Testing Claims About Proportions. In Exercises 9–32, 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 “Holidays, Birthdays, and Postponement of Cancer Death,” 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?

Step by step solution

01

Given information

There are 6062 deaths in the week before Thanksgiving and 5938 deaths in the week after Thanksgiving.

02

Hypotheses

The null hypothesis is written as follows.

The proportion of deaths in the week before Thanksgiving is equal to 0.5.

$H_0:p=0.5$

The alternative hypothesis is written as follows.

The proportion of deaths in the week before Thanksgiving is less than 0.5.

$H_1:p<0.5$

The test is left-tailed.

03

Sample size, sample proportion, and population proportion

The sample size is computed below.

$$\begin{gathered} n=6062+5938 \\ =12000 \end{gathered}$$

The sample proportion of deaths in the week before Thanksgiving is computed below.

$$\begin{gathered} \hat{p}=\frac{6062}{12000} \\ =0.505 \end{gathered}$$

The population proportion of deaths in the week before Thanks giving is equal to 0.5.

04

Test statistic

The value of the test statistic is computed below.

$$\begin{gathered} z=\frac{\hat{p}-p}{\sqrt{\frac{pq}{n}}} \\ =\frac{0.505-0.5}{\sqrt{\frac{0.5\left(1-0.5\right)}{12000}}} \\ =1.095 \end{gathered}$$

Thus, z=1.095.

05

Critical value and p-value

Referring to the standard normal table, the critical value of z at $\alpha =0.05$ for a left-tailed test is equal to -1.645.

Referring to the standard normal table, the p-value for the test statistic value of 1.095 is equal to 0.8632.

As the p-value is greater than 0.05, the null hypothesis is not rejected.

06

Conclusion of the test

There is not enough evidence to support the claim that the proportion of deaths in the week before Thanksgiving is less than 0.5.

As the proportion of deaths in the week before Thanksgiving is approximately 50.5% (greater than 50%), it appears that people can temporarily postpone death to survive the Thanksgiving holiday.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!