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500y0712 1/8/07
SECOND EXAM
ECO500 Business Statistics
Name: _____________________
Student Number : _____________________
Remember – Neatness, or at least legibility, counts. In all questions an answer needs a calculation or
explanation to count. Show your work!
Part I. (12 points) Show your work! Make Diagrams!
I. (12 points) Do all the following.
x ~ N 3,11
11  3 
  11  3
z
 P 1.27  z  0.73 
1. P11  x  11  P 
11 
 11
 P1.27  z  0  P0  z  0.73  .3980  .2673  .6653
For z make a diagram. Draw a Normal curve with a mean at 0. Indicate the mean by a vertical line!
Shade the area between -1.27 and 0.73. Because this area is on both sides of zero, we must add the area
between -1.27and zero to the area between zero and 0.73. If you wish, make a completely separate diagram
for x . Draw a Normal curve with a mean at 3. Indicate the mean by a vertical line! Shade the area
between -11 and 11. These numbers are on either side of the mean (3), so we add.
11  3 
0  3
z
 P 0.27  z  0.73 
2. P0  x  11  P 
11
11 

 P0.27  z  0  P0  z  0.73  .1064  .2673  .3737
For z make a diagram. Draw a Normal curve with a mean at 0. Indicate the mean by a vertical line!
Shade the area between -0.18 and 0.73. Because this area is on both sides of zero, we must add the area
between -0.18and zero to the area between zero and 0.73. If you wish, make a completely separate diagram
for x . Draw a Normal curve with a mean at 3. Indicate the mean by a vertical line! Shade the area
between zero and 11. These numbers are on either side of the mean (3), so we add
43  3 
3  3
z
 P0  z  3.64   .4999
3. P3  x  43  P 
11 
 11
For z make a diagram. Draw a Normal curve with a mean at 0. Shade the area between zero and 3.64.
Because this is completely on the left of zero and touches zero, we can simply look up our answer on the
standardized Normal table. If you wish, make a completely separate diagram for x . Draw a Normal curve
with a mean at 3. Shade the area between 3 and 3.64. This area includes the mean (3), but does not include
any points to the right of the mean, so that we neither add nor subtract
12  3 

 Pz  0.82   Pz  0  P0  z  0.82   .5  .2939  .2061
4. Px  12   P  z 
11 

For z make a diagram. Draw a Normal curve with a mean at 0. Indicate the mean by a vertical line!
Shade the area between above 0.82. Because this is completely on the left of zero, we must subtract the area
between 0.82 and zero from the entire area above zero. If you wish, make a completely separate diagram for
x . Draw a Normal curve with a mean at 3. Indicate the mean by a vertical line! Shade the area above 12.
This area is entirely to the left of the mean (3), so we subtract.
2  3
  20  3
z
 P 2.09  z  0.09 
5. P20  x  2  P 
11 
 11
 P2.09  z  0  P0.09  z  0  .4817  .0359  .4458
For z make a diagram. Draw a Normal curve with a mean at 0. Indicate the mean by a vertical line!
Shade the area between -2.09 and -0.09. Because this is completely on the left of zero, we must subtract. If
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500y0712 1/8/07
you wish, make a completely separate diagram for x . Draw a Normal curve with a mean at 3. Indicate the
mean by a vertical line! Shade the area between -20 and 2. These numbers are both to the left of the mean
(3), so we subtract.
6. x.125 (Do not try to use the t table to get this.) For z make a diagram. Draw a Normal curve with a
mean at 0. z .125 is the value of z with 12.5% of the distribution above it. Since 100 – 12.5 = 87.5, it is also
the 87.5th percentile or the .875 fractile. Since 50% of the standardized Normal distribution is below zero,
your diagram should show that the probability between z .125 and zero is 87.5% - 50% = 37.5% or
P0  z  z.125   .3750 . The closest we can come to this is P0  z  1.15   .3749 . So z .125  1.15 . To
get from z .125 to x.125 , use the formula x    z , which is the opposite of z 
x
.

