Chapter 7
7.
108
754
646
999
861
933
290
201
602
641
292
531
10.
a) Infinite
b) Infinite
c) Finite
d) Infinite
e) Infinite
15.
a)
4376 45500/10=4550 is the point estimate mean
4798
5578
6446
2717
4119
4920
4237
4495
3814
b)
4376 4550 -174
4798 4550 248
5578 4550 1028
6446 4550 1896
2717 4550 -1833
4119 4550 -431
4920 4550 370
4237 4550 -313
4495 4550 -55
3814 4550 -736 9068620/10=906862
√906862=952.29
43.
a) Mean=6883 Standard deviation=2000
2000/√50=282.84
b) 6883-300=6538 6883+300=7183
6538-6883=-345 7183-6883=300
-345/(2000/√50)=-1.22 300/(2000/√50)=1.06
-1.22= 0.1112 1.06=0.8554
.8554-.1112=.7442Probability
c)
2000/√50=282.84
7500-6883=617/282.84=2.18=.9854
1-.9854=.0146 Probability
45.
a) 14,16
14-15=-1/(4/√60)=-1.94 16-15=1/(4/√60)=1.94
-1.94= 0.9738 1.94=0.0262
0.9738 -0.0262=0.9476Probability
b)
45/60=.75
-.75+15=14.25 .75+15=15.75
14.25-15=-.75/(4/√60)=-1.45 15.75-15=.75/(4/√60)=1.45
-1.45=0.9265 1.45=0.0735
0.9265 -0.0735= 0.853Probability
52.
a)
.40/√380=.0205
√.40(1-.40)/380=.0251
.40+.04=.44 .40-.04=.36
.44-.36=.08/.0251=3.19
b)

Chapter 8

3.
a)
n=60 standard deviation=15 mean=80 CI=.95
1-.95=.05/2=.025
80+1.96(15/√60)=83.8 80-1.96(15/√60)=76.2
b)
1-.95=.05/2=.025
80+1.96(15/√120)=82.68 80-1.96(15/√120)=77.32
c) The effect of a larger sample size on an interval estimate caused the minimum population to go up and the max population went down.