Question

chapter 28 3

David thinks the IQ of husbands and wives differ. At the 5 % level of significance, test his hypothesa. Husband's IQ Wife's IQ 3. 102 110 115 110 120 95 100 100 115 108 105 100 125 100 I5 13 -15 -25 2 15 L

Answer 1

Let us denote the difference

d = Husband's IQ - wife's IQ

Let ud denotes the population mean of difference. To test null hypothesis H. :Md =0.0 against alternative hypothesis H1 :Md + 0.0 Let d denotes the sample mean of difference , Sd denotes the sample standard deviation of difference and n denotes the sample size. Here, d = -0.1429, sa = 228.810, sa = 15.126, n = 7 The test statistic can be written as: 1.Vn(d – 0.0) sd which under Ho follows a t distribution with n - 1 df. We reject H, at 0.05 level of significance if P-value < 0.05 or if tobs > to.025.n-1 Now, The value of the test statistic = -0.024987 -0.02 associated degrees of freedom = 6 The critical value =+ 2.447 P – value = P(|tn–1] > tobs ) = 2 * P(to < -0.024987) = 2 * 0.490438 = 0.980876 Since p-value > 0.05 and tobs * tcritical = 2.446912, so We fail to reject Ho at 0.05 level of significance Hence, we can conclude that there is no significant difference between two population means.