x  3  1.1511  15.65 . If you wish, make a completely separate diagram for x . Draw a Normal curve
with a mean at 3. Show that 50% of the distribution is below the mean (3). If 12.5% of the distribution is
above x.125 , it must be above the mean and have 37.5% of the distribution between it and the mean.
15 .65  3 

 Pz  1.15   Pz  0  P0  z  1.15   .5  .3749  .1251
Check: Px  15 .65   P  z 
11 

 .1250
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Part II. (At least 40 points. Parentheses give points on individual questions. Brackets give cumulative
point total.) Exam is normed on 50 points.
1. Find P12  x  17  for the following distributions (Use tables in c, d, f and h. All probabilities should
show 4 places to the right of the decimal point. Find the mean and standard deviation of the distribution.
(10)
a. Continuous Uniform with c  1, d  14 (Make a diagram!).
b. Continuous Uniform with c  15, d  25 (Make a diagram!).
c. Binomial Distribution with p  .45, n  25 .
d. Binomial Distribution with p  .85, n  25 .(2)
e. Geometric Distribution with p  .15.
f. Poisson Distribution with parameter of 15.
g. Show how you would do this for a Hypergeometric Distribution with p  .45, n  25 ,
N  80. Remember M  Np .
h. (Extra credit) Hypergeometric Distribution with p  .45, n  25 , N  520 .
i. (Extra credit) Exponential distribution with c  .01 .
j. Assume that the average number of workers logging onto a system every hour is 750. What is
the chance that none will log on in a given minute?
k. What is the chance that over 800 will logon in one hour?
Solution: Find P12  x  17  for the following distributions (Use tables in c, d, f and h. All probabilities
should show 4 places to the right of the decimal point. Find the mean and standard deviation of the
distribution.
a. Continuous Uniform with c  1, d  14 (Make a diagram!).
1
1
1


 .07692 . In the diagram below, shade the area between 12 and 14. (There is
d  c 14  1 13
no area between 14 and 16.) The horizontal line has a height of 113 .
0
1
12
14 17
The area in the box between 12 and 17 is 14  12  1  .1538 .
13
Another way to do a problem of this type is to remember that for any continuous distribution, we
can use differences between cumulative distributions, Pa  x  b  F b  F a  where the
xc
cumulative distribution is F x0   Px  x0  and F x  
for c  x  d ,
d c
F x   0 for x  c and F x   1 for x  d . If we use the cumulative distribution method,
remember F x   1 for x  d . So P12  x  17   F 17   F 12   1 
12  1
11
 1
14  1
13
 1  .8462  .1538 .

d  c2  14  12  169  14.0833
c  d 1  14

 7.5000 ,  2 
2
2
12
12
12
So   14.0833  3.7528
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500y0712 1/8/07
b. Continuous Uniform with c  15, d  25 (Make a diagram!).
1
1
1


 .1000 . In the diagram below, shade the area between 15 and 17. There is
d  c 25  15 10
no probability between 12 and 15. The horizontal line has a height of 110 .
0
17
25
17  15
So P12  x  17  
 .2000 . If we use the cumulative distribution method,
25  15
xc
remember F x   0 for x  d and F x  
for c  x  d , . So
d c
17  15
2
0 
 .2000 .
P12  x  17   F 17   F 12  
25  10
10

