An experiment was performed to compare the fracture toughness of high-purity 18 Ni maraging steel with commercial-purity steel of the same type. For 32 specimens, the sample average toughness was 65.6 for the high-purity steel, whereas for 38 specimens of commercial steel the sample average was 59.8. Because the high-purity steel is more expensive, its use for a certain application can be justified only if its fracture toughness exceeds that of commercial-purity steel by more than 5. Assume that both toughness distributions are normal, σ = 1.2 and σ = 1.1.
(a) Test the relevant hypothesis using α = .001.
(b) Compute β for the test conducted in part (a) when μ1 - μ2 = 6.

(a) Fail to reject the null hypothesis.

(b) The value of β is 0.00015.

Step-by-step explanation:

(a)

Here we need to test whether the difference between the two population means exceeds 5 or not.

The hypotheses are:

H₀: µ₁ - µ₂ ≤ 5 vs. Hₐ: µ₁ - µ₂ > 5.

Since the population standard deviations are provided we will use a z-test.

The information provided is:

Compute the value of the test statistic as follows:

The test statistic value is 2.89.

Decision rule:

Reject the null hypothesis if p-value is less than the significance level, α = 0.001.

Compute the p-value as follows:

The p-value of the test is 0.00193.

p-value = 0.00193 > α = 0.001.

Thus, we fail to reject the null hypothesis at α = 0.001.

Conclusion:

The fracture toughness does not exceeds that of commercial-purity steel by more than 5.

(b)

A type II error is a statistical word used within the circumstance of hypothesis testing that defines the error that take place when one is unsuccessful to discard a null hypothesis that is truly false. It is symbolized by β i.e.  

β = Probability of accepting H₀ when H₀ is false.

β = P (µ₁ - µ₂ ≤ 5 | µ₁ - µ₂ = 6)

*Use a z-table for the probability.

Thus the value of β is 0.00015.

the third picture

step-by-step explanation:

the solution is h=10 because you add 9 to each side of the equation.