Question

The quality control team of a chemical factory is comparing two types of alloys (Alloys A & B) for their sulphur content. According to industry standards, it is suggested that the mean sulphur content of Alloy A should be 2mg/cm' less than the mean sulphur content of Alloy B. The Team obtained the following random samples of sulphur content measurements for Alloys A and B (in mg/cm). Alloy 1 Sample Data from HW Dataset.xls: column A7-A Alloy 2 Sample Data from HWDataset.xls: column A7-B a) Test whether Alloy A contains on average 2mg/cm' less sulphur than Alloy B. State alle the steps of the testing procedure and your conclusion and any necessary assumptions. Use the significance level of a-5%. hi Construct a 95% two-sided confidence interval for the mean difference of the sulphur content values of Alloy A and Alloy B. Is the confidence interval compatible with the conclusion of the test in (a) above? Explain your answer.

Answer 1

Answer a) Date: -11/7/2019 Two sample t test is used to compare the means. The test is performed in following steps, Step 1: The Null and Alternative Hypotheses 73 I — &nf : Он На : Н2 — Н1 >2 Step 2: The t statistic is used to compare the two population means and the significance level for the test is a=0.05 The decision rules state the conditions that if, P-value a0.05, reject the null hypothesis, Step 3: The tstatistic is obtained using the formula, (Х, — X) — (12 — m) (n1-1)8)+(n2-1) s3) 1 + n1tn2-2 From the samples data value,

A7-A A7-B 14.76 16.1 14.21 18.06 14.39 14.87 13.17 16.14 15.25 17.29 14.12 14.78 14.05 16.78 14,45 16.53 14.23 16.13 14,83 15.7 13.14 15.65 6 16,94 14.64 15,48 15,48 17.55 15.45 15.52 14.92 16,21

16.91 14,2 17.05 17.53 Mean 14.3594 16.2710 Std Dev 0.8167 1.0181 16 20 16.2710 14.3594 - 2 (16-1)0.81672)+(20-1)1.01812) 16+20-2 16 20 (뚜 + 뿌) ( t= -0.2819 The P-value for the tstatistic is obtained from t distribution table for degree of freedom of (n1 n2 -2) 16+20-2=34 P-value 0.3898

The corresponding P-value is 0.3898 which is greater than 0.05 at the 5% significance level for the one sided alternative hypothesis. Hence, it can be concluded that the null hypothesis is not rejected. b) The confidence interval for difference in two means is obtained using the formula, 95% Confidence Interval = (X2 - X\) ttx SEx,-X Where, Х, — X — 16.271 — 14.359 1.9116 _ (п1 — 1)s7 + (пg - 1)3 1 1 SEX2-X 2 п1 + п2 n2 SEX-X, = 0.3135 2.032 for 95% significance level, degree of freedom 34 t _ 95% Confidence Interval =1.9116± 2.032 x 0.3135 95% Confidence Interval = [1.275,2.549