Question

The Rockwell hardness test. Was performed on two different materials. The results are in the following table: Material 1 ny = 40 Vi = 1.7 S = 0.25 Material 2 n2 = 45 72 = 1.4 S, = 0.23 Construct a 95% confidence interval for the difference of the hardness means. b) Do a hypothesis test for the difference of the hardness means, use a = 0.05. Is your result in agreement with part a)? explain.

Answer 1

a)

We need to construct the 95% confidence interval for the difference between the population means μ1​−μ2​, for the case that the population standard deviations are not known. The following information has been provided about each of the samples:

Sample Mean 1 (X1) = 1.7 Sample Standard Deviation 1 (81) = 0.25 Sample Size 1 (Ni) = 40 Sample Mean 2 (X2) = 1.4 Sample Standard Deviation 2 (82) = 0.23 Sample Size 2 (N2) = 45 Based on the information provided, we assume that the population variances are equal, so then the number of degrees of freedom are df = n1 + n2 – 2 = 40 +45 – 2 = 83. The critical value for a = 0.05 and df = 83 degrees of freedom is to = tı-a/2:n-1 = 1.989. The corresponding confidence interval is computed as shown below: Since the population variances are assumed to be equal, we need to compute the pooled standard deviation, as follows: 8p _ = 1 (n1 – 1)s +(n2 – 1)s ni+n2 - 2 (40 – 1) < 0.252 + (45 – 1) < 0.232 40+ 45 - 2 = 0.24 Since we assume that the population variances are equal, the standard error is computed as follows: 1 1 se = SpV Vn na = 0.24 x V10 + 1 = 0.052

Now, we finally compute the confidence interval:

CI = (X1 - X2 - te X se, X1 - X2 + tc X se) = (1.7 - 1.4 - 1.989 x 0.052, 1.7 - 1.4 +1.989 x 0.052) = (0.196, 0.404)

b)

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ1​ = μ2​

Ha: μ1​ ≠ μ2​

This corresponds to a two-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.

(2) Rejection Region

Based on the information provided, the significance level is α=0.05, and the degrees of freedom are df=83. In fact, the degrees of freedom are computed as follows, assuming that the population variances are equal.

Hence, it is found that the critical value for this two-tailed test is tc​=1.989, for α=0.05 and df=83.

The rejection region for this two-tailed test is R={t:∣t∣>1.989}.

(3) Test Statistics

Since it is assumed that the population variances are equal, the t-statistic is computed as follows:

XL - X2 (n-1) +(12-1) (1 + 1) 12+ -2 1.7 – 1.4 = (40-1)0.25% +(45-1)0.232 (1 123 (+) 1 40+45-2 = 5.762 po per + 4) 5,762

(4) The decision about the null hypothesis

Since it is observed that ∣t∣=5.762>tc​=1.989, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p=0, and since p=0<0.05, it is concluded that the null hypothesis is rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that population mean μ1​ is different than μ2​, at the 0.05 significance level.

Agreement

YES, the results are in agreement with part (a) since 0 does not lie in the above confidence interval so we would have concluded that the null hypothesis is rejected.