12 15
d  c  25  15   100  8.3333
c  d 15  25

 20  2 
2
2
12
12
12
2
2
So   8.3333  2.8868
c. Binomial Distribution with p  .45, n  25 .
P12  x  17   Px  17   Px  11  .99417  .54257  .4516
(Remember that q  1  p and that both p and q must be between 0 and 1.)
q  1  p  1  .45  .55,   np  25.45   11.25 ,  2  npq  11.25 .55  6.1875 so
  npq  6.1875  2.4875 . The average number of successes in 25 tries, when the
probability of a success in an individual try is 45% is 11.25.
d. Binomial Distribution with p  .85, n  25 .
P12  x  17  can't be done directly with tables that stop at p  .5 , so try to do it with
failures. (The probability of failure is 1 - .85 = .15.) 12 successes correspond to 25 – 12 = 13
failures out of 25 tries. 17 successes correspond to 8 failures. So try 8 to 13 successes when
p  .15 and n  25 . P8  x  13  Px  13  Px  7  1.00000  .97453  .0255
(Remember that q  1  p and that both p and q must be between 0 and 1.)
q  1  p  1  .85  .15,   np  25.85   21.25 ,  2  npq  21.25 .15  141.6667 so
  npq  141 .6667  11 .9024 . The average number of successes in 25 tries, when the
probability of a success in an individual try is 85% is 21.25.
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500y0712 1/8/07
e. Geometric Distribution with p  .15.
Remember that F c  Px  c  1  q c , because success at try c or earlier implies
that there cannot have been failures on the first c tries. q  1  p  1  .15  .85

P12  x  17  Px  17  Px  11  F 17  F 11  1  .8517  1  .8511
 .85  .85
11
.85  .85
11
17
17

 .16734  .06311  .1042 or


 .8511 1  .856  .167341  .37715  .1042
If q  1  p  1  .15  .85,  
q
1
1
.85
.85

 6.6667 ,  2  2 

 37 .77778 so
2
.0225
p .15
p
.15
  37.7778  6.1464 . When the probability of a success in an individual try is 15% and we
play a game repeatedly, on the average our first success will occur between the 6th and 7th try.
f. Poisson Distribution with parameter of 15.
P12  x  17   Px  17   Px  11  .74886  .18475  .5641
  m  15 ,  2  m  15 , so   m  15  3.8730
g. Show how you would do this for a Hypergeometric Distribution with p  .45, n  25 ,
N  80. Remember M  Np .
We have no cumulative tables or cumulative distribution formula for this distribution, so the only
available method is to add together probabilities over the range 10 to 15.
C 36 C 44
C M C N M
Px   x Nn x and M  Np  80.45   36 so Px   x 8025 x
Cn
C 25
17
44 
 C x36 C 25
1
44
x 


C x36 C 25
x
 C 80  C 80
25
25 x 12
x 12

We could take this further using the cumulative distribution formula at the end of the outline. The
M  x 1  n  x 1 
formula is Px  
 N  M  n  x  Px  1 and with n  25 , N  80 and M  36 this
x


and P12  x  17  
becomes Px  
P14  
67
14
17



80  x  1  25  x  1 
68  13 
P12  ,
Px  1 . So P13  


x
13  32 
 19  x 
66
 12 
 33  P13 , P15   15
 
 11 
 34  P14 , etc. So if we compute P12  , the rest is fairly
 
easy.
M
and q  1  p then   np  25.45   11.25 and
N
N n
80  25
55
11 .25 .55   .69620 6.1875   4.30774 so

npq 
25.45 .55  
N 1
80  1
79
If p 
2
  .69620 6.18755  .83438 2.4875   4.30774  2.07551 . If we take a sample of 25 from
a population of 80 of which 45% are successes, on the average we will get 11.25 successes.
h. (Extra credit) Hypergeometric Distribution with p  .45, n  25 , N  520 .
Since N  520 is more than 20 times n  25 , we can use the binomial distribution with p  .45
and n  25 . P12  x  17   Px  17   Px  11  .99417  .54257  .4516
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500y0712 1/8/07
i. (Extra credit) Exponential distribution with c  .01 .
In ‘Great Distributions I Have Known, we have the following information.
x is usually the
Exponential
f x   ce cx and
amount of time
1
1
cx


you have to wait F x  1  e
c
c
when x  0 and
until a success.
the mean time to a
1
success is . Both
c
are zero if x  0 .
Since this is a continuous distribution, P12  x  17   F 17   F 12 



 1  e 0.17  1  e .12  e .12  e .17  .88692 .84366  .0433
Note: F 17   .15634 and F 12   .11308 .
  
