Question

The percentage of titanium in an alloy used in aerospace castings is measured in 51 randomly selected parts. The sample deviation in these parts is 0.37. Use this information to test for the hypothesis that the true deviation in the percentage of titanium different from 0.25 using a 5% level of significance. Show all work for this problem!

Answer 1

Let o denotes the population standard deviation. Hypotheses : To test null hypothesis H:0 =0.25 against alternative hypothesis H1 :0 +0.25 Let s denotes the sample standard deviation and n denotes the sample size. Here, sa = 0.1369, s = 0.3700, n = 51 The test statistic can be written as: (n-1)s 0.252 which under H, follows a x2 distribution with n - 1 df. Decision rule / Rejection region : We reject H, at 0.05 level of significance if P-value < 0.05 or if xãos > x6.025,1–1 or if xãos < xố.975,0–1 0.04 - 0.03 - 0.02- 0.01 - 0 80 0.00 - 20 40 60 Now, chisa The value of the test statistic (51 – 1) *0.372 0.252 = 109.520000 109.520 associated degrees of freedom = 50 The smaller critical value is = xô.975,50 =32.357364 The bigger critical value is = x2.025,50 =71.420195 P-value = 2 * P(xn-1 > Xəbs) = 2 * P(Xžo > 109.52) = 2 * (1 – 0.999998) = 0.000005 Conclusion: Since p-value < 0.05 and xəbs > Xéritical = 71.420195, so we reject H, at 0.05 level of significance Hence, we can conclude that the the population standard deviation is significantly different from 0.25.