Stat 240 Homework 7 Key
12.15
a. s.e. ( x ) 
s

2
12.16
a. s.e. ( x ) 
s

2
 0.25 cm. Over all possible samples of n = 64 from this population, the
n
64
average difference between the sample mean and the population mean is about 0.25 cm.
b. The interval is 27 to 28 cm, computed as 27.5  2  0.25.
Calculate the interval as Sample estimate  2  Standard error, which here is x  2  s.e.( x )
Interpretation: With approximate 95% confidence, we can say that in the population of men
represented by this sample the mean foot length is between 27 and 28 cm.
 0.2 . This standard error is smaller than in the previous exercise
n
100
because the sample size is larger (in this exercise).
b. s.e. ( x1  x 2 ) 
s12 s 22


n1 n 2
22 22

 .0625  .04  0.32 cm.
64 100
c. The interval is 2.86 to 3.14 cm, computed as 3.5  2  0.32 cm.
Parameter is 1   2  difference between mean population foot lengths of men and women.
Calculate interval as x1  x 2  2  s.e. ( x1  x 2 ) .
12.24
12.27
12.28
x1  x 2  27 .5  24  3.5 cm. Standard error was found in part (a).
Interpretation: With 95% confidence, we can say that in the population(s) represented by these
sample(s), the difference in mean foot lengths of men and women is between 2.86 cm and 3.14
cm.
6
a. 76  2.31
or 71.38 to 80.62; df = n  1 =91 = 8 for t*.
9
6
b. 76  1.86
or 72.28 to 79.72
9
6
c. 76  1.75
or 73.375 to 78.625; df = n 1 = 161 = 15 for t*.
16
8
d. 76  2.13
or 71.74 to 80.26
16
8
e. 100  2.13
or 95.74 to 104
16
s
6

 2 . Roughly, this is the average difference between the sample and
a. s.e. ( x ) 
n
9
population means over all random samples of n = 9 that can be selected from this population.
b. 26.2  (2 2), or 22.2 to 30.2 situps.
c. With 95% confidence, we can estimate that in the population of men represented by this sample
the mean number of situps in a minute is between 22.2 and 30.2.
1.6
a. The confidence interval is about 3.84 to 5.14 days, computed as 4.5  2.06 
.
25
s
Compute interval as Sample estimate  Multiplier  Standard error, which here is x  t *
.
n
s
1.6

 0.32 .
Sample estimate is x = 4.5 days.
Standard error is s.e. ( x ) 
n
25
Multiplier is t *  2.06 . In Table A.2, use df = n1 = 251=24.
12.28 continued:
b. The confidence interval is about 6.32 to 8.84 days, computed as 8.1  2.07 
1.8
.
23
Compute interval as Sample estimate  Multiplier  Standard error, which here is x  t *
s
.
n
Sample estimate is x = 4.5 days.
Standard error is s.e. ( x ) 
s
n

1.6
 0.32 .
25
Multiplier is t  2.07 . In Table A.2, use df = n1 = 231=22.
*
c. Yes, it is reasonable to conclude that taking zinc lozenges reduces the duration of symptoms.
The confidence intervals computed in parts (a) and (b) do not overlap, so a reasonable conclusion
is that in a population of cold sufferers the mean duration of symptoms if zinc lozenges are taken
is less than the mean duration if placebo is taken.
d. For relatively small samples such as these, a necessary condition (in theory) is that the
population of measurements has a normal distribution. The researchers were verifying this
condition.
12.38
a. Females = 12.55; Males = 13.65; difference = 1.10 ( 1.10 if computed as Females  Males as
in output).
b. 3.24 to 1.03. With 95% confidence, we can say that the difference (Females  Males) in
number of letters printed in the populations of females and males is between 3.24 and 1.03
c. Cannot say there is a difference because the confidence interval includes the value 0.
d.
[ SE1 ] 2  [ SE2 ] 2  0.74 2  0.77 2  1.067 . Equivalently, compute
4.01 2 4.46 2
.

29
34
e. 2.00
13.7
a. H0:  = 200, Ha:  < 200 where  is the mean cholesterol level for the population of people
similar to those in the control group.
b. 193.13
c. 30
d. s  22.30  4.07
n
30
e. test statistic = 1.69 and p-value = 0.051.
f. t = (Mean  null value)/(SE Mean) = (193.13  200)/4.07 = 1.688 or about 1.69.
g. Because the p-value is 0.051 > 0.05, do not reject the null hypothesis. There is not sufficient
evidence to conclude that the mean cholesterol level for the population of people like the control
patients is less than 200.
13.8
a. Null standard error is s.e. ( x ) 
s

15
 2.121
n
50
Sample statistic - Null value 102  100

 0.943
b.
Null standard error
2.121
c. No, to find the proper p-value we must know whether the alternative hypothesis is one-sided or
two-sided.