Stats 244– Assignment 14- Solutions

1. a)

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b)

n Σx Σy Σx2 Σy

2 Σxy12 2549.7 2173.6 543653.8 397953 462959.7

S xx S yy S xy

1906.26 4242.01 1123.99

b 0.58963a 55.8516

RSS 3579.27

s 2 357.927s 18.919

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Stats 244 – Assignment 14 - Solutions

c) Use the test statistic 0 0 .58961.361

18.9191906.26

b

xx

b bt

ssS

β− −= = = =

and reject H0 if |t| > t0.025 = 2.228 (df = n -2 = 10). We accept H0 , the slope is not significantly different from zero.

d) Prediction Limits for y at x = x 0

x 0 Prediction Lower Upper

220 185.6 141.1 230.0195 170.8 123.8 217.8

2. A.

n Σx Σy Σx2 Σy2 Σxy 20 401.8 91.73 8577.18 448.981 1926.97

Sxx Syy Sxy 505.018 28.2613 84.1113

b 0.16655a 1.24049

RSS 14.2524

s2 0.7918s 0.88983

B.

t0.025 2.101 Confidence Limits for β

Lower Upper 0.0834 0.2497

Confidence Limits for α -0.4823 2.9633

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Stats 244 – Assignment 14 - Solutions

D. Conclusions - Infant mortality rate for cities with 0 pollution index is a = 1.24 - Infant mortality increases at a rate of b = 0.167 units per unit increase in the

pollution index E.

Prediction Limits for y at x = x0

x0 Prediction Lower Upper 10.70 3.023 0.954 5.091

3. a)

n Σx Σy Σx2 Σy2 Σxy 20 2786 546 500088 15810 68719

Sxx Syy Sxy 111998 904.2 -7338.8

b -0.0655 a 36.4278

RSS 423.317

s2 23.5176 s 4.8495

b)

t0.025 2.101Confidence Limits for β

Lower Upper -0.0960 -0.0351

Confidence Limits for α 31.6136 41.242

Stats 244– Assignment 14 - Solutions

c)

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d) 95% Prediction Limits

x0 Prediction Lower Upper 80 31.19 20.59 41.78

100 29.88 19.37 40.38 120 28.56 18.11 39.02 140 27.25 16.81 37.69 160 25.94 15.48 36.40 180 24.63 14.12 35.15 200 23.32 12.72 33.93

4. It is claimed that the average July rainfall in Saskatchewan is 9.8 cm.In the last n = 10 years the amount of rainfall in July was:

7.8 6.4 10.2 11.4 5.4 9.8 6.3 8.1 12.1 14.4 Test so see if the claim is true using a 5% significance level and

i. the t test.Solution:

9.8 9.19 9.89.19, 2.89653, 0.665972.8965310

xx s t sn

− −= = = = = −

Since |t| ≤ t0.025 = 2.262 for n – 1 = 9 d.f., H0 is accepted.

ii. the sign test.Solution: S = {number of observations ≥ 9.8} = 5 If H0 is true then S has a binomial distribution with n = 10 and p = ½.

x 0 1 2 3 4 5 6 7 8 9 10p (x ) 0.001 0.010 0.044 0.117 0.205 0.246 0.205 0.117 0.044 0.010 0.001

If the critical region is {0, 1, 9, 10} then α =0.001+ 0.010 + 0.010 + 0.001 = 0.022. With this critical region H0 is accepted. iii. Wilcoxon’s signed rank test.Solution:

x i x i - 9.8 signed rank 7.8 -2 -56.4 -3.4 -7

10.2 0.4 211.4 1.6 35.4 -4.4 -99.8 0 16.3 -3.5 -88.1 -1.7 -4

12.1 2.3 614.4 4.6 10

Now W+ = 22 and W- = 33. Tables for wilcoxon’s test for n = 10.

The two-tailed critical region for Wilcoxon’s test is reject H0 if W+ (or W-) ≤ W0.0245 = 8. Since neither W+ ≤ 8 nor W- ≤ 8 H0 is accepted.

Estimate the average rainfall in Saskatchewan with a 95% confidence interval i) assuming that rainfall is normally distributed.Solution:

( )0.0252.89653 or 9.19 2.262

10sx t

n⎡ ⎤± ± ⎢ ⎥⎣ ⎦

or 7.12 to 11.26

ii) not assuming that rainfall is normally distributed.Solution:

By examining the table for the binomial distribution with n = 10 and p = ½. x 0 1 2 3 4 5 6 7 8 9 10

p (x ) 0.001 0.010 0.044 0.117 0.205 0.246 0.205 0.117 0.044 0.010 0.001 We see that p(0) + p(1) = 0.001+ 0.010 = 0.011 and p(0) + p(1) + p(2) = 0.001+ 0.010 + 0.044 = 0.55. Thus the first smallest observation to the first largerst obsertvation (5.4 to 14.4) is a (1 – 2(0.011))100 = 97.8% confidence interval for the median Thus the 2nd smallest observation to the 2nd largerst obsertvation (6.3 to 12.1) is a (1 – 2(0.055))100 = 89.0% confidence interval for the median