1
1

 100 is the average time to a success.
c 0.01
j. Assume that the average number of workers logging onto a system every hour is 750. What is the
chance that none will log on in a given minute?
Since 750  12 .5 , we use the Poisson table with a parameter of 12.5. The relevant part of the
60
table is below, so we have a probability of essentially zero.
k P(x=k)
P(xk)
0 0.000004 0.00000
If we are talking about one minute   m  12.5 ,  2  m  12.5 , so   m  12.5  3.5355.
k. What is the chance that over 800 will logon in one hour?
The Poisson distribution with a mean of 750 can be approximated by the Normal distribution.

800  750 
Px  800   P  z 
  Pz  1.83   .5  .4664  .0336
750 

If we are talking about one hour   m  750 ,  2  m  750 , so   m  750  27.3861.
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500y0712 1/8/07
2. Assume that the income in an area is Normally distributed with a known population standard deviation of
$2000. A random sample of 15 households yields a sample mean of $25000. Test the null hypothesis that
the population mean is at least $26000. Use a test ratio. Find a p-value. (5) Make 3 diagrams.
a. Show the rejection region and test the null hypothesis if the significance level is 5% (3)
b. Show the rejection region and test the null hypothesis if the significance level is 1% (1)
c. Find a p-value for the null hypothesis and compare the p-value to the results of a) and b). Make a
diagram. (2)
Solution:
Interval for
Confidence
Hypotheses
Test Ratio
Critical Value
Interval
Mean (
x  0
  x  z 2  x
xcv   0  z 2  x
H0 :   0
z
known)

H1 :    0
xcv   0  z 2  x
x

x 
n
The statement of the hypothesis to be tested is   26000 . Because it contains an equality, this is a null
hypothesis. So we have H 0 :   26000 and H 1 :   26000 Given:  0  26000 ,   2000 , n  15 and
x  25000 , so that  x 


