Question

A researcher wanted to test his claim that the mean walking pace of business

travelers is different from that of leisure travelers at the airport. To test his claim he
obtained the following lists of 9 business and 8 leisure travelers. Test his claim at the
.05 = level of significance.
Business
Travelers
42 31 37 45 49 52 43 39 45
Leisure
Travelers
32 29 35 40 38 34 42 33

Answer 1

`t = \frac{42.56-35.375}{\sqrt{(6.327^2)/(9) +(4.34^2)/(8)}}=2.755`

The degrees of freedom are given by:

`df=n_1 +n_2-2 =9+8-2= 15`

Since we have a two tailed test the p value can be calculated like this:

`p_v=2* P(t_(15) >2.755) = 0.0147`

And since the p value is lower than the significance lvel given of 0.05 we have enough evidence to conclude that we have significant differences between the two groups on this case.

Explanation:

We have the following data given:

Business Travelers

42 31 37 45 49 52 43 39 45

Leisure Travelers

32 29 35 40 38 34 42 33

For this case we need to begin finding the sample mean and deviations for each group with the following formulas:

`\bar X =(\sum_(i=1)^n X_i)/(n)`

`s=\sqrt{(\sum_(i=1)^n (X_i -\bar X)^2)/(n-1)}`

And we got:

`\bar X_1 = 42.56` represent the sample mean for the Business travelers

`s_1 = 6.327` represent the sample deviation for the Business travelers

`n_1= 9` the sample size for the Business travelers

`\bar X_2 = 35.375` represent the sample mean for the Leisure travelers

`s_2 =4.34` represent the sample deviation for the Leisure travelers

`n_2= 8` the sample size for the Leisure travelers

The system of hypothesis for this case are:

Null hypothesis:
`\mu_1 =\mu_2`

Alternative hypothesis:
`\mu_1 \\eq \mu_2`

The statistic for this case is given by:

`t =\frac{\bar X_1 -\bar X_2}{\sqrt{(s^2_1)/(n_1)+(s^2_2)/(n_2)}}`

And replacing we got:

`t = \frac{42.56-35.375}{\sqrt{(6.327^2)/(9) +(4.34^2)/(8)}}=2.755`

The degrees of freedom are given by:

`df=n_1 +n_2-2 =9+8-2= 15`

Since we have a two tailed test the p value can be calculated like this:

`p_v= 2*P(t_(15) >2.755) = 0.0147`

And since the p value is lower than the significance lvel given of 0.05 we have enough evidence to conclude that we have significant differences between the two groups on this case.