A research department is investigating the duration of a tire.The department wants to show it exceeds a certain value. Formulate the hypothesis

Question

The research department of a tire manufacturer is investigating the duration of a tire using a new rubber component. 16 tires were produced and the duration was tested. The average duration and stdev were 60139.7 and 3645.94, respectively. The department wants to show that the average duration of the new tire exceeds 60000 km. Formulate the appropriate test and hypothesis, using alpha = 0.05

I did

$$H_0 : \mu>60000$$

I am not given the variance of the population, so I use the t-student distribution for the test:

$$T_0 = \frac{x-\mu_o}{s/\sqrt{n}} = \frac{60139.7-60000}{3645.94/4}=0.153266373$$

The upper bound:

$$T_{0.05,15}=2.947 $$

So... what do I conclude? Do I reject the hypothesis? I'm trying to imagine a t-student distribution where the upper bound is 2.947 and the acceptation area is everything above it. T_0 is below it, so is it false?

Answer 1

For the (one-sided) test here, you would only reject the null hypothesis if your calculated test statistic ($0.1532\ldots$) were greater than the critical value ($2.947$). Since this is not the case, you fail to reject the null hypothesis.