1 Solutions to review problems #1
Here are solutions to the …rst set of review problems:
4.77: The sampling distribution of yis normal with mean y= 60 and standard deviation y==pn=
5=p16 = 1:25:Since the distribution of yis normal, approximately 95% of the values of yshould fall within
y1:96 ywhich is 60 1.96(1.25), giving the interval (57.55, 62.45).
5.41 a) A 99% con…dence interval for is given by yt:005;14 s=pn= 31.47 2:977 (5:04)=p15 = 31.47
3:87, which gives the interval (27.6, 35.34).
b) The question is a bit unclear on exactly what a null hypothesis might be, but if we choose H0:= 35,
then at the =:01 signi…cance level we will fail to reject H0, since the 99% con…dence interval includes
= 35:
11.65 a) Yes, the plotted points seem to follow a line.
b) From the printout, byi= 12:51 + 35:83 xi.
11.66 a) b2
"=1
n2P(yibyi)2=MS E = 1:069:
b) From the printout, s:e:(b
1) = 6:96:
c) For this research hypothesis, H0:1= 0 and Ha:1>0since they are interested in detecting a
positive relationship. The pvalue in the printout for H0:1= 0 is p=:0004, but the printout is for
the two-sided alternative hypothesis Ha:16= 0:Thus, to get the pvalue for our one-sided alternative
hypothesis we divide the printed pvalue by 2, yielding p=:0004=2 = :0002:
11.30 a) The plot looks good, there could be one or more in‡uential points.
b) The estimated regression equation is byi= 99:78 + 51:92 xi, and the residual standard deviation is
b"=pMS E =p148:999 = 12:21:
c) A 95% con…dence interval for 1is given by b
1t:025;28 s:e:(b
1);yielding 51.92 2:048 (:586) or 51.9
1:2, giving an interval of (50.7, 53.1).
11.31 a and b) From the printout, t= 88:53;with a pvalue of p < :0001:
11.32 a) F= 7837:26;and p < :0001:
b) They are equal, because both tests are testing the same null hypothesis when we do simple linear
regression. The test statistics are related by t2=F:
1
