Question

Please do 4 and 5

MAT2572 Probability and Mathematical Statistics Final Exam Name o Show your work. Your answers must be neatly written and logically organized to receive full credit. 1. (10) The tributed with a standard deviation of 400 hours. Suppose that a randotn sample resulted in an average lifetime of 9000 hours. Obtain a 95 percent of the mean lifetime of such a tube. e life of a particular brand of television picture tube is known to be normally die 2. (10) T o estimate p, the proportion of all newborn babies who are male, the gender of 10,000 newborn babies was noted. If 5106 were male, determine a 90 percent confidence interva estimate of p. e mean response time of a species of pigs to a stimulus is 0.8 seconds. 28 pigs were given 2 ounces of alcohol and then tested. If their average response time was 1.0 seconds with a standard deviation of 0.3 seconds, can we conclude that alcohol affects the man response time? Use the 5 percent level of significance. 4. (15) A random sample of 100 student absences yielded the following data on the days of the week of which the absences occurred: Monday Tuesday Wednesday Thursday Friday 13 Day Observed Frequency 26 19 27 Test the hypothesis that an absence is equally likely to occur on any of the five days. Wh is your conclusions? Use the 5 percent level of significance. 5. (15') The side effects of a new drug are being tested against a placebo. A simple ran sample of 500 patients yields the results below. At a significance level of a 0.05, is enough evidence to conclude that the treatment is independent of the side effect of na Result Drug Placebo Nausea 55 45 No nausea 190 210 6. (10) Each ball baring produced is independently defective with probability 0.05 of 3 is inspected, find the probability that (a) None are defective. 7. (15) A continuous random variable Y has a probability density function 0 <y<1 b) Two or more are defective. (a) Find the value of c.

Answer 1

4 The null and alternative hypotheses are Ho: Absence is equally likely to occur on any of the five days H1: Absence is not equally likely to occur on any of the five days The calculation of expected frequency is Ected 27+19+13+15+26 20 The calculation of the test statistic is (Observed - Expected) Expected (27-20) (19-20) (13-20)2 (15-20) (26-20) 20 20 20 20 20 The degrees of freedom are 5 1-4 The calculation of p-value in Excel is fCHISQ.DIST.RT(8,4) 0.091578 The p-value is 0.0916 Since the p-value is greater than the level of significance, do not reject the null hypothesis and conclude that absence is equally likely to occur on any of the five days.

The null and alternative hypotheses are Ho: Treatment is independent of the si de effect H1: Treatment is not independent of the side effect The formulas and calculation in Excel are 1 Observed Placebo 3 Result Total 4 Nausea SUM(B4:CA) =SUM(B5:C5) -SUM(D4:D5) 45 5 No nausea 190 -SUM(B4:BS) SUM(C4:C5) 8 Expected Placebo 10 Result 11 Nausea 12 No nausea 13 Observed-Expected)A2/Expected 14 ((C4-C11) 2)/C11 ((C5-C12)A2)/C12 15 =(( B4-B11)^2)/B11 =( (B5-B12)^2)/B12 17 Chi Square Test Statistic SUM(B15:C16) 18 Degrees of freedom(2-1)*(2-1) 19 P-value CHISQ.DIST.RT(B17,1)

The output will be 1 Observed Placebo Total 3 Result Drug 45 4 Nausea 100 5 No nausea 190 210 245 255 7 8 Expected Placebo 10 Result Drug 11 Nausea 49.0 51.0 12 No nausea 196.0 204.0 13 (Observed Expected) 2/Expected 14 15 0.735 0.706 16 0.184 0.176 17 Chi Square Test Statistic 1.80 18 Degrees of freedom 19 P-value 0.180 Since the p-value is greater than the level of significance, do not reject the null hypothesis and conclude that treatment is independent of the side effect.