Question

Under specified conditions, an automobile manufacturer claims that its new compact car will get more miles per gallon (mpg) than other cars in its class. For cars of the same class, the average is 23 with a variance of 9.00 mpg. To investigate, the manufacturer tested 18 cars in which the average was 19.5 mpg. What can the car manufacturer conclude with α = 0.05?

a) What is the appropriate test statistic?
---Select--- na z-test one-sample t-test independent-samples t-test related-samples t-test

b)
Population:
---Select--- automobile manufacturer mpg specified conditions tested cars cars in same class
Sample:
---Select--- automobile manufacturer mpg specified conditions tested cars cars in same class

c) Compute the appropriate test statistic(s) to make a decision about H₀.
(Hint: Make sure to write down the null and alternative hypotheses to help solve the problem.)
critical value =  ; test statistic =  
Decision:  ---Select--- Reject H0 Fail to reject H0

d) If appropriate, compute the CI. If not appropriate, input "na" for both spaces below.
[  ,  ]

e) Compute the corresponding effect size(s) and indicate magnitude(s).
If not appropriate, input and select "na" below.
d =  ;   ---Select--- na trivial effect small effect medium effect large effect
r² =  ;   ---Select--- na trivial effect small effect medium effect large effect

f) Make an interpretation based on the results.

Under specified driving conditions, the new compact car gets significantly more mpg than cars in its class.

Under specified driving conditions, the new compact car gets significantly less mpg than cars in its class.    

Under specified driving conditions, the new compact car does not get significantly different mpg than cars in its class.

Answer 1

Solution:-

a) One sample z-test.

b) Population: All the other cars in same class.
Sample: The 18 cars tested by the manufacturer .
c)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u < 23
Alternative hypothesis: u > 23

Note that these hypotheses constitute a one-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample z-test.

Analyze sample data. Using sample data, we compute the standard error (SE), z statistic test statistic (z).

SE = s / sqrt(n)

S.E = 0.70711
z = (x - u) / SE

z = - 4.95

z_Critical = + 1.645

Rejection region is z > 1.645

where s is the standard deviation of the sample, x is the sample mean, u is the hypothesized population mean, and n is the sample size.

Interpret results. Since the z-value does not lies in the rejection region, hence we failed to reject the null hypothesis.

From the above test we do not have sufficient evidence in the favor of the claim that  its new compact car will get more miles per gallon (mpg) than other cars in its class.

d) 95% confidence interval for the mean is C.I = ( 18.114, 20.886).

C.I = 19.5 + 1.96 × 0.70711

C.I = 19.5 + 1.386

C.I = ( 18.114, 20.886)