2000

2000 2
 266667  516 .40 . This means
15
n
15
x   0 25000  26000
z

 1.936 . This is a left-sided test because we will reject the null hypothesis if
x
516 .40
x is too low.
a) If   .05 , the rejection region is the area below z.05  1.645 . Make a diagram of a Normal
distribution with z on the horizontal axis. Show a vertical line at zero and shade the area below -1.645.
Since z  1.936 is in this rejection zone, reject the null hypothesis.
b) If   .01, the rejection region is the area below z.01  2.327 . Make a diagram of a Normal
distribution with z on the horizontal axis. Show a vertical line at zero and shade the area below -2.327.
Since z  1.936 is not in this rejection zone, do not reject the null hypothesis.
c) This is a left-sided test and our p-value is Pz  1.94   Pz  0  P1.94  z  0  .5  .4738  .0262 .
Since this is between .01 and .05, we reject the null hypothesis when   .05 , but not when   .01 . .
Of a Normal distribution with z on the horizontal axis
The Minitab output follows with comments.
MTB > OneZ 15 25000;
SUBC>
SUBC>
SUBC>
Sigma 2000;
Test 26000;
Alternative -1.
The command OneZ sets up a confidence interval or hypothesis test
for a sample of 15 with a sample mean of 25000. A semicolon
indicates that the instruction is incomplete.
This subcommand sets the population standard deviation.
This subcommand sets up a test of the mean’s equaling 26000.
This subcommand makes the alternative hypothesis ‘<.’
One-Sample Z
Test of mu = 26000 vs < 26000
The assumed standard deviation = 2000
95%
Upper
N
Mean SE Mean
Bound
Z
P
15 25000.0
516.4 25849.4 -1.94 0.026
All computations are as in c) except for the 95%
one sided confidence interval   25849.4 .
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500y0712 1/8/07
3. The claimed mean weight of a batch of canned vegetables is 16 oz. You take a sample of 20 cans and
find the following weights.
x
x2
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
16.04 257.2816
15.90 252.8100
15.81 249.9561
15.94 254.0836
15.97 255.0409
16.05 257.6025
15.91 253.1281
16.03 256.9609
15.84 250.9056
15.93 253.7649
16.04 257.2816
15.93 253.7649
15.96 254.7216
16.00 256.0000
16.16 261.1456
15.79 249.3241
15.90 252.8100
16.03 256.9609
16.03 256.9609
15.74 247.7476
319.00 5088.2514
There are 9 parts to this problem. Note that the absolute value of the test ratio I got in part a) was 2.174, you
should be very close. Please do not round excessively or your answers will be way off. Use   .05 . Clearly
state your null and alternative hypothesis for each problem. Assume that the sample is taken from a
Normally distributed population.
a) Test the hypothesis that the mean is less than 16 using a test ratio. Make a diagram showing your
rejection regions. Find an approximate p-value for the test ratio. (2)
b) Test the hypothesis that the mean is less than 16 using a critical value for x . Make a diagram showing
your rejection regions. (2)
c) Test the hypothesis that the mean is less than 16 using a confidence interval for the population mean.
Make a diagram showing your confidence interval. (2)
d) Test the hypothesis that the mean is equal to 16 using a test ratio. Make a diagram showing your rejection
regions. Find an approximate p-value for the test ratio. (1)
e) Test the hypothesis that the mean is equal to 16 using a critical value for x . Make a diagram showing
your rejection regions. (1)
f) Test the hypothesis that the mean is equal to 16 using a confidence interval for the population mean.
Make a diagram showing your confidence interval. (1)
g) Test the hypothesis that the mean is greater than 16 using a test ratio. Make a diagram showing your
rejection regions. Find an approximate p-value for the test ratio. (1)
h) Test the hypothesis that the mean is greater than 16 using a critical value for x . Make a diagram showing
your rejection regions. (1)
i) Test the hypothesis that the mean is greater than 16 using a confidence interval for the population mean.
Make a diagram showing your confidence interval. (1)
Solution:
 0  16,
 x  319 .00 ,  x
 x  319 .00  15.9500
x
n
20
s  0.0106  0.10296 s x 
2
s
 5088 .2514 , n  20 , df  n  1  19 and   .05 .
2
 x  nx

n 1
2

5088 .2514  20 15 .9500 2
 0.0106
19
0.0106
 0.000530  0.0230
20
8
500y0712 1/8/07
Interval for
Mean (
unknown)
Confidence
Interval
  x  t 2 s x
DF  n 1
Hypotheses
Test Ratio
H0 :   0
t
H1 :    0
x  0
sx
Critical Value
xcv   0  t  2 s x
sx 
s
n
a) H 0 :   16 , H 1 :   16 . This is a left sided test because we reject the null hypothesis if x is too small.
Make a diagram showing an approximately Normal distribution with t on the horizontal axis and a mean
19
indicated by a vertical line at zero. We reject the null hypothesis if the t-ratio is below  t .05
 1.729 .
Shade the area below -1.729. Since t 
x   0 15 .95  16

