Question

The treadwear index provided on car tyres helps prospective buyers make their purchasing decisions by indicating a tyre's resistance to tread wear. A tyre with a treadwear grade of 200 should last twice as long, on average, as a tyre with a grade of 100. A consumer advocacy organisation wishes to test the validity of a popular branded tyre that claims a treadwear grade of 200. A random sample of 19 tyres indicates a sample mean treadwear index of 195.2 and a sample standard deviation of 21.4.

(a) Using 0.05 level of significance, is there evidence to conclude that the tyres are not meeting the expectation of lasting twice as long as a tyre graded at 100? Show all your workings.

(b) What assumptions are made in order to conduct the hypothesis test in (a)?

Answer 1

Sample size = n = 19

Sample mean = = 195.2

Standard deviation = s = 21.4

Claim: The tyres are not meeting the expectation of lasting twice as long as a tyre graded at 100.

The null and alternative hypothesis is

Level of significance = 0.05

Here population standard deviation is unknown so we have to use t-test statistic.
Test statistic is

195.2 - 200 ー4. 8 t=T =4.909497 =-0.98 4.909497-0.98 2

Degrees of freedom = n - 1 = 19 - 1 = 18

Critical value = 2.101 ( Using t table)

Test statistic | t | < critical vaue we fail to reject null hypothesis.

Conclusion: There is not sufficient evidence to conclude that the tyres are not meeting the expectation of lasting twice as long as a tyre graded at 100.

b) Assumptions:

i. Data is normally distributed.

ii. Here population standard deviation is unknown so we have to use t-test statistic.