (using Student鈥檚 t distribution) is equal to 0.1049.

The same conclusion is drawn as above, that there isn鈥檛 sufficient evidence to reject the claim that the population mean student evaluation course is equal to 4.00.

Moreover, it can be observed that the test statistic is exactly the same for the two tests: Test using normal distribution whenrole="math" localid="1649068219563" $蟽$is known and test using student鈥檚 t distribution whenis unknown.