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=65386883+300=7183
6538-6883=-3457183-6883=300
-345/(2000/√50)=-1.22 300/(2000/√50)=1.06
-1.22=0.11121.06=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=0.97381.94=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=0.92651.45=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)=82.6880-1.96(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.

16.
n=100 standard deviation=8.5 mean=49 CI=.95
1-.95=.05/2=.025
8.5/√100=.85 standard error
100-1=99 degrees of freedom
t-score= 1.984
a)1.984*.85=1.68
b)49-1.68=47.3249+1.68=50.68
c) The pilot for continental airlines flies no less than 47 hours ,which is higher than the pilots from united airlines who average just 36 hours. One reason that the labor costs are high could be due to the number of pilots they have on payroll. If pilots average 36...

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