“An Aggie does not lie, cheat, or steal or tolerate those who do.”
Answer Key Homework 9 (Due date: April 20, 2005, Wednesday in class)

1.
If the  is not given numerically in any question use 0.05.
Exercise 7.25 (Hint: Notice that you are computing conservative sample size in part (b). Check the
large sample conditions to use the large sample formula. If they are not satisfied, just tell me that you
cannot compute them because conditions are not satisfied. Do not use the formula 7.10 in the textbook
to answer.)
Answer part (a) and as a part (b) calculate the 99% confidence interval for the true proportion of the
electrote aware of her position on using the state funds to pay for abortions when the sample
proportion is 0.6 for 100 residents.
(a) 1- =0.95   =0.05. /2=0.025  z / 2 =1.96 and w0.1 then
^
^


4 z2 / 2 p1  p 
^
4(1.96) 2 (0.5)(1  0.5)


n 
=
=
384.16

385
.
p
 0.5 is used because it was
w2
(0.1) 2
asking the conservative sample size.
^
(b)
p =0.6, n=100
^
Since
^
n p  60  10 and n(1  p)  40  10 ,
^
^
^
^ 
^

p(1  p) ^
p(1  p) 
99% C.I for :  p  z / 2
, p  z / 2

n
n





0.6(0.40)
0.6(0.40) 
 = (0.4739,0.7262)
  0.6  2.575
,0.6  2.575

100
100


2.
Exercise 7.37 (Hint: sample mean and standard deviation are given in the output for you to use it. Do
not need to interpret. Only calculate)
(a) 95% confidence interval for  is
_
s _
s  
0.0809 
 x  t / 2;n 1
, x  t / 2;n 1
   0.9255  2.093
 =(0.8876,0.9634) where
n
n 
20 

t / 2;n 1  t 0.025;19 =2.093
(b) 95% prediction interval for a single individual randomly selected student from this population is
_
1 _
1 
1 
 x  t / 2;n 1 s 1  , x  t / 2;n 1 s 1     0.9255  2.093(0.0809) 1 
 =(0.7520,1.0

 

n
n
20

 

990)
(c) Tolerance interval which includes at least 99% of the cadences in the population distribution using
_

_


a confidence level of 95% is  x  critical ( s ), x  critical ( s )   0.9255  3.615(0.0809) 
=(0.6331,1.2180)
3.
Exercise 7.44 (Exercise 7.42, 5 th edition) (Hint: you do not need to derive, really need to calculate it
because formulas are given)
n=9, s=2.81 then 95% confidence interval for 2 is
“An Aggie does not lie, cheat, or steal or tolerate those who do.”
 (n  1)  s 2 (n  1)  s 2

, 2
 2
1 / 2;n 1

/
2
;
n

1

  8(2.81) 2 8(2.81) 2

,
  17.534
2.18



  (3.6027,28.9765)

95% confidence interval for  is (1.8981, 5.3830)
4.
Exercise 7.47 (Exercise 7.45, 5th edition) (Hint: in part (a) it is simply asking the confidence interval
and use the significance level 0.05. in part (b), do not use formula 7.10 again. Check the large sample
conditions and use the large sample confidence interval)
(a) 95% confidence interval for  is
_
x  z / 2
s
n

(4247.08  (387.8) 2 / 48) / 47
387.8
=(6.7019 , 9.4565)
 1.96
48
48
(b) There are 13 values exceeding 10 out of 48 then 13/48=0.2708 is the point estimate for the
proportion of all such bonds whose strength would exceed 10.
Since n
^
^


p =13 ≥ 10 and n 1  p  =35 ≥10, we have a large sample


^
^
p(1  p)
0.2708(0.7292)
 0.2708  1.96
95% confidence interval for p is p  1.96
n
48
^
=(0.1451 , 0.3965)
5.
Exercise 8.35 (Hint: Make sure to write the hypothesis, calculate the test statistics and write the
decision and conclusion)
n=200, x: number passed=124, =0.05 then
^
p  124 / 200  0.62
H 0 : p  0.7 versus H a : p  0.7 (differs from 0.7)
Is the large sample conditions satisfied? Since np0=140 10 and n(1-p0)=6010, YES
^
Test statistics:
z
p  p0
p

