Question

I will be testing the population mean number of miles driven to work (one way) by my circle of friends and coworkers. I believe that in this sample group that the average miles driven from their homes to their offices is less than 15 miles. My data will be a survey of the 10 closest friends and coworkers.

The 10 responses were as follows for the average miles driven from their homes to their employer/office: 12,10,18,25,5,8,15,4.5 & 10. I will test this hypothesis at the .05 significance level. Assume a random sample.

Answer 1

Solution:

Here, we have to use one sample t test for the population mean.

The null and alternative hypotheses are given as below:

Null hypothesis: H0: The average mile driven from homes to offices is 15 miles.

Alternative hypothesis: Ha: The average miles driven from homes to offices is less than 15 miles.

H0: µ = 15 versus Ha: µ < 15

This is a lower tailed test.

The test statistic formula is given as below:

t = (Xbar - µ)/[S/sqrt(n)]

From given data, we have

µ = 15

Xbar = 11.2

S = 6.6131

n = 10

df = n – 1 = 9

α = 0.05

Critical value = -1.8331

(by using t-table or excel)

t = (11.2 – 15)/[6.6131/sqrt(10)]

t = -1.8171

P-value = 0.0513

(by using t-table)

P-value > α = 0.05

So, we do not reject the null hypothesis

There is not sufficient evidence to conclude that the average miles driven from homes to offices is less than 15 miles.