Solutions to Assignment 1

• Problem 11-6
• Problem 11-9
• Problem 11-10
• Problem 11-17
• Problem 11-21
• Problem 11-23

• Problem 11-14
• Problem 11-24

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• 11-6
• a) 12
• b) 5
• c) 12
• d) 9


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• 11-9.

Actual (Expected) Frequencies

	High	Medium	Low	Total
Peak	20 (15)	7 (9)	3 (6)	30
Trough	30 (50)	40 (30)	30 (20)	100
Rising	20 (15)	8 (9)	2 (6)	30
Falling	30 (20)	5 (12)	5 (8)	40
Total	100	60	40	200

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• 11-10
• a) Chip sales are independent of where the economy is with respect to business cycles.
• b) Observed Chi-square = 34.597
• c) at alpha = .10 Chi-square with 6 degrees of freedom = 10.645
Thus since the observed value is bigger, reject the null hypothesis.

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• 11-17

Ho: mu = 14, sigma=5, normal

H₁: Some other distribution.

Following the hint we must determine 5 equ-probable intervals for a normal distribution and see if indeed all the observations fall rather evenly into these intervals.
For any distribution the 20, 40,60,and 80th percentiles divide the distribution into 5 equally probable intervals. The z-values for these percentiles are -0.84, -0.25, 0.25, and 0.84 (from the z-table or e.g. NORMSINV(.20). The corresponding percentiles for a normal distribution with mean = 14 and standard deviation = 5 are resp:

• 14 - 0.84 * 5 = 9.8
• 14- 0.25 * 5 = 12.75
• 14 + 0.25 * 5 = 15.25
• 14 + 0.84 * 5 = 18.20
(These could have been directly obtained from NORMINV(prob, 14,5,1) where prob is 0.2, 0.4, 0.6, and 0.8)
Since there are 40 observations, if the null is true, we would have expected a fifth of all observations (8) to fall in each of the categories:

Mu =14, sigma = 5

Range	fo	fe	stat
9.8 or less	9	8	0.125
9.8-12.75	8	8	0.000
12.75-15.25	6	8	0.500
15.25-18.20	5	8	1.125
18.20 or more	12	8	2.000

Observed Chi-square is 3.750;
Critical Chi-square (5%) with 4 df = 9.488
Note: since we did not calculate any estimates of mu or sigma df = number of categories minus 1 = 4.
We therefore accept the null hypothesis
The p-value for 3.75 = CHIDIST(3.75,4) = 0.44

 Alternatively we could have defined other ranges, say by dividing the entire range of values (roughly from 1 to 21) equally into 5 categories of width 4 resulting in intervals: 1-5; 5-9; 9-13; 13-17; 17-21 and calculated the Chi-square on the basis of the observed and the expected frequencies falling into these intervals.

Alternative: with equally spaced intervals

Note: In calculating the Cum_P use either e.g.,  =NORMDIST(5,14,5,1) which returns the probablity of the random variable to assume a value up to 5.
Or if you want to use the tables; calculate the z -value (e.g. (5-14)/5 = -1.8 )  and look up the probability for z=1.8 which is 0 .4641.  This is the probablity from 0 to 1.8 (the blue shaded area) or, due to the symmetry of normal distribution, to the left,  from 0 to -1.8 . Therefore the probability up to  z = -1.8 is 0.5 - 0.4641= 0.0359.
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• 11-21.

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• 11-23.
 
• a) Since the sample size is large (n=200) the test on the mean of a population does not depend on the normality of the population. Central limit theorem assures that the distribution of the sample means will tend to normality regardless of the shape of the distriubution of the population. Therefore for the test of the mean checking the normality is not critical.
• b)


H_o: Sample comes from a normal distribution (with some mean and some standard distribution)
H₁: From some other distribution

Sales	z(UL)	Cum Prob	Int Probability	Observed(Expected)	Stat
65.5 or less	-1.3889	.0824	.0824	10 (16.49)	2.552
66.5 - 70.5	-0.8333	.2023	.1199	20 (23.98)	0.660
70.5 - 75.5	-0.2778	.3906	.1883	40 (37.65)	0.146
75.5 - 80.5	0.2778	.6094	.2188	50 (43.76)	0.889
80.5 - 85.5	0.8333	.7977	.1883	40 (37.65)	0.146
85.5 or more	infinity	1.000	.2023	40 (40.47)	0.005

Note: Cum probablities can be obtained from the normal table or from e.g.,  =NORMSDIST(-1.3889) or NORMDIST(65.5,78,9,1)
 Interval probabilities are obtained from the  difference of cumulative category probablities
          for instance P(66.5-70.7) = .2023-.0824 = .1199, etc
Observed Chi-square = 4.399
Crit Chi-square (5%, 3 df) CHIINV(.05,3) = 7.815 ( the student estimated mu and sigma from the data hence df = 6 - 2 -1 = 3)
the p-value for 4.399 (CHIDIST(4.399,3) = 22%)
accept the null.

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Additional Problems:
• 11-14.
• a) and b)

 

Mu = 5 ; sigma = 1.5

Range	fo	z	Cum Prob	Int Prob	fe
<2.6	6	-1.6	.0548	.0548	8.22	.5995
2.6 - 3.79	30	-.8	.2119	.1571	23.59	1.7613
3.8 - 4.99	41	.0	.5000	.2881	43.22	.1142
5.0 - 6.19	52	.8	.7881	.2881	43.22	1.7829
6.2 - 7.39	12	1.6	.9452	.1571	23.59	5.5709
7.40 - and up	9	Inf	1.0000	.0548	8.22	.0740
Total	150			1.0000	150	10.0028

• c) Observed Chi-square = 10.0028
• d) Critical Chi-square with 5 degrees of freedom at 10% level of significance = 9.236. Since the computed value exceeds this we reject the hypothesis that the sample comes from a normal distribution with mean 5 and standard deviation of 1.5
NOTE: Since the parameters of the distribution were given as part of the hypothesis we did not need to estimate them from the data and thus lost only one degree of freedom ( with 6 categories we have 5 degrees of freedom).
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• 11-24

Poisson with lambda = 3

x	fo	P(x)	fe	Stat
0	22	.04979	24.89	0.33633393
1	74	.14936	74.68	0.00620268
2	115	.22404	112.02	0.07922641
3	95	.22404	112.02	2.58622415
4	94	.16803	84.02	1.18652484
5	80	.10082	51.41	17.3698376
6 or more	20	.08392	41.96	11.492
	500		500	33.0564

Critical value of Chi-square at .05 with 6 degrees of freedom = 12.592. Since the observed Chi-square exceeds this, the null hypothesis that the sample came from a poisson distribution with mean 3 is rejected.


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