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 random independent 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 observed value of the test statistic for the difference in mean mpg? About 0.72 About 2.18 About 25.45 mpg None of the above

Answer 1

Answer:

About 0.72

Step-by-step explanation:

Given the data:

With device : 22.6, 23.4, 28.4, 29, 29.3, 20.0

Without device :26.9, 24.4, 20.8, 20.8, 20.2, 26.0, 28.1, 25.6

Using calculator:

With device data:

Sample size (n1) = 6

Degree of freedom (df1) = 6 - 1 = 5

Mean(m1) = 25.45

Variance (s²1) = 15.631

Without device data:

Sample size (n2) = 8

Degree of freedom (df2) = n - 1 = 7

Mean (m2) = 24.1

Variance (s²2) = 9.54

s²t = ((df1/(df1 + df2)) * s²1) + ((df2/(df1 + df2)) * s²2)

s²t = ((5/(5+7)*15.63) + ((7/(5+7))*9.54)

= ((5/12) * 15.63) + ((7/12) * 9.54) = 12.0775

s²m1 = s²t/n1 = 12.0775/6 = 2.0123

s²m2 = s²t/n2 = 12.0775/8 = 1.511

T - statistic :

Tstat = (m1 - m2)/√(s²m1 + s²m2)

Tstat = (25.45 - 24.10) / √(2.0123 + 1.511)

Tstat= 1.350/√3.5233

Tstat = 1.350 / 1.8770455

Tstat = 0.7192153 = 0.72