Question

In: Statistics and Probability

Two car types underwent fuel efficiency tests. The table below summarizes the descriptive statistics generated from...

Two car types underwent fuel efficiency tests. The table below summarizes the descriptive statistics generated from two samples of independent cars. Assume that both samples are from normal distributions with equal population variance: ? 1 ~?(? 1 , ? 2 ) and ? 2 ~?(? 2 , ? 2 ).

Car type

Notation

Sample size

Mean (miles per gallon)

Standard deviation (miles per gallon)

1

?1

9

20

6.4

2

?2

9

30

6.1

a) Is there sufficient evidence to claim that the first type of car runs on average below 25 miles per gallon (? 0 : ? 1 = 25 versus ? 1 : ? 1 < 25)? Conduct the test using p-value with

? = 0.1.

b) Find if ? for the test in part (a) is greater than or less than 0.1 when the true ? 1 = 20.

c) Can we claim that the two types of cars have the same fuel efficiency (? 0 : ? 1 = ?2 versus ? 1 : ? 1 ≠ ? 2 )? Conduct the test using p-value with ? = 0.05.

d) Can we claim that the first type is less fuel efficient than the second (? 0 : ? 1 = ? 2 versus ? 1 : ? 1 < ? 2 )? Conduct the test using critical region with ? = 0.05.

Solutions

Expert Solution

a)

test statistic :

df = n-1=9-1=8

P-value for left taled test = 0.0236

b) This is a left tailed test.

We will fail to reject the null (commit a Type II error) if we get a Z statistic greater than -1.2816.

This -1.2816 Z-critical value corresponds to some X critical value ( X critical), such that

So I will incorrectly fail to reject the null as long as a draw a sample mean that greater than 22.27.

To complete the problem what I now need to do is compute the probability of drawing a sample mean greater than 22.27 given µ = 20. Thus, the probability of a Type II error is given by

c)

significance level :

Test statistic :

degree of freedom = n_1+n_2-2=9+9-2=16

P-value for two-tailed test = 0.0037

The P-value is less than the significance level hence we reject the null hypothesis.

d) here the alternate hypothesis changes

The test statistic is the same t = -3.3931

df = 16

The P-value for the left tailed test is 0.0019

the test is significant, we can reject the null hypothesis.


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