113 Name:
Section:
STAT Midterm 3
1 Otis1 1979 interviewed people waiting to see the space aliens lm Close Encounters of the Third Kind." Each person was asked to state his or her degree of agreement with the statement Life on Earth is being observed by intelligent aliens," on a scale from 1 strongly disagree to 7 strongly agree. Denote with x the i-th response. Assume that x N ; 2 , with known standard deviation is = 1. The purpose of the study was to test Otis' assertion that individuals selected movies that they were predisposed to believe in intelligent aliens. Thus we want to test
i i
H0 : = 4:0 vs. H1 :
4:0
1a 4pts The test adopted can be described as follows: If the sample mean of n = 25 responses is larger than 4.4, reject H0 . Find for the test. Solution: Note that x N 4:0; 2 =5 with 2 =5 = 0:2, and hence P 4:4 = P z 4:4 , 4:0 = P z 2:0 = 0:0228: x
0:2
Note, since
2
=1
is know you can use the z statistic instead of the t statistic.
1b 3pts Find the power 1 , against H1 : = 5: Solution: Power = Prreject false H0 . If = 5 then x N 5; 1=5 and: P 4:4 = P z 4:4 , 5 = P z ,3:0 = 0:9987: x
1=5
1c 3pts If the observed X for n = 25 was in fact 4.5, what is p-value observed signi cance level? Solution: The alternative H1 is one-sided to the right. Thus p = P 4:5 = P z 4:5 , 4 = P z 2:5 = 0:0062: x
1=5
2 10pts Let y1 ; y2; : : : ; y be a random sample of n observations from a normal distribution with known mean and unknown variance 2 . Find the maximum likelihood estimator of 2 . p 2 Solution: Remember pyi = 1= 2 e, 1 yi ,2 = 2 . Hence L = f y1 f y2 : : : f yn = P
n
=
2 ,n=2
,n e, 2 12
n y ,2 i=1 i
2
logL
=
,n=2 log2 , n log , 21
+ 1
3
and hence
^ 2 = 1=n
P
@logL = ,n= @ n y , 2 . i=1 i
Xyi ,
n Xyi , i=1
2
2
=0
1
Otis, L. 1979. Selective exposure to the lm Close Encounters".
Journal of Psychology, 101, 293-295.
3. Here are annual US data on beer production in mio gallons X and the number of married people Y : 1960 1965 1970 1975 1980 1985 X 95 108 135 158 193 200 Y 84.4 92.2 95.0 95.7 104.6 107.5 Sample standard deviations and covariances are computed to be s = 43.4, s = 8.4, and r = 0:96. You might also need S = 9407, S = 354, S = 1761. 3a 2pts Does the high correlation of r = 0:96 mean that drinking leads to marriage, or marriage leads to drinking, or what? explain in no more than 10 words. Solution: no, correllation is not the same as causation 3b 3pts The following gure shows a scatter plot of marriages against beer consumption. In the gure, indicate the distances residuals, deviations whose squares are minimized to obtain the least squares regression line.
x y xx yy xy
105
MARRIED 95 100
85
90
100 120 140 160 BEER PROD. 180 200
3c 3pts If beer consumption is measured in mio liters, i.e. z = 4 x, how does the correlation coe cient change? Solution: Denote with rzy and rxy the correllation coe cient for z; y and x; y, respectively. Then P Pzi , zyi , y 4 xi , xyi , y rzy = p SSzy = p P == pP P P = rxy SSzz SSyy zi , z 2 yi , y2 42 xi , x2 yi , y 2 3d 2pts Using the method of least squares, the estimate slope Solution:
^1 =
1
of the regression line y =
0
+ 1 x.
SSxy SSxx
= 0:19:
Name:
Section:
STAT 113 Midterm 2 Problem 4
part of the exam on Friday. No group work on this problem. 4 Prostate cancer is one of the most virulent forms of cancer. Generally, it has spread before being detected and is usually fatal. One dietary factor that has been studied for its relationship with prostate cancer is fat consumption. The following table lists fat consumption in g day, rst column and prostate cancer death rates per 100,000, second column for several countries.
El Salvador Philippines Japan Mexico Greece Colombia Bulgaria Yugoslavia Poland Panama Isreal Romania Venezuela Czechoslovakia Italy 38 29 42 57 96 47 67 72 93 58 95 67 62 96 86 0.9 1.3 1.6 4.5 4.8 5.4 5.5 5.6 6.4 7.8 8.4 8.8 9.0 9.1 9.4 Spain Portugal Finland Hungary UK Germany Canada Austira France Netherlands Australia Denmark US Norway Sweden 97 73 112 100 143 134 142 119 137 152 129 156 147 133 132 10.1 11.4 11.1 13.1 12.4 12.9 13.4 13.9 14.4 14.4 15.1 15.9 16.3 16.8 18.4
Take home problem. Work on this problem at home and hand in your solution together with the in-class
The data are available on the STA 113 homepage click exams". Or copy the data set by typing from your acpub account:
cp afs acpub project sta215 fat.data fat.data
4a 2pts Construct a scattergram for the data attach your plot on a seperate sheet. Solution:
15
40 60 80 100 120 FAT CONS.
5
PROST. CANC. 10
140
160
4b 2pts Assuming the relationship between the variabels is best described by a straight line, use the method of least squares to estimate the intercept and the slope of the line. Solution:
The regression equation is C2 = - 1.06 + 0.113 C1 Predictor Constant C1 s = 2.295 Coef -1.063 0.11336 Stdev 1.170 0.01126 t-ratio -0.91 10.06 p 0.371 0.000
R-sq = 78.3
R-sqadj ...