Question

Exhibit 9-3 n = 49 Ho: u = 50 X54.8 Hu#50 o= 14 Refer to Exhibit 9-3. Part 1: If the test is done at a 5% level of significance, the null hypothesis should not be rejected а. b. be rejected C. Not enough information is given to answer this question. d. None of the answers is correct. Part 2:The p-value is equal to a. .0164 b. .3849 . .2698 d. 2302 Part 3: The test statistic equals

Answer 1

Let u denotes the population mean. Hypotheses : To test null hypothesis Ho :je =50 against alternative hypothesis H1 :+ 50 Let denotes the sample mean, o denotes the population standard deviation and n denotes the sample size. Here, i = 54.8000,02 = 196.0000, 0 = 14.00, n = 49 The test statistic can be written as: Vn(– 50) which under H, follows a standard normal distribution. Decision rule / Rejection region : We reject H, at 0.05 level of significance if P-value < 0.05 or if |zobs > 20.025 = 1.959964 0.4 - 0.3 density 0.2 0.1 0.0 -2 0 Z Now, Value of test statistic: The value of the test statistic is zobs = 49 (54.8000 – 50) 14.00 = 2.400000 ~ 2.40 The critical value =+ 1.959964 +1.960 P-value = P(Z > Zobs) = 2* P(Z > 2.400000) = 2* (1 – 0.991802) = 0.016395 – 0.0164 Conclusion : Since p-value < 0.05 and Zobs > Zcritical = 1.96, so we reject H, at 0.05 level of significance Hence, we can conclude that the the population mean is not equal to 50.

1. ans-> b) be rejected

2. ans-> a) 0.0164

3. ans-> 2.40