252solngr1 2/05/04
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Graded Assignment 1
1. (Keller & Warrack) A random sample of 11 young adult men was asked how many minutes of sports they
watched daily. Results are below.
50
48
65
74
66
37
45
68
64
65
58
Compute a 99% confidence interval for the mean. Does the mean from your interval differ significantly
from 50 minutes? Why? What about 40 minutes?
If there are 0.9 million young adult men in the metropolitan area , do a confidence interval for the total
amount of time they watch in a day. It might help to convert your result to hours.
2. Find z .075
3. Using the additional information that the population standard deviation is known to be 10 minutes, make
an 85% confidence interval for the mean. Does this result differ significantly from 50 minutes? 40 minutes?
4. Assume that the sample mean and standard deviation you calculated in part 1 are correct but that the
survey of 11 men was taken in Nome, Alaska, where there are only 68 young men. How would your results
in 1) differ?
Solution:
1)
index
1
2
3
4
5
6
7
8
9
10
11
sum
sx 
sx
x
x2
50
48
65
74
66
37
45
68
64
65
58
640
2500
2304
4225
5476
4356
1369
2025
4624
4096
4225
3364
38564

n
  .05
 x  640 ,  x  38564 ,
 x  640  58.1818
x
2
n
s x2 
x
n  11
11
2
 nx 2

38564  1158 .1818 2
10
n 1
1327 .6596

 132 .7660
10
s x  132 .7660  11 .5224 .
s x2
132 .7660

 12 .06696  3.4741
n
11

2
 .025
10
t n1  t.005
 3.169
2
Confidence interval:   x  tn1 s x is the formula for a two sided interval.
2
  x  t n1 s x  58.1818  3.1693.4741  58.1818  11.0094 or 47.17 to 69.19. If we ask if the mean is
2
significantly different from 50, our null hypothesis is H 0 :   50 and since 50 is between the top and the
bottom of the confidence interval, do not reject H 0 and say that the mean is not significantly different from
50.
If we ask if the mean is significantly different from 40, the null hypothesis is H 0 :   40 and 40 is not
between the top and the bottom of the confidence interval. So we reject H 0 or we can say that the mean is
significantly different from 40. If there were 0.9 million adult men in the population, multiply both sides of
the interval by 0.9 million to estimate the total time young adult men watch the tube.. The interval we
252solngr1 2/05/04
got was 47.17    69.19 , so if we multiply both sides by 0.9 million the interval for the sum is
42 .45 million 
 x  62.27 million minutes or 0.708 million   x  1.036 million minutes.
2) Find z .075 . Please don’t tell me that because P0  z  075   .0259 , that z .075 is .0259.
(i) The diagram: The diagram for z will be a Normal curve centered at zero and will show one point,
z .075 , which has 7.5% above it (and 92.5% below it!) and is above zero because zero has 50% below it.
Since zero has 50% above it, the diagram will show 42.5% between zero and z .075 .
From the diagram, we want one point z .075 so that Pz  z .075   .075 or P0  z  z.075   .4250 . From
the interior of the Normal table the closest we can come to .4250 is P0  z  1.44   .4251 . This means that
z.075  1.44 .
Check: Pz  1.44   Pz  0  P0  z  1.44   .5  .4251  .0749  .075 .
3) We know that x  58.1818 , n  11 and   10 . So  x 


10

10 2
 9.0909  3.0151
11
n
11
The 85% confidence interval has 1    .85 or   .15 , so z  z.075  1.44 and the interval is
2
  x  z 2  x  58.1818  1.44 3.0151   58.18  4.34 or 53.84 to 62.52. If we test the null hypothesis
H 0 :   50 against the alternative hypothesis H 0 :   50 , since 50 is not on the confidence interval, we
reject the null hypothesis. The result is significantly different from 40 for the same reason.
4) If N  68 , the sample of 11 is more than 5% of the population, so use
sx 
N n
68  11
 3.4741
 3.4741 0.93443  3.4741 0.96657   3.3583 .
N 1
68  1
sx
n
Recall that x  58.1818 ,   .01 ,