0.62  0.70
0.7(0.30) / 200
 2.47
Then P-value=2P(z<-2.47)=2(0.0068)=0.0136.
P-value  =0.05 then reject H0.
Yes. The true proportion for this county during the current year differs from the previous year)
6.
Exercise 8.77 (Exercise 7.75, 5th edition) (Hint: Make sure to write the hypothesis, calculate the test
statistics and write the decision and conclusion)
n=10, s=0.58
H 0 :   0.50 (uniformity specification) versus H a :   0.50
Test statistics:
2 
(n  1)  s 2
 02

9(0.58) 2
=12.1104
0.50
In the upper tailed test, reject H0 if 
2
 2;n1   02.01;9  21.665 . 12.1104<21.665 then fail to
reject H0. No, it does not contradict with the uniformity specification.
7.
Exercise 9.26 (Hint: Make sure to write the hypothesis, determine the test statistics and write the
decision and conclusion using the MINITAB output with both assumptions)
H 0 : 1   2  0 versus H a : 1   2  0 OR
“An Aggie does not lie, cheat, or steal or tolerate those who do.”
H 0 : 1   2 versus H a : 1   2
In both equal and unequal true variances, test statistics, t=3.6362 and the P-value=0.0008/2=0.0004.
Very small P-value. We will reject H0 with =0.05.
Yes. The data suggest that the true average potential drop for type 1 is higher than type 2.
We rejected H0 when H0 was false (as stated in the article). No error.
8.
Exercise 9.28 (This question is not in the 5th edition) (Hint: Make sure to write the hypothesis,
determine the test statistics and write the decision and conclusion using the MINITAB output with
both assumptions)
Two-sample T for YF vs OF
N
Mean
StDev
YF 10
30.70
2.75
OF
5
16.20
4.44
SE Mean
0.87
2.0
Difference = mu YF - mu OF
Estimate for difference: 14.50
T-Test of difference = 0 (vs not =)
95% CI for difference assuming unequal true variances: (8.93, 20.07)
assuming unequal true variances, T-Value = 6.69 P-Value = 0.001 DF =5
T-Test of difference = 0 (vs not =)
95% CI for difference assuming equal true variances: (10.52, 18.48)
assuming equal true variances, T-Value = 7.88 P-Value = 0.000 DF = 13
Pooled StDev = 3.36
T-Test of difference = 10 (vs not =)
assuming unequal true variances, T-Value = 2.08
T-Test of difference = 10 (vs not =)
assuming equal true variances, T-Value = 2.44
Both use Pooled StDev = 3.36
P-Value = 0.092
P-Value = 0.030
DF = 5
DF = 13
H 0 :  OF 10  YF versus H a :  OF 10  YF OR
H 0 : YF   OF  10 versus H a : YF   OF  10
In unequal true variances, test statistics, t=2.08, the degrees of freedom=5 and the Pvalue=0.092/2=0.046. Reject H0 with =0.10.
Yes. The data suggest that the true average maximum lean angle for older females is more than 10
degrees smaller than younger females.
In equal true variances, test statistics, t=2.44, the degrees of freedom=13 and the Pvalue=0.03/2=0.015. Reject H0 with =0.10.
Yes. The data suggest that the true average maximum lean angle for older females is more than 10
degrees smaller than younger females.
9.
Exercise 9.38 (b) (Hint: Assume differenced data are normally distributed. Make sure to write the
hypothesis, determine the test statistics and write the decision and conclusion using the MINITAB
output)
Test of mu difference = 0 vs mu difference not = 0
Variable
Normal-High
Variable
Normal-High
n
15
(
Mean
-42.23
95.0% CI
-44.63, -39.83)
StDev
4.34
SE Mean
1.12
T
-37.72
P
0.000
“An Aggie does not lie, cheat, or steal or tolerate those who do.”
test statistics, t=-37.72, the degrees of freedom=14 and the P-value=0.000/2=0.000. Very small Pvalue. We will reject H0 with any .
OR using the 95% confidence interval for the difference in the true means for Normal and High,
(-44.63, -39.83), notice that zero does not fall between the limits.
No. The data suggest that the true average peak stresses for the two types of concrete are not identical.