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?

There are6062 deaths in the week before Thanksgiving and 5938 deaths in the week after Thanksgiving.

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.

The sample size is computed below.

\(\begin{array}{c}n = 6062 + 5938\\ = 12000\end{array}\).

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

\(\begin{array}{c}\hat p = \frac{{6062}}{{12000}}\\ = 0.505\end{array}\).

The population proportion of deaths in the week before Thanksgivingis equal to 0.5.

The value of the test statistic is computed below.

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

Thus, z=1.095.

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.

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

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

Asthe 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.