Question

USA Today reported that about 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 1009 Chevrolet owners and found that 482 said they would buy another Chevrolet. Does this indicate that the population proportion of consumers loyal to Chevrolet is more than 47%? Use α = 0.01.

(a) What is the level of significance?

State the null and alternate hypotheses.

(b) What sampling distribution will you use?

What is the value of the sample test statistic? (Round your answer to two decimal places.)
(c) Find the P-value of the test statistic. (Round your answer to four decimal places.)

(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level α?

(e) Interpret your conclusion in the context of the application.

Answer 1

Solution:

Part a

Level of significance = α = 0.01

Null hypothesis: H₀: the population proportion of consumers loyal to Chevrolet is 47%.

Alternative hypothesis: H_a: the population proportion of consumers loyal to Chevrolet is more than 47%.

Part b

We use normal distribution.

Test statistic formula for this test is given as below:

Z = (p̂ - p)/sqrt(pq/n)

Where, p̂ = Sample proportion, p is population proportion, q = 1 - p, and n is sample size

x = number of items of interest = 482

n = sample size = 1009

p̂ = x/n = 482/1009 = 0.477700694

p = 0.47

q = 1 - p = 0.53

Z = (p̂ - p)/sqrt(pq/n)

Z = (0.477700694 – 0.47)/Sqrt[0.47*0.53/1009]

Z = 0.4901

Test statistic = 0.4901

Part c

P-value = 0.3120

(by using z-table)

Part d

We fail to reject the null hypothesis because P-value is greater than alpha value 0.01.

Data is not statistically significant at level α=0.01.

Part e

There is not sufficient evidence to conclude that the population proportion of consumers loyal to Chevrolet is more than 47%.