Question

A new machine on the market claims to cut the trees, on average, in less than 9.8 minutes. A random sample of 15 test runs on the new machine yielded a mean of 8.5 minutes with a standard deviation of 1.5.

Answer 1

Given that, sample size (n) = 15, sample mean = 8.5 minutes

and sample standard deviation (s) = 1.5 minutes

The null and alternative hypotheses are,

H0 : μ = 9.8 minutes

Ha : μ < 9.8 minutes (claim)

This hypothesis test is a left-tailed test.

Test statistic is,

=> Test statistic = t = -3.357

Degrees of freedom = 15 - 1 = 14

Using t-table we get, t-critical value at significance level of 0.05 with 14 degrees of freedom is, tcrit = -1.761 (negative, since it is left-tailed test).

Since, test statistic = -3.357 is less than -1.761, we​​​​​ reject the null hypothesis (H0).

Conclusion : There is sufficient evidence to support the claim that the average cutting time of trees on average is less than 9.8 minutes.