2
10
 .005 and t n1  t .005
 3.169 .
2
Confidence interval:   x  tn1 s x is the formula for a two sided interval.
2
  x  t n1 s x  58.1818  3.1693.3583  58.1818  10.6425 or 47.54 to 68.82. The interval is smaller,
2
but it doesn’t change anything – the mean is still significantly different from 40 but not 50.
How I got these results
‘MTB >’ is the Minitab prompt. The retrieval is done using the ‘file’ pull-down menu and the ‘open worksheet’ command followed
by finding where I put the data. Other instructions were typed in the ‘session’ window.
I put the data in column 1 in Minitab and used the ‘describe’ command to get the mean and standard deviation.
—————
2/5/2004 6:57:26 PM
————————————————————
Welcome to Minitab, press F1 for help.
MTB > Retrieve "C:\Documents and Settings\RBOVE.WCUPANET\My Documents\Drive
D\MINITAB\2gr1-041.MTW".
Retrieving worksheet from file: C:\Documents and Settings\RBOVE.WCUPANET\My
Documents\Drive D\MINITAB\2gr1-041.MTW
# Worksheet was saved on Fri Jan 30 2004
Results for: 2gr1-041.MTW
MTB > print c1
Data Display
C1
50
48
58
65
74
66
37
45
68
64
65
252solngr1 2/05/04
MTB > describe c1
Descriptive Statistics: C1
Variable
N
Mean
C1
11
58.18
Variable
Minimum
Maximum
C1
37.00
74.00
I put the square of column 1 in column 2.
MTB > let c2=c1*c1
Median
64.00
Q1
48.00
TrMean
58.78
Q3
66.00
StDev
11.52
SE Mean
3.47
I summed both columns.
MTB > sum c1
Sum of C1
Sum of C1 = 640.00
MTB > sum c2
Sum of C2
Sum of C2 = 38564
I printed out both columns.
MTB > print c1 c2
Data Display
Row
Row
C1
C1
C2
C2
1
50
2500
2
48
2304
3
65
4225
4
74
5476
5
66
4356
6
37
1369
7
45
2025
8
68
4624
9
64
4096
10
65
4225
11
58
3364
I cleared column 2 and saved column 1, using it to do 2 confidence intervals.
MTB > Erase c2.
This is the instruction for a 99% interval using t.
MTB > tinterval 99.0 c1
One-Sample T: C1
Variable
C1
N
11
Mean
58.18
StDev
11.52
SE Mean
3.47
(
99.0% CI
47.17,
69.20)
This is the instruction for a 99% interval using t. currently preferred by the Minitab establishment.
MTB > oneT c1;
SUBC> conf 99.0.
One-Sample T: C1
Variable
C1
N
11
Mean
58.18
StDev
11.52
SE Mean
3.47
(
99.0% CI
47.17,
69.20)
This is the instruction for an 85% interval using z and a population standard deviation of 10.
MTB > zinterval 85.0 10 c1
One-Sample Z: C1
The assumed sigma = 10
Variable
C1
N
11
Mean
58.18
StDev
11.52
SE Mean
3.02
(
85.0% CI
53.84,
62.52)
252solngr1 2/05/04
This is the instruction for an 85% interval using z and a population standard deviation of 10 currently preferred by the
Minitab establishment.
MTB > oneZ c1;
SUBC> sigma 10;
SUBC> conf 85.0.
One-Sample Z: C1
The assumed sigma = 10
Variable
N
Mean
StDev
SE Mean
85.0% CI
C1
11
58.18
11.52
3.02 (
53.84,
62.52)
I used ‘select all’ and ’copy’ to get my results into this document.
Note that Minitab does not have provision for a finite population correction.