Question

Hypothesis Testing_03 (two independent samples) The diameter of steel rods manufactured on two different extrusio steel rods manufactured on two different extrusion machines is being investigated. Two random samples of sizes , 15 and 17 are selected and the sample means and sample variances are sf = 0.35, 12 = 8.68, and s} = 0.40, respectively. e sample means and sample variances are *; -8.73 d. Write down null and alternative hypotheses to test if the machines produce rods with different mean diameters. b. What is the type of statistical test that is appropriate? Explain. c. Test the hypotheses at a 0.05, using the critical value approach. What is your decision on null hypothesis(H)? Interpret the test results. d. What is the P Value for the test? Test the hypotheses using the P Value. Do you get the same answer as part (c)? e. Construct a 95% confidence interval for the mean difference of diameters. Test the hypotheses using the confidence interval approach. Do you get the same answer as in parts (c) and (d)?

Answer 1

Sciulion. Given that Suuple sizes 11:15 12=14 si'=0.35' so = 0.40 = 5:43 , Z = 8.68 The null hypothesis is : Ho : HICHI alternative hypothesis is ' Hi Ho Mo and 30.05 The test statistic is 7 (21-57.7 -CH2-H2) T =- compute the pooled Estivelor spt em (ni-u)s,'t ini) si Sp' = Sint-2 sp= (1-5-1) (0.35)+(1+-1)(-ų) 15+17-22 = 0.3767 sp = To: 3767 sp = 0.6137

Now, To - Tito 8.75 -5.65 0.6137 Ett 15 To = 0.93 degrees of freedom of net1.-2 15 +14.2 - So Decision rule : Reject to if 'To > t itt or Toc-truth at x = 0.05 levd of significance. 1. teleff = 470.05(2130) = 2.048 Since, To= 0.23 talardf (critically D..) - 9.6442 so we ofil to neject null hypothesis. Hince , conclude that there is no significant Difference between the two means. The value of p is = 0.80 i p-value 0.05 since the p-value is greater than N=0.05 I fail to reject the ne4 Mayptiesis . There is strong evidence to conclude that there is no significant different bloco the two wrans.

The 95% confidence Interval is 27 = = tfall doephe lowerlinit = 18-43-8 .6)-( = -0.394 upper limit = (9.73 -8.68) + 28.042 (0.6197) list = 0-494 .: The 951. Confidence interval is -0.394 = Mi 42 % 0-4941
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