Question

12 Suppose the mean IQ score of people in a certain country is 105. Suppose the director of a college obtains a simple random sample of 40 students from that country and finds the mean IQ is 105 2 with a standard deviation of 13.7. Test the claim that the mean IQ score of people is more than 105 at the a = 0.05 level of significance. a. Write both null and alternative. Indicate the tail. b. Find test statistics (show formula and final answer with two decimals) c. Find P-value (round to 4 decimal places) d. Write the conclusion in the contest of the problem Find a 95% confidence interval for mean (write formula and final answer): 1. Interpret your intervals.

Answer 1

Aim: To test whether the ICL score of people to more than 105 or not Given : The simple_random_sample = 0 = 40 oample mean = X = 105.2 sample standard deviation Ep = 13.7. da = 0.05 Parameter : Let M be the population meanso M is the mean to score of people in a certain countly let Ho be some specified of value of so Mo 105 Hypothesis ; Hö: The mean Iq ocore of people in a certain country is equal to 105 VIS Hi: The mean 19 score of people in a certain country more than 105 . ia Mathematically Ho: vis Hi: Y > Yo. M = Mo M = 105 Ho: VIS H1: M > 105 Here population standard deviation. io not given , we use t-test for te testing hypothesis the Test statistic: tal = Ň - M S nten-1,2) ce because the testis one sided) vn 00 under Ho, H = 105. Frale 105.2 - 105 13.7 40

tral 0.2 2:1662 tral 0.0923 Decision : Reject Hoit hal z tun-t,®) triticale tiñ-l, d) F(40-1, 0.05 739, 0.05) 1.6849 60 here tal = 0.0923 <ten-1,0) 1.6 849. so we fail to reject HO - . value <a > P = 0·05 - Also we reject to if f do p value = 0.4635 00 we fail to zeject Ho Condusion: The mean IQ ocore of people in a certain country may be equal to 105. There is no oufficient evidence to conclude that the mean TG poore of people is more than 105.

P = The 95% confidence interval can be given as- 1-a tral <tin-1.a 2)] PL X-M port < tin-1,a) SI X-M < tin-12) X S ñ Pr 7h-1, 1) x 3 un Here the alternative hypothesis is right twled so we we get a lower bound in the confidence interval We know o <4 co. oo here lower bound X - 720-1,2) XS 56 - 1] 105.2 - 1.684 9 x 13.7 140 105.2 - 3.6498 = 101.5502 Lower bound so here the 95% confidence pointerval for mean can be given [ 101.5502 < 4 < 00). as-

anower = 105 a) || Hypothesis : Ho: 4 vis Test statistic tral X-M siro HI: M > 105. (Right tail). 0:09 b2 c). p-value O: 4635 d) e2 as - Conclusion : There io no su sufficient evidence to conclude that the mean IQ score of people is more than 105 The 95% confidence interval can be given X ten-1, 2) x S in [ 101: 5502 SM Sol We can be 95.6. confident that the mean IQ score will be greater than 101.5502 Here wsing compidence interval, we fail to reject Ho under Hog H= 105 # this value u lies in the obtained confidence interval We teject Ho if the value of Hundes Ho do not lie in the obtained confidence internal. Ho as