Question

A researcher wanted to know whether people drive faster in sports cars or pick up trucks. He went out on the freeway with a radar gun and track the speed of people in both types of cars. A sample of n=30 sport car drivers went an average of M=69 (s=5.35) whereas a sample n=30 pick up truck drivers wet an average of M=67 9 (s=5.30). What statistics should be run? Calculate Harlteys F-Max, one-tailed to two-tailed test, calculate confidence interval of 95%, calculate appropriate statistic

Answer 1

We have to compare the mean speed of sports car with the mean speed of pick up trucks on the basis of sample observation

As sample mean and sample standard deviation are given for two indepdndent samples  

We shall use two sample t test for mean

Harltey's F max is given by

Fmax = Larger variance / Smaller variance = 5.352 / 5.302 = 1.02

As the value is close to 1 , we can conclude  that there is homogeinity of variance between two groups

(Though if the value comes close to 1 , we need to compare it with table value

degrees of freedom = n-1 = 30-1 =29 (in each group) and there are k=2 groups

From Fmax table , for k=2 and df = 29

table value is 2.07

As calculated value , Fmax < 2.07

we can conclude  that there is homogeinity of variance between two groups )

Now we conduct two sample t test for equal variance

This is one tailed test (as we want to find out which car drives faster , and on the basis of sample mean , we claim that people with pickup van drives faster)

The null and alternative hypothesis

The 95% confidence interval for true difference between population means

where

degrees of freedom = n1+n2 -2 = 58

For 95% confidence with df = 58 , two tailed critical value of t is

tc = 2.00

Thus 95% confidence interval is

=( -0.75 , 4.75)

The test statistic is

=1.45

For 95% confidence with df = 58 , right  tailed critical value of t is

tc = 1.67

Since calculated value of t < 1.67

We fail to reject H0

There is not sufficient evidence to conlcude that pick up cars faster than sports car .