Answer to Question #251391 in Statistics and Probability for Carrie

Question

The following data give the distance travelled to work by a sample of employees.

14:8 12:9 13:2 15:0 13:5 12:1 14:8 16:1 15:1 16:2

Test at the 5% level of significance the claim that the average distance travelled is less than 15

kilometres if you are given a population standard deviation of 2:5 kilometers.

(a) State the null and alternative hypothesis. (2)

(b) Calculate the sample mean average distance travelled. (3)

(c) State and calculate the appropriate test statistic. (7)

(d) Determine the critical value of the test. (2)

(e) State whether or not you reject the null hypothesis, giving the reason. (3)

(f) Draw an appropriate conclusion. (3)

Answer 1

(a)

$H_0: \mu = 15 \\ H_1: \mu< 15$

(b)

$n=10 \\ \bar{x} = \frac{14.8+12.9+...+15.1+16.2}{10} = 14.37 \\ \sigma = 2.5$

(c)

Test-statistic:

$Z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}} \\ Z = \frac{14.37-15}{2.5 / \sqrt{10}} \\ = \frac{-13.563}{0.790} \\ = -17.168$

(d) α= 0.05

One-tail test

$Z_c = 1.64$

(e) Reject H₀ if $Z < -Z_c$

$Z = -17.168 < Z_c = -1.64$

Reject null hypothesis.

(f) There is enough evidence to conclude that the average distance travelled is less than 15 kilometres at the 5% level of significance.