Question

[40] QUESTION 4 Assume that X is normally distributed and that a random sample of size n = 20 from this distribution yielded the following statistics: X = 31 and S=7. (a) Is there strong evidence to conclude that Ho : q2 = 30 against H:02 #30 at the a = 0.05 level of significance? (b) Construct a 90% confidence interval for the population standard deviation. [15] [100]

Answer 1

Given that distributed. X is normally Sample s → » - Sample Size (n)= 20 Sample mean (*) = 31 Sample S.D (8) = 7 Null hypothesis (Ho) o? = 30 Alternative hypothesis (Hi) * 30 Level of significance d = 0.05 Rejection Region : Based on the information the significance leve a=0.05, and the rejection region for this two-tailed test is REX : X < 8.907 or X> 39.862} Test statistic : To test the above hy bothe sig, the chi-square test for one population variance will be used x (0-1) S'

X . (n-1) s? - under Ho 2 x = (20-1) 49 30 19x49 = 30 x= 31.033 .. The X statistic is X=31.033 . Decision about the null hypothesis Since it is observed that X = 8.907 < X=31.033 < X 332.882 it is then concluded that the null hypothesis is not rejected. - Conclusion . It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population variance a2 is different than 30, at 0.05 significan of levd. 2

(6) Construct a 90% confidence interval for the population standard deviation We know the loo (1.2)% confidence interval bor the population Standard deviation in 1-4, -1 C.Lo ( ,519 - (17 (9 ,741 (5.5525, 9.5922) It means is it we repeated experiments the proportion of such intervals that 1 Population Contain the standard devalton (@) would be 907. .. The 90% confidence interval o for o is G.L = (5.5575 , 9.5929) 2