Weather-Related Deaths Review Exercise 5 involved weather-related U.S. deaths. Among the 450 deaths included in that exercise, 320 are males. Use a 0.05 significance level to test the claim that among those who die in weather-related deaths, the percentage of males is equal to 50%. Provide an explanation for the results.

Short Answer

Expert verified

There is enough evidence to warrant rejection of the claim that the percentage of male deaths out of the given sample ofweather-related deaths is equal to 50%.

It can be said that a comparatively greater number of males participate in outdoor activities like golfing, fishing and surfing during May, June and July. Thus, their proportion is greater than 0.5.

Step by step solution

01

Given information

It is given that out of 450 weather-related deaths, 320 were males.

02

Hypotheses

Letp is the population proportion of males who died.

The null hypothesis is as follows:

The proportion of male deaths is equal to 0.50.

Symbolically,

\({H_o}:p = 0.50\)

The alternative hypothesis is as follows:

The proportion of male deaths is not equal to 0.50.

Symbolically,

\({H_o}:p \ne 0.50\)

It is a two-tailed test.

03

Test statistic

Let\(\hat p\)denote the sample proportion of male deaths.

The value of\(\hat p\)is computed as shown below:

\(\begin{aligned}{c}\hat p = \frac{{{\rm{Number}}\;{\rm{of}}\;{\rm{male}}\;{\rm{deaths}}}}{{{\rm{Total}}\;{\rm{number}}\;{\rm{of}}\;{\rm{deaths}}}}\\ = \frac{{320}}{{450}}\\ = 0.711\end{aligned}\)

Here, p = 0.50.

Thus,

\[\begin{aligned}{c}q = 1 - p\\ = 1 - 0.50\\ = 0.50\end{aligned}\]

Since the sample size (n) equal to 450 is large, the value of the z-score is computed as follows:

\[\begin{aligned}{c}z = \frac{{\hat p - p}}{{\sqrt {\frac{{pq}}{n}} }}\;\;\; \sim N\left( {0,1} \right)\\ = \frac{{0.711 - 0.50}}{{\sqrt {\frac{{\left( {0.50} \right)\left( {0.50} \right)}}{{450}}} }}\\ = 8.957\end{aligned}\]

Thus, the test statistic, z is 8.957.

04

Obtain the critical value and the p-value

Refer to the standard normal table,

The critical value of z at 0.05 level of significance for a two-tailed test is equal to 1.96.

The corresponding p-value obtained using the test statistic is equal to 0.000.

05

Decision and conclusion of the test

Since the absolute value of the z-score is greater than the critical value and the p-value is less than 0.05, the null hypothesis is rejected.

There is enough evidence to conclude that the percentage of male deaths out of the given sample ofweather-related deaths is not equal to 50%.

06

Possible explanation for the results

It can be said that a comparatively greater number of males participate in outdoor activities like golfing, fishing and surfing during the months of May, June and July. Thus, their proportion is greater than 0.5.

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Most popular questions from this chapter

In Exercises 5–20, conduct the hypothesis test and provide the test statistic and the P-value and , or critical value, and state the conclusion.

World Series Games The table below lists the numbers of games played in 105 Major League Baseball (MLB) World Series. This table also includes the expected proportions for the numbers of games in a World Series, assuming that in each series, both teams have about the same chance of winning. Use a 0.05 significance level to test the claim that the actual numbers of games fit the distribution indicated by the expected proportions.

Games Played

4

5

6

7

World Series Contests

21

23

23

38

Expected Proportion

2/16

4/16

5/16

5/16

Benford’s Law. According to Benford’s law, a variety of different data sets include numbers with leading (first) digits that follow the distribution shown in the table below. In Exercises 21–24, test for goodness-of-fit with the distribution described by Benford’s law.

Leading Digits

Benford's Law: Distributuon of leading digits

1

30.10%

2

17.60%

3

12.50%

4

9.70%

5

7.90%

6

6.70%

7

5.80%

8

5.10%

9

4.60%

Tax Cheating? Frequencies of leading digits from IRS tax files are 152, 89, 63, 48, 39, 40, 28, 25, and 27 (corresponding to the leading digits of 1, 2, 3, 4, 5, 6, 7, 8, and 9, respectively, based on data from Mark Nigrini, who provides software for Benford data analysis). Using a 0.05 significance level, test for goodness-of-fit with Benford’s law. Does it appear that the tax entries are legitimate?

Cybersecurity When using the data from Exercise 1 to test for goodness-of-fit with the distribution described by Benford’s law, identify the null and alternative hypotheses.

Questions 6–10 refer to the sample data in the following table, which describes the fate of the passengers and crew aboard the Titanic when it sank on April 15, 1912. Assume that the data are a sample from a large population and we want to use a 0.05 significance level to test the claim that surviving is independent of whether the person is a man, woman, boy, or girl.


Men

Women

Boys

Girls

Survived

332

318

29

27

Died

1360

104

35

18

Is the hypothesis test left-tailed, right-tailed, or two-tailed?

In a study of the “denomination effect,” 43 college students

were each given one dollar in the form of four quarters, while 46 other college students were each given one dollar in the form of a dollar bill. All of the students were then given two choices: (1) keep the money; (2) spend the money on gum. The results are given in the accompanying table (based on “The Denomination Effect,” by PriyaRaghubir and Joydeep Srivastava, Journal of Consumer Research,Vol. 36.) Use a 0.05 significance level to test the claim that whether students purchased gum or kept the money is independent of whether they were given four quarters or a \(1 bill. Is there a “denomination effect”?

Purchased Gum

Kept the Money

Students Given Four Quarters

27

16

Students Given a \)1 Bill

12

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