Question

The owner of a large car dealership believes that the financial crisis decreased the number of customers visiting her dealership. The dealership has historically had 800 customers per day. The owner takes a sample of 100 days and finds the average number of customers visiting the dealership per day was 750. Assume that the population standard deviation is 350. The value of the test statistic is Multiple Choice z = 1.429 o t99 = 1.429 t99=1429 t99 = -1.429 z=-1.429

Answer 1

Let u denotes the population mean. To test null hypothesis H. :f =800 against alternative hypothesis H1 :M + 800 Letī denotes the sample mean, o denotes the population standard deviation and n denotes the sample size. Here, w = 750,02 = 122500,0 = 350, n = 100 The test statistic can be written as: | Vn (3 – 800.0) = which under Ho follows a standard normal distribution. We reject Ho at 0.05 level of significance if P-value < 0.05 or if |zobs > 20.025 Now, The value of the test statistic = -1.428571 The critical value == 1.959964 P– value = P(Z > Zobs) = 2* P(Z < -1.428571) = 2 * 0.076564 = 0.153127 Since p-value > 0.05 and (ZobsZcritical = 1.959964, so We fail to reject Ho at 0.05 level of significance Hence, we can conclude that the population mean is not different from 800.

ans-> iv) z = -1.429