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Homework Assignment 4Problem MBS-2. Rush Hour Traffic in ActivStats (Tests for a Mean Homework)In parliamentary procedure to reduce the average tot up of vehicles that use the Lincoln Tunnel, the Port Authority in New York is experimenting a peak-hour pricing,. The mean outlet of cars waiting in a queue is 1220 (population mean m = 1220). A random sample of 10 days (n = 10) is drawn and the data representing the number of cars is analyzed. The data Desk tool in ActivStats reports that the sample mean is 989.8 cars (sample mean y = 989.8) and the sample standard deviation is 160.676 cars (s = 160.676). In order to determine whether the peak-hour pricing has contributed for reducing the number of cars using the tunnel, we have to assume that the numerical data are independently drawn and represent a random sample from a population that is commonly distributed. Therefore, as the sampling distribution of the mean is normally distributed and the population standard deviation (s) is not known, it is appropriate to use the t- try out. The null and the alternative hypothesis for this test are Ho m=1220 (or less) and Ha m 1220. For this problem we can use 1% level of significance (a = 0.01). Because s is not known, we chose a t-test with a test statistic t, given by the formulat = (y - m) / sxt = (y - m) / (s / &8730n)This test is one-sided and a critical value (t*) is needed for identifying the value of the test statistic that is ask to reject the null hypothesis. t* (df = n-1, a)t* (df = 10-1, a = 0.01)t* (df = 9, a = 0.01)The critical value t*, which I obtained from the t-distribution table in Kazmier, and corresponds to 9 degrees of exemption and 0.01 level of significance is t* = 2.821 On the basis of the collected data, we can compute the t-test statistict = (y - m) / (s / &8730n)t = (989.8 1220) / (160.676/&873010) (after replacing for y, m, s and n) t = - 4.5306Therefore, in order to reject the null hypothesis the sample mean must have a value that is bigger than the critical value ( reject Ho if t t*, otherwise do not reject Ho). Because t = -4,5306 falls within the non-rejection region (Fig. 1) below the critical value t* = 2.