Question

USA Today reported that about 47% (.47) of the general consumer population in the United States...

USA Today reported that about 47% (.47) of the general consumer population in the United States is loyal to the automobile manufacturer of their choice. Suppose Chevrolet did a study of a random sample of 987 Chevrolet owners and found that 490 said they would buy another Chevrolet.

Does this indicate that the population proportion of consumers loyal to the Chevrolet is more than 47%? Use α = 0.01. Solve the problem by calculating the value of z and the probability associated with it. Find the difference in the two proportions (population and sample) and divide by the standard error - which is 0.016 for this problem - to get the value of the test statistic. Since the sampling distribution of is the normal distribution, you can use critical values from the standard normal distribution as shown in the table of critical values of the z distribution. (Round the test statistic and the critical value to two decimal places. Round the p-value to four decimal places.)

A: test statistic =

B: critical value =

C: p-value=

2. An Independent Samples t test produces the following results: t(29) = 5.15, p < .05.

1) How many participants were included in the study?

2) What was the decision abut the Null Hypothesis?

Homework Answers

Answer #1

1)

Claim: The population proportion of consumers loyal to the Chevrolet is more than 47%.

The null and alternative hypothesis is

H0: P 0.47

H1: P > 0.47

Level of significance = 0.01

Sample size = n = 987

x = 490

Test statistic is

Critical value = 2.32

Test statistic z < critical value we fail to reject the null hypothesis.

Conclusion:

The population proportion of consumers loyal to the Chevrolet is NOT more than 47%.

2)

1) t(29) = 5.15 , p < 0.05

n - 1 = 29

n = 30

2) P-value < 0.05 we reject null hypothesis.

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