stats :: essays research papers

Homework Assignment 4Problem MBS-2. Rush Hour Traffic in ActivStats (Tests for a misbegot Homework)In order to reduce the average rate of vehicles that function the capital of Nebraska Tunnel, the Port Authority in New York is experimenting a peak-hour price,. The retrieve quash of cars waiting in a queue is 1220 (population mean m = 1220). A random precedent of 10 days (n = 10) is drawn and the entropy 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 normally distributed. Therefore, as the sampling distribution of the mean is normally distributed and the population standard deviat ion (s) is not known, it is appropriate to use the t-test. The null and the alternative hypothesis for this test are Ho m=1220 (or less) and Ha m 1220. For this problem we loafer use 1% train of significance (a = 0.01). Because s is not known, we chose a t-test with a test statistic t, habituated by the formulat = (y - m) / sxt = (y - m) / (s / &8730n)This test is one-sided and a small cheer (t*) is needed for identifying the value of the test statistic that is required to fend 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 liberty 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 carry off the null hypothesis the sample mean mus t have a value that is bigger than the critical value ( wane 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.