Question

It is claimed that a random sample of 100 tyres with a mean life of 15269 km is drawn from a population of tyres which has a mean life of 15200 km and a standard deviation of 1248 km test the validity of a claim 5% LOS

Answer 1

[AI-generated answer — Bot GPT 3.5]

To test the validity of the claim using a level of significance of 5%, we will conduct a hypothesis test.

Hypotheses:
Null hypothesis (H0): The mean life of the sample of 100 tyres is equal to the mean life of the population of tyres (μ = 15200 km).
Alternative hypothesis (H1): The mean life of the sample of 100 tyres is not equal to the mean life of the population of tyres (μ ≠ 15200 km).

Significance level (α) = 0.05

Given information:
Sample mean (x̄) = 15269 km
Population mean (μ) = 15200 km
Population standard deviation (σ) = 1248 km
Sample size (n) = 100

Calculate the test statistic:
t = (x̄ - μ) / (σ / √n)
t = (15269 - 15200) / (1248 / √100)
t = 69 / (1248 / 10)
t ≈ 0.5544

Calculate the critical t-value:
Degrees of freedom = n - 1 = 100 - 1 = 99
Using a t-distribution table or software, find the critical t-values at a 5% significance level for a two-tailed test with 99 degrees of freedom.
The critical t-values are approximately -2.626 and 2.626.

Make a decision:
Since the calculated t-value (0.5544) is within the range of critical t-values (-2.626, 2.626), we fail to reject the null hypothesis.

Conclusion:
There is not enough evidence to support the claim that the mean life of the sample of 100 tyres is different from the mean life of the population of tyres at a 5% level of significance.