Question

17. A random sample of n, 133 individuals results in xy +39 successes. An independent sample of ny - 147 individuals results in X - 58 successes. Does this represent sufficient evidence to conclude that P1 P2 at the a=0.01 level of significance? Click here to view the standard normal distribution table (page 1). Click here to view the standard normal distribution table (page 2). Click here to view the table of critical t-values. Click here to view the chi-square critical values table. The given situation is about a (1) Write the hypotheses for the test. Ho (2) H (3) Calculate the test statistic. (4) (Round to two decimal places as needed.) Identify the critical region. Select the correct choice below and fill in all answer boxes within your choice. (Type an integer or decimal rounded to two decimal places as needed.) A. test statistic OB test statistic > OC. test statistic or test statistic > What is the conclusion? (5) the null hypothesis and conclude there (6) sufficient evidence that the parameter of interest of population 1 is (7) the parameter of interest of population 2 at the a= 0.01 level of significance 3: Standard Normal Distribution Table (page

Answer 1

5. Fail to reject, is not, less than

Let p1 denotes the true proportion of first population and p2 denotes the true proportion of second population Hypotheses : To test null hypothesis H. : P1 – P2 = 0 against alternative hypothesis H1 :P1 - P2 < 0 Let P1, P2 denote the respective sample proportions and n1, n2 denotes the sample sizes respectively. Here, P1 39 133 0.293, n1 = 133, P2 58 147 0.395, n2 = 147 The test statistic can be written as: (p1 - 2) More P.*(1-) Py*(1-P) + ni n2 which under Ho follows a standard normal distribution. Decision rule / Rejection region : We reject H, at 0.01 level of significance if P-value < 0.01 or if Zobs <-20.01 = -2.326348 0.4 0.3 density 0.2 0.1 0.0 Z Now, The value of the test statistic Zobs -1.795878 -1.796 (0.293 – 0.395) 0.2932*(1-0.2932) 0.3946*(1-0.3946) 133 147 The critical value Zcritical = -2.326348 -2.326 P-value = P(Z < Zobs) P(Z < -1.795878) = 0.036257 0.0363 Conclusion : Since p-value > 0.01 and Zobs Zcritical = -2.326348, so We fail to reject H, at 0.01 level of significance Hence, we can conclude that the the first population proportion is not significantly lesser than second population proportion.