Question

2.10 It is often assumed that the L.Q. scores of human beings are Normally distributed. On the basis of the following data, test this claim at level of significance α 0.05 by using the appropriate χ 2 goodness-of-fit test. Specifically, if X is the r.v. denoting the I.Q. score of an individual chosen at random, then: (i) Set P1 = P(X-90), P2 = P(90 < X 100), p = P(100 < X 110), Pi = P(110 < X < 120), p5 = P(120 < X < 130), Ps = P(X 130). (ii) Calculate the probabilities pi,i-1,..., 6 under the assumption that X ~ N(100, 152) and call tem po, i = 1, . .., 6. Then set up the hypothesis Ho: Pi = Pio, 2 = 1, . . . , б. (ii) Use the appropriate χ2 goodness-of-fit test to test the hypothesis at level of significance α-0.05. The available data are given below, where r denotes the observed number of individuals lying in a given interval. x:90 | 90<xs100 | 100 x 110 | 110cx 120 | 120cxs130 | χ> 130 10 18 23 18

Answer 1

lower	upper	p	Oi	EI	(Oi- Ei)^2/Ei
0	90	0.252493	10	25.24925	9.209767
90	100	0.247507	18	24.75075	1.841261
100	110	0.247507	23	24.75075	0.123839
110	120	0.161281	22	16.12813	2.137807
120	130	0.068461	18	6.846109	18.17226
130	10000	0.02275	9	2.275013	19.8792
		1	100	100	51.36414

Formula in Excel

lower	upper	p	Oi	EI	(Oi- Ei)^2/Ei
0	90	=NORMDIST(B2,100,15,1)-NORMDIST(A2,100,15,1)	10	=100*C2	=(D2-E2)^2/E2
=B2	=10+B2	=NORMDIST(B3,100,15,1)-NORMDIST(A3,100,15,1)	18	=100*C3	=(D3-E3)^2/E3
=B3	=10+B3	=NORMDIST(B4,100,15,1)-NORMDIST(A4,100,15,1)	23	=100*C4	=(D4-E4)^2/E4
=B4	=10+B4	=NORMDIST(B5,100,15,1)-NORMDIST(A5,100,15,1)	22	=100*C5	=(D5-E5)^2/E5
=B5	=10+B5	=NORMDIST(B6,100,15,1)-NORMDIST(A6,100,15,1)	18	=100*C6	=(D6-E6)^2/E6
=B6	10000	=NORMDIST(B7,100,15,1)-NORMDIST(A7,100,15,1)	9	=100*C7	=(D7-E7)^2/E7
		=SUM(C2:C7)	=SUM(D2:D7)	=SUM(E2:E7)	=SUM(F2:F7)

df = 5

critical value = =CHISQ.INV.RT(0.05,5)

11.0705

since TS = 51.364 > critical value (11.07)

we reject the null hypothesis

we conclude that there is evidence that data does not follow N(100,15^2)