# An engineer measured the Brinell hardness of 25 pieces of ductile iron that were subcritically annealed. The resulting data were: 170, 167, 174, 179, 187, 179, 183, 179, 156, 163, 156, 187, 156, 174, 170, 183, 179, 174, 179, 170, 159, 187, 167, 156. Based on these results, the engineer hypothesized that the mean Brinell hardness of all such ductile iron pieces is greater than 170. Is there enough evidence at the 5% significance level to support the engineer’s hypothesis? Assume data is normally distributed.

No, there is insufficient evidence to support the claim that the mean is greater than 170.

We are given a set of data such that `n=24,s~~10.44,bar(x)=172.25`, where n is the sample size, s is the sample standard deviation, and the mean of the sample was 172.25. An engineer hypothesized that the true mean was greater than 170, and we are asked to test this claim with 5% significance.

I. `H_0: mu=170`

`H_1:mu>170`, which is the claim.

II. This is a one-tailed test. We do not know the population standard deviation, so we must use the student's t-table with degrees of freedom 23 and `alpha=.05` to get a critical value of 1.714.

III. We compute the test value `t=(bar(x)-mu)/(s/sqrt(n))=(172.25-170)/(10.44/sqrt(24))~~1.0558`.

IV. Since 1.0558<1.714, it is not in the critical region, so we do not reject the null-hypothesis. Alternatively, we can try to use the t-table to estimate the probability of getting this type of sample or one more extreme. From a t-table with degrees of freedom 23, we find 1.060 (close to 1.055) with `p~~.15` (my calculator gives `p~~.1510864077`). With `p>alpha=.05`, we do not reject the null-hypothesis.

V. There is insufficient evidence to support the claim that the true mean is greater than 170.

Note that we have not disproved the claim. We only found that there isn't enough evidence to support the claim. We also haven't showed that `mu<=170`, as we cannot prove the null-hypothesis.