 2.174 is in the ‘reject’ region, reject the null
sx
0.0230
hypothesis.
A p-value in this case would be Px  15.95   Pt  2.174  . If we look at the 19 line of the t-table we find
19
19
t .025
 2.093 and t .01
 2.539 . Since 2.174 is between them, we can say .01  pvalue  .025 . Note that
these are both below 5%, confirming our rejection.
b) H 0 :   16 , H 1 :   16 . This is a left sided test because we reject the null hypothesis if x is too small.
We thus want a critical value for x that is below 16. The critical value formula xcv   0  t  s x , becomes
2
xcv   0  t s x  16  1.729 0.0230   15.96. Make a diagram showing an approximately Normal
distribution with x on the horizontal axis and a mean indicated by a vertical line at 16. Shade the rejection
region below 15.96. Since x  15 .95 is in the ‘reject’ region, reject the null hypothesis.
c) H 0 :   16 , H 1 :   16 . The confidence interval goes in the direction of the alternate hypothesis.
  x  t 2 s x becomes   x  t s x  15.95  1.729 0.0230   15.99. Make a diagram showing an
approximately Normal distribution with  on the horizontal axis and a mean indicated by a vertical line at
x  15.95. Represent the confidence interval   15.99 by a shaded area. Note that  0  16 is (barely)
outside of the shaded area. The confidence interval thus contradicts the null hypothesis, which we reject.
d) H 0 :   16 , H 1 :   16 This is a two-sided test because we reject the null hypothesis if x is either too
large or too small. Make a diagram showing an approximately Normal distribution with a mean indicated by
19
 2.093 or above
a vertical line at zero. We reject the null hypothesis if the t-ratio is below  t .025
19
t .025
 2.093 . Shade the area below -2.093 and the area above 2.092. Since t 
x   0 15 .95  16

sx
0.0230
 2.174 is in the lower ‘reject’ region, reject the null hypothesis.
A p-value in this case would be 2Px  15.95   2Pt  2.174  . If we look at the 19 line of the t-table we
19
19
find t .025
 2.073 and t .01
 2.539 . Since 2.174 is between them, we can say .01  Pt  2.174   .025 . If
we double this we get .02  pvalue  .05 Note that these are both below 5%, confirming our rejection.
e) H 0 :   16 , H 1 :   16 This is a two sided test because we reject the null hypothesis if x is too small
or two large. We thus want a two critical values for x . The critical value formula xcv   0  t  s x ,
2
becomes xcv  16  2.093 0.0230   16  .0481 . Make a diagram showing an approximately Normal
distribution with x on the horizontal axis and a mean indicated by a vertical line at 16. Shade a rejection
region below 15.9519 and a rejection region above 16.0481. Since x  15 .95 is in the lower ‘reject’ region,
reject the null hypothesis.
f) H 0 :   16 , H 1 :   16 The confidence interval   x  t s x becomes
2
  15.95  2.093 0.0230   15.95  .0481 . Make a diagram showing an approximately Normal distribution
with  on the horizontal axis and a mean indicated by a vertical line at x  15.95. Represent the
confidence interval 15.9019    15.9991 by a shaded area. Note that  0  16 is (barely) outside of the
shaded area. The confidence interval thus contradicts the null hypothesis, which we reject.
9
500y0712 1/8/07
g) H 0 :   16 , H 1 :   16 This is a right sided test because we reject the null hypothesis if x is too large.
Make a diagram showing an approximately Normal distribution with a mean indicated by a vertical line at
19
zero. We reject the null hypothesis if the t-ratio is above t .05
 1.729 . Shade the area above 1.729. Since
t
x   0 15 .95  16

