Hypothesis Test with Known\(\sigma \)How do the results from Exercise 14 “Speed Dating” change if\(\sigma \)is known to be 1.99? Does the knowledge of\(\sigma \)have much of an effect?

A sample of 199 male attractiveness ratings is considered with a sample mean value equal to 6.19. The population standard deviation is equal to 1.99.

It is claimed that the mean value of the population of male attractiveness ratings is less than 7.00.

The null hypothesis is written as follows:

The mean of the population of male attractiveness ratings is equal to 7.00.

\({H_0}:\mu = 7.00\)

The alternative hypothesis is written as follows:

The mean of the population of male attractiveness ratings is less than 7.00.

\({H_1}:\mu < 7.00\)

The test is left-tailed.

The sample size is equal to n=199.

The sample mean value of the male attractiveness ratings is equal to 6.19.

The population mean value of the male attractiveness ratings is equal to 7.00.

The population standard deviation is equal to 1.99.

The value of the test statistic is computed below:

\(\begin{array}{c}z = \frac{{\bar x - \mu }}{{\frac{\sigma }{{\sqrt n }}}}\\ = \frac{{6.19 - 7.00}}{{\frac{{1.99}}{{\sqrt {199} }}}}\\ = - 5.742\end{array}\)

Thus, z=-5.742.

Referring to the standard normal table, the critical value of z at\(\alpha = 0.01\)for a left-tailed test is equal to -2.3263.

Referring to the standard normal table, the p-value for the test statistic value of -5.742 is equal to 0.000.

Since the p-value is less than 0.01, the null hypothesis is rejected.

There is enough evidence to support the claim that the population mean value of the male attractiveness ratings is less than 7.00.

When \(\sigma \) is unknown, the test statistic value is equal to -5.742, and the corresponding p-value (using student’s t distribution) is equal to 0.000.

The same conclusion is drawn as above, that there is sufficient evidence to support the claim that the population mean value of the male attractiveness ratings is less than 7.00.

Furthermore, the test statistic value is identical for the two tests: the test using normal distribution when sigma is known and the test using Student's t distribution when sigma is unknown.

There does not seem to be a significant effect of \(\sigma \) for conducting the hypothesis test.