Do the following using Mintab (1) Make a normal probability plot for each of the two curing times.

Transcription

1 SMAM 314 Computer Assignment 4 1. An experiment was performed to determine the effect of curing time on the comprehensive strength of concrete blocks. Two independent random samples of 14 blocks were prepared the same way except for curing time. The blocks in one sample were cured for 2 days, while the blocks in the other were cured for 6 days. The compressive strengths of the blocks in MPa are given below. Row 2 days 6 days Do the following using Mintab (1) Make a normal probability plot for each of the two curing times. (2) Perform a test of hypothesis for equality of the variances of the compressive strengths for the two curing times. MTB > VarTest c1 c2; SUBC> Unstacked.

2 Test for Equal Variances: 2 days, 6 days 95% Bonferroni confidence intervals for standard deviations N Lower StDev Upper 2 days days F-Test (Normal Distribution) Test statistic = 0.67, p-value = Perform an apprpopriate test of hypothesis to determine whether the blocks cured for six days have a higher average compressive strength than those cured for two days. Test for Equal Variances for 2 days, 6 days MTB > TwoSample c1 c2; SUBC> Pooled; SUBC> Alternative -1. Two-Sample T-Test and CI: 2 days, 6 days

3 Two-sample T for 2 days vs 6 days N Mean StDev SE Mean 2 days days Difference = mu (2 days) - mu (6 days) Estimate for difference: % upper bound for difference: T-Test of difference = 0 (vs <): T-Value = P-Value = DF = 26 Both use Pooled StDev = MTB > Answer the following questions of a sheet stapled to the computer output. A. Based on the normal probability plot and the p values given there is it reasonable to assume that the compressive strengths for each curing time are normally distributed?explain. The points lie reasonably close to a straight line. The p values are greater than.05. The data is probably normally distributed. B. Based on the test of hypothesis for equality of the variances does it appear reasonable to use the pooled two sample t test? Explain. The p value of the F statistic is.480. Thus you would fail to reject the hypothesis of equality of the variances so the pooled test would be appropriate. C. For the test performed in (3) above (1) What is the null and the alternative hypothesis? H 0 µ 1 = µ 2 H 1 µ 1 < µ 2 (2) What is the region of rejection to perform this test at α=.05? T < (3) What is the value of the test statistic? T= 5.87 (4) What is the p value? pvalue =0 (5) Would you reject H 0 at α=.05? Explain. The null hypothesis would be rejected at α=.05. (6) Would you recommend to your boss that there is a definite advantage to a longer curing time? Explain. Yes I would. 2. Two rubber specimens from each of eight stocks were subject to an abrasion test. One specimen from each stock was cured at 90 degrees C and the other was cured at 140 degrees C. The weight loss for each specimen in each stock was measured. The results are presented in the table below.

4 Use Minitab to Stock 90 degrees C 140 degrees C (1) Make a normal probability plot of the difference. (2) Perform an appropriate test of hypothesis to determine whether there is a statistically significant difference in the weight losses at the two temperatures. (Observe that the specimens cured at the different temperatures are paired within the stocks.) Paired T-Test and CI: 90, 140 Paired T for N Mean StDev SE Mean Difference

5 95% CI for mean difference: (-0.229, 0.581) T-Test of mean difference = 0 (vs not = 0): T-Value = 1.03 P-Value = MTB > Answer the following questions on a page stapled to the computer output. A. Based on the normal probability plot does it appear that the differences are normally distributed? Yes it does the points lie reasonably close to a straight line. B. For the test of hypothesis performed in (2) (1) What is the null and the alternative hypothesis? H 0 µ d = 0 H 1 µ d 0 (2) What are the assumptions necessary to perform this test? Paired data Differences are normally distributed. (3) What is the region of rejection to perform this test at α=.10? Τ>1.895 Τ< (4) What is the value of the test statistic in the Minitab output? T=1.03 (5) What is the p value? Pvalue =.338 (6) Would you reject H 0 at α=.10? Explain. Fail to reject H 0 at α=.10 because T is not in the rejection region and p value >.10 (7) What conclusion do you come to about the difference in weight losses at the two temperatures? There does not appear to be a significant difference in weight losses at the two temperatures.