Question 6.11: A new concrete mix is being evaluated. The plan is to sample......
A new concrete mix is being evaluated. The plan is to sample 100 concrete blocks made with the new mix, compute the sample mean compressive strength \overline{X}, and then test H_{0} : \mu ≤ 1350 versus H_{1} : \mu > 1350, where the units are MPa. It is assumed from previous tests of this sort that the population standard deviation 𝜎 will be close to 70 MPa. Find the critical point and the rejection region if the test will be conducted at a significance level of 5%.
We will reject H_{0} if the P-value is less than or equal to 0.05. The P-value for this test will be the area to the right of the value of \overline{X}. Therefore, the P-value will be less than 0.05, and H_{0} will be rejected, if the value of \overline{X} is in the upper 5% of the null distribution (see Figure 6.11). The rejection region therefore consists of the upper 5% of the null distribution. The critical point is the boundary of the upper 5%. The null distribution is normal, and from the z table we find that the z-score of the point that cuts off the upper 5% of the normal curve is z_{.05} = 1.645. Therefore, we can express the critical point as z = 1.645 and the rejection region as z ≥ 1.645. It is often more convenient to express the critical point and rejection region in terms of \overline{X}, by converting the z-score to the original units. The null distribution has mean 𝜇 = 1350 and standard deviation \sigma_{\overline{X}} = \sigma ∕ \sqrt{n}\ ≈\ 70 ∕ \sqrt{100} = 7. Therefore, the critical point can be expressed as \overline{X} = 1350+(1.645)(7) = 1361.5. The rejection region consists of all values of \overline{X} greater than or equal to 1361.5.
