Question

In: Statistics and Probability

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

1.

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 996 Chevrolet owners and found that 503 said they would buy another Chevrolet. Does this indicate that the population proportion of consumers loyal to the car company is more than 47%? Use α = 0.01. Solve the problem using both the traditional method and the P-value method. 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.)

test statistic =
critical value =
P-value =


State your conclusion in context of the application.

There is sufficient evidence at the 0.01 level to conclude that the true proportion of consumers loyal to the car company is more than 47%.

There is insufficient evidence at the 0.01 level to conclude that the true proportion of consumers loyal to the car company is more than 47%.     


Compare your conclusion with the conclusion obtained by using the P-value method. Are they the same?

The conclusions obtained by using both methods are the same.

We reject the null hypothesis using the P-value method, but fail to reject using the traditional method.  

   We reject the null hypothesis using the traditional method, but fail to reject using the P-value method.

2.

In this problem, assume that the distribution of differences is approximately normal. Note: For degrees of freedom d.f. not in the Student's t table, use the closest d.f. that is smaller. In some situations, this choice of d.f. may increase the P-value by a small amount and therefore produce a slightly more "conservative" answer.

Suppose that at five weather stations on Trail Ridge Road in Rocky Mountain National Park, the peak wind gusts (in miles per hour) for January and April are recorded below.

Wilderness District 1 2 3 4 5
January 139 120 126 64 78
April 101 110 108 88 61

Does this information indicate that the peak wind gusts are higher in January than in April? Use α = 0.01. Solve the problem using the critical region method of testing. (Let d = January − April. Round your answers to three decimal places.)

test statistic =
critical value =


Interpret your conclusion in the context of the application.

Reject the null hypothesis, there is sufficient evidence to claim average peak wind gusts are higher in January.

Fail to reject the null hypothesis, there is sufficient evidence to claim average peak wind gusts are higher in January.    

Fail to reject the null hypothesis, there is insufficient evidence to claim average peak wind gusts are higher in January.

Reject the null hypothesis, there is insufficient evidence to claim average peak wind gusts are higher in January.


Compare your conclusion with the conclusion obtained by using the P-value method. Are they the same?

We reject the null hypothesis using the critical region method, but fail to reject using the P-value method.

We reject the null hypothesis using the P-value method, but fail to reject using the critical region method.     

The conclusions obtained by using both methods are the same.

Solutions

Expert Solution

1. Given : n=996 , X=503

The estimate of the sample proportion is ,

The null and alternative hypothesis is ,

The test is right tailed test.

The test statistic is ,

The critical value is , ; From Z-table

The p-value is ,

p-value= ; The Excel function is , =1-NORMDIST(2.21,0,1,TRUE)

Decision : From critical value : Here , Z-stat=2.21<2.33 Therefore , fail to reject Ho

From p-value : Here , p-value>0.01 Therefore , fail to reject Ho.

Conclusion :

There is insufficient evidence at the 0.01 level to conclude that the true proportion of consumers loyal to the car company is more than 47%.     


Compare your conclusion with the conclusion obtained by using the P-value method.

The conclusions obtained by using both methods are the same.


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