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 1006 Chevrolet owners and found that 486 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.

H₀: p = 0.47; H₁: p < 0.47H₀: p = 0.47; H₁: p > 0.47    H₀: p = 0.47; H₁: p ≠ 0.47H₀: p > 0.47; H₁: p = 0.47

(b) What sampling distribution will you use?

The standard normal, since np > 5 and nq > 5.The standard normal, since np < 5 and nq < 5.    The Student's t, since np > 5 and nq > 5.The Student's t, since np < 5 and nq < 5.

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.)


Sketch the sampling distribution and show the area corresponding to the P-value.

(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?

At the α = 0.01 level, we reject the null hypothesis and conclude the data are statistically significant.At the α = 0.01 level, we reject the null hypothesis and conclude the data are not statistically significant.    At the α = 0.01 level, we fail to reject the null hypothesis and conclude the data are statistically significant.At the α = 0.01 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.

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

There is sufficient evidence at the 0.01 level to conclude that the true proportion of customers loyal to Chevrolet is more than 0.47.There is insufficient evidence at the 0.01 level to conclude that the true proportion of customers loyal to Chevrolet is more than 0.47.

Answer 1

a)

Answer:

The level of significance = 0.01

H0: p = 0.47; H1: p > 0.47

Explanation:

The null hypothesis is defined as the population proportion of consumers loyal to Chevrolet is 0.47 and the alternative hypothesis test the claim that this proportion is greater than 0.47.

b)

Answer:

The standard normal, since np > 5 and nq > 5

The sample test statistic, z = 0.83

Explanation:

Normality condition

Since the sample data values satisfy the normality condition,

n*p = 1000*0.47 = 472.8 > 5 and

n*q = 1000*0.53 = 533.2 > 5

the standard normal distribution is used.

Test statistic

The z-statistic is obtained using the formula,

where p = 486/1006 = 04831, p0 = 0.47

c)

Answer:

Explanation:

The p-value is obtained from z-distribution table for z = 0.83

d)

Answer:

Fail to reject the null hypothesis. The data are not statistically significant at a 1% significance level

Explanation:

Since the p-value = 0.2025 is greater than 0.01 at a 1% significance level, the null hypothesis is not rejected.

e)

Answer:

There is insufficient evidence at the 0.01 level to conclude that the true proportion of customers loyal to Chevrolet is more than 0.47.

Explanation:

Since the null hypothesis is failed to reject, there is not sufficient evidence to conclude that the true proportion of customers loyal to Chevrolet is more than 0.47.