A company in Maryland has developed a device that can be attached to car engines

Question

A company in Maryland has developed a device that can be attached to car engines, which it believes will increase the miles per gallon that cars will get. The owners are interested in estimating the difference between mean mpg for cars using the device versus those that are not using the device. The following data represent the mpg for independent random samples of cars from each population. The variances are assumed equal and the populations normally distributed.

With Device    Without Device
22.6                26.9
23.4                24.4
28.4    20.8
29.0    20.8
29.3                20.2
20.0    26.0
          28.1
                      25.6

Given this data, what is the critical value if the owners wish to have a 90 percent confidence interval estimate?

t = 2.015

t = 1.7823

z = 1.645

z = 1.96

t = 2.015

t = 1.7823

z = 1.645

z = 1.96

Explanation / Answer

n = 14 samples

df = 14 - 2 = 12

looking up the t table with 12 df and (1 - 0.90)/2 = 0.05 level, the critical value is 1.7823.

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