 2.174 is not in the ‘reject’ region, do not reject the null hypothesis.
sx
0.0230
A p-value in this case would be Px  15.95   Pt  2.174  . If we look at the 19 line of the t-table we find
19
19
t .025
 2.093 and t .01
 2.539 . Since 2.174 is between them, we can say .01  Pt  2.174   .025 . But this
also implies, as we know .01  Pt  2.174   .025 . If we subtract both of these from 1 we get
.99  Pt  2.174   .975 or .99  pvalue  .975 Note that these are both above 5%, confirming our nonrejection.
h) H 0 :   16 , H 1 :   16 This is a left sided test because we reject the null hypothesis if x is too large.
We thus want a critical value for x that is above 16. The critical value formula xcv   0  t  s x , becomes
2
x cv   0  t s x  16  1.729 0.0230   16.04 . Make a diagram showing an approximately Normal
distribution with a mean indicated by a vertical line at 16. Shade the rejection region above 16.04. Since
x  15 .95 is not in the ‘reject’ region, do not reject the null hypothesis.
i) H 0 :   16 , H 1 :   16 The confidence interval goes in the direction of the alternate hypothesis.
  x  t 2 s x becomes   x  t s x  15.95  1.729 0.0230   15.91. Make a diagram showing an
approximately Normal distribution with  on the horizontal axis and a mean indicated by a vertical line at
x  15.95. Represent the confidence interval   15.9 by a shaded area. Note that  0  16 is inside the
shaded area. The confidence interval thus does not contradict the null hypothesis, which we cannot reject.
Annotated Minitab output follows.
————— 1/12/2007 6:52:05 PM ————————————————————
Welcome to Minitab, press F1 for help.
MTB > WOpen "C:\Documents and Settings\RBOVE\My Documents\Minitab\500x06021000.MTW".
Retrieving worksheet from file: 'C:\Documents and Settings\RBOVE\My
Documents\Minitab\500x06021-000.MTW'
Worksheet was saved on Fri Jan 12 2007
Results for: 500x06021-000.MTW
MTB > print c1 c2 c3
Data Display
Row
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
C1
16.04
15.90
15.81
15.94
15.97
16.05
15.91
16.03
15.84
15.93
16.04
15.93
15.96
16.00
16.16
15.79
15.90
16.03
16.03
15.74
C2
257.282
252.810
249.956
254.084
255.041
257.602
253.128
256.961
250.906
253.765
257.282
253.765
254.722
256.000
261.146
249.324
252.810
256.961
256.961
247.748
C1 is the original data ( x ), C2 is x 2 and C3 is 1000x 2
C3
2572816
2528100
2499561
2540836
2550409
2576025
2531281
2569609
2509056
2537649
2572816
2537649
2547216
2560000
2611456
2493241
2528100
2569609
2569609
2477476
10
500y0712 1/8/07
MTB > sum c1
Sum of C1
Sum of C1 = 319
MTB > sum c2
Sum of C2
Sum of C2 = 5088.25
MTB > sum c3
Sum of C3
Sum of C3 = 50882514
This was gotten to get one more decimal place in x 2 .
MTB > Onet C1;
The command Onet sets up a confidence interval or hypothesis test
for a sample of 15. The data is in C1. A semicolon indicates that the
instruction is incomplete.
This subcommand sets up a test of the mean’s equaling 16.
SUBC>
Test 16.
One-Sample T: C1
Test of mu = 16 vs not = 16
Variable
N
Mean
StDev
C1
20 15.9500 0.1030
MTB > Onet C1;
SUBC>
Test 16;
SUBC>
Alternative -1.
SE Mean
0.0230
95% CI
(15.9018, 15.9982)
T
-2.17
P
0.043
This subcommand makes the alternative hypothesis ‘<.’
One-Sample T: C1
Test of mu = 16 vs < 16
Variable
C1
N
20
Mean
15.9500
StDev
0.1030
MTB > Onet C1;
SUBC>
Test 16;
SUBC>
Alternative 1.
SE Mean
0.0230
95%
Upper
Bound
15.9898
T
-2.17
P
0.021
This subcommand makes the alternative hypothesis ‘>.’
One-Sample T: C1
Test of mu = 16 vs > 16
Variable
C1
N
20
Mean
15.9500
StDev
0.1030
SE Mean
0.0230
95%
Lower
Bound
15.9102
T
-2.17
P
0.979
11
500y0712 1/8/07
4. a) Use a test ratio and the sample standard deviation you computed in problem 3 to test if the population
standard deviation is 0.15 (3))
b) Confirm your results by creating a confidence interval for the variance. (1)
c) Repeat the test assuming that the sample size is 40. (2)
d) Confirm your results by creating a confidence interval for the variance. (1)
Solution: a) s 2  0.0106 , s  0.0106  0.10296, n  20 , df  n  1  19 and   .05 .
Interval for
Confidence
Hypotheses
Test Ratio
Critical Value
Interval
VarianceH 0 :  2   02
n  1s 2
n  1s 2
 .25 .5 2  02
2
2 
2  2
Small Sample
s cv 
.5 .5 2 
 02
n 1
H1: :  2   02
VarianceLarge Sample
 
