Question

Hypothesis Test with Known\(\sigma \)How do the results from Exercise 13 “Course Evaluations” change if\(\sigma \)is known to be 0.53? Does the knowledge of\(\sigma \)have much of an effect?

Short answer

Null Hypothesis: The population mean student course evaluation is equal to 4.00.

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

Test Statistic: -1.638

Critical Value: 1.96

P-Value: 0.1014

The null hypothesis fails to reject the claim.

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

There are no major differences in the test statistic value as well as in the conclusion drawn from the two hypothesis tests: Using normal distribution when\(\sigma \)is known and using student’s distribution when\(\sigma \)is unknown.

Thus, the knowledge of the population mean\(\left( \sigma \right)\) does not have much effect.

Step-by-step solution

Given information

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.

Hypotheses

The null hypothesis is written as follows:

The population mean student course evaluation is equal to 4.00.

\({H_0}:\mu = 4.00\)

The alternative hypothesis is written as follows:

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

\({H_1}:\mu \ne 4.00\)

The test is two-tailed.

Important values

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.

Test statistic

The value of the test statistic is computed below:

\(\begin{array}{c}z = \frac{{\bar x - \mu }}{{\frac{\sigma }{{\sqrt n }}}}\\ = \frac{{3.91 - 4.00}}{{\frac{{0.53}}{{\sqrt {93} }}}}\\ = - 1.638\end{array}\)

Thus, z=-1.638.

Critical value and p-value

Referring to the standard normal table, the critical value of z at\(\alpha = 0.05\)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.

Conclusion of the test

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

Step 7: \(\sigma \) known vs \(\sigma \) unknown

Refer to Exercise 13, when \(\sigma \) is unknown, the test statistic value is equal to -1.638, and the corresponding p-value (using Student’s t distribution) is equal to 0.1049.

The same conclusion is drawn as above, that there isn’t 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 when\(\sigma \)is known and test using student’s t distribution when\(\sigma \)is unknown.