Q:

# 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 1004 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.What is the sample test statistic?What is the p value of the test statistic?

Accepted Solution

A:
Answer:$$z=\frac{0.48 -0.47}{\sqrt{\frac{0.47(1-0.47)}{1004}}}=0.635$$   $$p_v =P(z>0.635)=1-P(z<0.635)=1-0.737=0.263$$   Step-by-step explanation:1) Data given and notation n   n=1004 represent the random sample taken X=482 represent the Chevrolet owners who said they would buy another Chevrolet.$$\hat p=\frac{482}{1004}=0.480$$ estimated proportion of Chevrolet owners who said they would buy another Chevrolet. $$p_o=0.47$$ is the value that we want to test $$\alpha$$ represent the significance level   z would represent the statistic (variable of interest) $$p_v$$ represent the p value (variable of interest)   2) Concepts and formulas to use   We need to conduct a hypothesis in order to test the claim that the proportion of Chevrolet owners who said they would buy another Chevrolet is higher than 47%:   Null hypothesis:$$p\leq 0.47$$   Alternative hypothesis:$$p > 0.47$$   When we conduct a proportion test we need to use the z statisitc, and the is given by:   $$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$$ (1)   The One-Sample Proportion Test is used to assess whether a population proportion $$\hat p$$ is significantly different from a hypothesized value $$p_o$$. 3) Calculate the statistic   Since we have all the info requires we can replace in formula (1) like this:   $$z=\frac{0.48 -0.47}{\sqrt{\frac{0.47(1-0.47)}{1004}}}=0.635$$   4) Statistical decision   P value method or p value approach . "This method consists on determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.   If we use the significance level provided, $$\alpha=0.01$$. The next step would be calculate the p value for this test.   Since is a one side upper test the p value would be:   $$p_v =P(z>0.635)=1-P(z<0.635)=1-0.737=0.263$$   So based on the p value obtained and using the significance level assumed $$\alpha=0.01$$ we have $$p_v>\alpha$$ so we can conclude that we FAIL to reject the null hypothesis, and we can said that at 1% of significance the proportion of Chevrolet owners who said they would buy another Chevroletis not significantly higher than 0.47 or 47% .