Question

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?

Short answer

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

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

Test Statistic: -5.742

Critical Value: -2.3263

P-Value: 0.000

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.

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

There is not a major impact of the knowledge of the population standard deviation \(\left( \sigma \right)\) on the result of the hypothesis test.

Step-by-step solution

Given information

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.

Hypotheses

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.

Important values

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.

Test statistic

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.

Critical value and p-value

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.

Conclusion of the test

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

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

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.