Question

(6 points) Golf-course designers have become concerned that old courses are becoming obsolete since new technology has given golfers the ability to hit the ball so far. Designers, therefore, have proposed that new golf courses need to be built expecting that the average golfer can hit the ball more than 240 yards on average. Suppose a random sample of 101 golfers be chosen so that their mean driving distance is 239.6 yards. The population standard deviation is 43.5. Use a 5% significance level. Calculate the followings for a hypothesis test where Ho : u = 240 and H : u<240: (a) The test statistic is (b) The P-value is The final conclustion is A. There is sufficient evidence to warrant rejection of the claim that the mean driving distance is equal to 240 B. There is not sufficient evidence to warrant rejection of the claim that the mean driving distance is equal to 240

Answer 1

Ans: Given that Sample mean: X= 239.6 population standard deviation = 43.5 sample size - Job Test of Hypothesis: He: 4-240 Versus Hy: <240 a) Test statistic: X-ll OND z = = 239.6 - 240 43.5 Vior z =-0.0924 The test statistic is 7 = -0.0924

b) Here wie test one sided a-test level of stgnificance = 0.05 P-value = P(ZK0.09) = 0.4633 Since b-value >0.05 then we fail to reject Hoot 0.05 stgnificance level c) The final conclusion is A) There is sufficient evidence to warrant rejection of the claim that the mean digiving distance is equal to 240

Ans! X =2.2 3,-1.8984 ni-10 X2 = 6.6 Sz: 1.1738 Here we assume that population variance are equal 97. I 4. Confidence interval for the difference between mean female and male elink consumption is (x,-*z-tn, 22% Vot han fort att man 2245 hit the Here 1420.977 => 2 20.023 & df = n, th2-2=10710-2-18 where (914)s? +(12-1352 (10-1)1.487(10-1)1.172 nith 2-2 notlo-2 Sa S-1.66 & we calculate tn, h2-2,412 t18,0.023% = 2.81

9774 Confidence interval is (2-2-6-6 – 2.8150 (1.66) Viotto,2.2-6-6+2-81(166) Vio+T6) (9.4-2.8) (1•66)(0-45) 9-4.4+2-81(166) (0-46)) (-4.4 - 2.10 ,-4.4 +2.10) (-6.5, -2.3) 9 |-- Confidence Integral - (6.5,-2-3)