Q28BSC Testing Claims About Proportions... [FREE SOLUTION]

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Chapter 8: Q28BSC (page 356)

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?

Short Answer

Expert verified

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

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

Test statistic: 1.095

Critical value: -1.645

P-value: 0.8632

The null hypothesis is 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.

Asa sufficient proportion of deaths occur in the week before Thanksgiving, it appears that people can temporarily postpone death to survive the Thanksgiving holiday.

Step by step solution

Given information

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

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.

Sample size, sample proportion,and population proportion

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.

Test statistic

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.

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.

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

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.

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.