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Chapter 8: Q30BB (page 356)

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

Short Answer

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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鈥檚 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

01

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.

02

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.

03

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.

04

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.

05

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.

06

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.

07

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鈥檚 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.

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Most popular questions from this chapter

In Exercises 1鈥�4, use these results from a USA Today survey in which 510 people chose to respond to this question that was posted on the USA Today website: 鈥淪hould Americans replace passwords with biometric security (fingerprints, etc)?鈥� Among the respondents, 53% said 鈥測es.鈥� We want to test the claim that more than half of the population believes that passwords should be replaced with biometric security.

Requirements and Conclusions

a. Are any of the three requirements violated? Can the methods of this section be used to test the claim?

b. It was stated that we can easily remember how to interpret P-values with this: 鈥淚f the P is low, the null must go.鈥� What does this mean?

c. Another memory trick commonly used is this: 鈥淚f the P is high, the null will fly.鈥� Given that a hypothesis test never results in a conclusion of proving or supporting a null hypothesis, how is this memory trick misleading?

d. Common significance levels are 0.01 and 0.05. Why would it be unwise to use a significance level with a number like 0.0483?

Final Conclusions. In Exercises 25鈥�28, use a significance level of = 0.05 and use the given information for the following:

a. State a conclusion about the null hypothesis. (Reject H0or fail to reject H0.)

b. Without using technical terms or symbols, state a final conclusion that addresses the original claim.

Original claim: More than 58% of adults would erase all of their personal information online if they could. The hypothesis test results in a P-value of 0.3257.

The Ericsson method is one of several methods claimed to increase the likelihood of a baby girl. In a clinical trial, results could be analysed with a formal hypothesis test with the alternative hypothesis of p>0.5, which corresponds to the claim that the method increases the likelihood of having a girl, so that the proportion of girls is greater than 0.5. If you have an interest in establishing the success of the method, which of the following P-values would you prefer: 0.999, 0.5, 0.95, 0.05, 0.01, and 0.001? Why?

Technology. In Exercises 9鈥�12, test the given claim by using the display provided from technology. Use a 0.05 significance level. Identify the null and alternative hypotheses, test statistic, P-value (or range of P-values), or critical value(s), and state the final conclusion that addresses the original claim.

Body Temperatures Data Set 3 鈥淏ody Temperatures鈥� in Appendix B includes 93 body temperatures measured at 12 虏鲁 on day 1 of a study, and the accompanying XLSTAT display results from using those data to test the claim that the mean body temperature is equal to 98.6掳F. Conduct the hypothesis test using these results.

Testing Claims About Proportions. In Exercises 9鈥�32, test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, or critical value(s), then state the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim. Use the P-value method unless your instructor specifies otherwise. Use the normal distribution as an approximation to the binomial distribution, as described in Part 1 of this section.

Postponing Death An interesting and popular hypothesis is that individuals can temporarily postpone death to survive a major holiday or important event such as a birthday. In a study, it was found that there were 6062 deaths in the week before Thanksgiving, and 5938 deaths the week after Thanksgiving (based on data from 鈥淗olidays, Birthdays, and Postponement of Cancer Death,鈥� by Young and Hade, Journal of the American Medical Association, Vol. 292, No. 24). If people can postpone death until after Thanksgiving, then the proportion of deaths in the week before should be less than 0.5. Use a 0.05 significance level to test the claim that the proportion of deaths in the week before Thanksgiving is less than 0.5. Based on the result, does there appear to be any indication that people can temporarily postpone death to survive the Thanksgiving holiday?
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