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Chapter 10: Q. 27 (page 625)

Many new products introduced into the market are targeted toward children. The choice behavior of children with regard to new products is of particular interest to companies that design marketing strategies for these products. As part of one study, randomly selected children in different age groups were compared on their ability to sort new products into the correct product category (milk or juice). Here are some of the data:

Age groupNNumber who sorted correctly4- to 5-year-olds50106- to 7-year-olds5328

Are these two age groups equally skilled at sorting? Use information from the Minitab output below to support your answer.

Short Answer

Expert verified

There is sufficient evidence to reject the claim that the two age groups are equally skilled.

Step by step solution

01

Given Information

Given

x1=10

n1=50

x2=28

n2=53

Determine the hypothesis

H0:p1-p2=0

Ha:p1-p2≠0

02

Explanation

The sample proportion is the number of successes divided by the sample size:

p^1=x1n1=1050=0.2

p^2=x2n2=2853≈0.528

p^p=x1+x2n1+n2=10+2850+53=38103≈0.369

Determine the value of the test statistic:

localid="1650451317669" z=p^1-p^2p^p1-p^p1n1+1n2=0.2-0.5280.369(1-0.369)150+153≈-3.4

The p-value is the probability of obtaining the value of the test statistic, or a value more extreme. Determine the p-value using table A:

localid="1650451340981" P=P(Z<-3.45orZ>3.45)=2×P(Z<-3.45)=2×0.0003=0.0006

If the p-value is smaller than the significance level, reject the null hypothesis:

P<0.05⇒RejectH0

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Min Jae carried out the significance test shown below to answer this question. Unfortunately, he made some mistakes along the way. Identify as many mistakes as you can, and tell how to correct each one.

State: I want to perform a test of

H0:p1-p2=0

Ha:p1-p2>0

where p1=the proportion of Instructor A's students that passed the state exam and p2=the proportion of Instructor B's students that passed the state exam. Since no significance level was stated, I'll use σ=0.05

Plan: If conditions are met, I’ll do a two-sample ztest for comparing two proportions.

Random The data came from two random samples of 50students.

- Normal The counts of successes and failures in the two groups -30,20,22, and 28-are all at least 10.

- Independent There are at least 1000 students who take this driving school's class.

Do: From the data, p^1=2050=0.40and p^2=3050=0.60. So the pooled proportion of successes is

p^C=22+3050+50=0.52

- Test statistic

localid="1650450621864" z=(0.40-0.60)-00.52(0.48)100+0.52(0.48)100=-2.83

- p-value From Table A, localid="1650450641188" P(z≥-2.83)=1-0.0023=0.9977.

Conclude: The p-value, 0.9977, is greater than α=0.05, so we fail to reject the null hypothesis. There is no convincing evidence that Instructor A's pass rate is higher than Instructor B's.
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