Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

lz https%2A%2F%2FconnectmedK Findehtm232 ×To Module 9 Slides Business Sta... . M

ID: 3362370 • Letter: L

Question

lz https%2A%2F%2FconnectmedK Findehtm232 ×To Module 9 Slides Business Sta... . Module 7 Slides Business Tools Help Suggested Sites . Web Sice Gallery . Saved Two-Sample Procedures: Problem 9-98 (algo) atn tere rmachmey sendd oned to prodisce ta oome rodasreld eateent pron dtins of dsiecavre p parts were found to be defective. Similarly. 385 parts were randomly selected from Machine B, and 36 of these parts were found to be defective. T significance level using Mega Stat (or Excet a. Select the correct symbol to replace in the null hypothesis Hpa-pe ? 0 O> O% O 2 b. Select the correct symbol to replace .?-in the alternative hypothesis Hi 4-' o 2 02 O

Explanation / Answer

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

a) Null hypothesis: P1 = P2
b) Alternative hypothesis: P1 P2

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the proportion from population 1 is too big or if it is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.01. The test method is a two-proportion z-test.

Analyze sample data. Using sample data, we calculate the pooled sample proportion (p) and the standard error (SE). Using those measures, we compute the z-score test statistic (z).

p = (p1 * n1 + p2 * n2) / (n1 + n2)

p = 0.06775

SE = sqrt{ p * ( 1 - p ) * [ (1/n1) + (1/n2) ] }
SE = 0.0195

c)

z = (p1 - p2) / SE

z = - 3.05

where p1 is the sample proportion in sample 1, where p2 is the sample proportion in sample 2, n1 is the size of sample 1, and n2 is the size of sample 2.

Since we have a two-tailed test, the P-value is the probability that the z-score is less than - 3.05 or greater than 3.05.

d) Thus, the P-value = 0.0011

Interpret results. Since the P-value (0.0011) is less than the significance level (0.01), we cannot accept the null hypothesis.