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Chocolate and Happiness Use the results from part (b) of Cumulative Review Exercise 2 to test the claim that when asked, more than 80% of women say that chocolate makes them happier. Use a 0.01 significance level.

Short Answer

Expert verified

There is enough evidence to conclude that the percentage of women who say that chocolate makes them happier is more than 80%.

Step by step solution

01

Given information

It is given that a survey was sponsored by a chocolate company.

Out of 1708 women who were surveyed, 85% of them said that chocolate made them happier.

02

Hypotheses

The null hypothesis for testing the claim is as follows:

The population percentage of women who say that chocolate makes them happier is equal to 80%.

\({H_0}:p = 0.80\)

The alternative hypothesis for testing the claim is as follows:

The population percentage of women who say that chocolate makes them happier is more than 80%.

\({H_0}:p > 0.80\)

The test is right-tailed.

03

Test statistics

Let\(\hat p\)denote the sample proportion ofwomen who said that chocolate makes them happier.

Here,

\(\begin{aligned}{c}\hat p = 85\% \;\\ = 0.85\end{aligned}\)

Here, p=0.80.

Thus,

\[\begin{aligned}{c}q = 1 - p\\ = 1 - 0.80\\ = 0.20\end{aligned}\]

Since the sample size (n) equal to 1708 is large, the value of the z-score is computed as follows:

\[\begin{aligned}{c}z = \frac{{\hat p - p}}{{\sqrt {\frac{{pq}}{n}} }}\;\;\; \sim N\left( {0,1} \right)\\ = \frac{{0.85 - 0.80}}{{\sqrt {\frac{{\left( {0.80} \right)\left( {0.20} \right)}}{{1708}}} }}\\ = 5.166\end{aligned}\]

Thus, the test statistics, z is 5.166.

04

Critical value and p-value

Referring to standard normal table,

The critical value of z at 0.01 level of significance for a right-trailed test is equal to 2.3263.

The corresponding p-value 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 0.01, the null hypothesis is rejected.

There is enough evidence to conclude that the percentage of women who say that chocolate makes them happier is more than 80%.

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Most popular questions from this chapter

Clinical Trial of Lipitor Lipitor is the trade name of the drug atorvastatin, which is used to reduce cholesterol in patients. (Until its patent expired in 2011, this was the largest-selling drug in the world, with annual sales of $13 billion.) Adverse reactions have been studied in clinical trials, and the table below summarizes results for infections in patients from different treatment groups (based on data from Parke-Davis). Use a 0.01 significance level to test the claim that getting an infection is independent of the treatment. Does the atorvastatin (Lipitor) treatment appear to have an effect on infections?


Placebo

Atorvastatin 10 mg

Atorvastatin 40 mg

Atorvastatin 80 mg

Infection

27

89

8

7

No Infection

243

774

71

87

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

Police Calls Repeat Exercise 11 using these observed frequencies for police calls received during the month of March: Monday (208); Tuesday (224); Wednesday (246); Thursday (173); Friday (210); Saturday (236); Sunday (154). What is a fundamental error with this analysis?

Bias in Clinical Trials? Researchers investigated the issue of race and equality of access to clinical trials. The following table shows the population distribution and the numbers of participants in clinical trials involving lung cancer (based on data from โ€œParticipation in Cancer Clinical Trials,โ€ by Murthy, Krumholz, and Gross, Journal of the American Medical Association, Vol. 291, No. 22). Use a 0.01 significance level to test the claim that the distribution of clinical trial participants fits well with the population distribution. Is there a race/ethnic group that appears to be very underrepresented?

Race/ethnicity

White

non-Hispanic

Hispanic

Black

Asian/

Pacific

Islander

American Indian/

Alaskan Native

Distribution of

Population

75.6%

9.1%

10.8%

3.8%

0.7%

Number in Lung

Cancer Clinical Trials

3855

60

316

54

12

Do World War II Bomb Hits Fit a Poisson Distribution? In analyzing hits by V-1 buzz bombs in World War II, South London was subdivided into regions, each with an area of 0.25\(k{m^2}\). Shown below is a table of actual frequencies of hits and the frequencies expected with the Poisson distribution. (The Poisson distribution is described in Section 5-3.) Use the values listed and a 0.05 significance level to test the claim that the actual frequencies fit a Poisson distribution. Does the result prove that the data conform to the Poisson distribution?

Number of Bomb Hits

0

1

2

3

4

Actual Number of Regions

229

211

93

35

8

Expected Number of Regions

(from Poisson Distribution)

227.5

211.4

97.9

30.5

8.7

In Exercises 1โ€“4, use the following listed arrival delay times (minutes) for American Airline flights from New York to Los Angeles. Negative values correspond to flights that arrived early. Also shown are the SPSS results for analysis of variance. Assume that we plan to use a 0.05 significance level to test the claim that the different flights have the same mean arrival delay time.

Flight 1

-32

-25

-26

-6

5

-15

-17

-36

Flight 19

-5

-32

-13

-9

-19

49

-30

-23

Flight 21

-23

28

103

-19

-5

-46

13

-3

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