s 2DF 
H 0 :  2   02
 z 2  2DF 
z 
2  2DF   1
2
H1 :  2   02
s cv 
 2 DF
 z  2  2 DF
H 0 :   0.15 , H 1 :   0.15  0  .15  02  0.15 2  .0225
2 
n  1s 2
 02

19 0.0106 
 8.9511 According to the table for 19 degrees of freedom  .2025  32.8523
.0225
and  .2975  8.9065 . Make a diagram centered near 19 with  2 on the horizontal axis showing one
rejection region below 8.9065 and another rejection region above 32.8523. Since the computed value of
 2 lies between the two table values do not reject the null hypothesis.
b)
n  1s 2
 .2025
2 
n  1s 2
 .2975
means that
19 0.0106
32 .8523
 2 
19 0.0106 or
8.9065
0.00613   2  0.02261 or
0.03156    0.15038 . Since this interval contains 0.15, do not reject the null hypothesis.
c) s 2  0.0106 , s  0.0106  0.10296,  0  .15 , n  40 , df  n  1  39 and
  .05 .  2 
n  1s 2
 02

39 0.0106 
 18 .3733
.0225
z  2  2  2 DF  1
 218 .3733   239   1  36 .7467  77  6.0619  8.7750  2.7131 . For a two-sided test make a
diagram with z on the horizontal axis. Show a vertical line at zero and shade the area below
z.025  1.960 and the area above z .025  1.960 . Since z  2.7131 is in this rejection zone, reject the
null hypothesis.
d)
s 2 DF
z   2 DF
2
 
s 2 DF
 z   2 DF
becomes
0.10296 239 
1.960  239 
 
0.10296 239 
 1.960  239 
or
2
0.10296 8.8318 
0.10296 8.8318 
or 0.00954    0.1323 . Since this interval does not contain
 
1.960  8.8318
 1.960  8.8318
0.15, reject the null hypothesis.
12
500y0712 1/8/07
5. If out of a sample of 100 parts, ten are found defective, test the hypothesis that the proportion of
defective parts in the population is no more than 5%.
a) Test the hypothesis that using a test ratio. Make a diagram showing your rejection regions. Find an
approximate p-value for the test ratio. (2)
b) Test the hypothesis using a critical value for p . Make a diagram showing your rejection regions. (2)
c) Test the hypothesis using a confidence interval for the population proportion. Make a diagram showing
your confidence interval. (2)
Solution: H 0 : p  0.05 , H 1 : p  0.05 . p 0  0.05 , q 0  1  p 0  1  0.05  .95, n  100 , x  10,
p  x  10
 .10 q  1  p  1  .10  .90   .05 . Note that because we will reject the null
n
100
hypothesis if the observed proportion is too far above 5%, this is a right-sided test.
Interval for
Confidence
Hypotheses
Test Ratio
Critical Value
Interval
Proportion
p  p0
p  p  z 2 s p
pcv  p0  z 2  p
H 0 : p  p0
z

H
:
p

p
p
1
0
pq
p0 q0
sp 
p 
n
n
q  1 p
q0  1  p0
For the first two parts  p 
a) z 
p  p0
p

p0 q0
.05.95 

 0.000475  0.02179 .
100
n
.10  .05
 2.2946 . Make a diagram with z on the horizontal axis and zero indicated by a
0.02179
vertical line. The rejection region is the area above z .05  1.645 . Since the computed value of z is in the
rejection region, reject the null hypothesis.
The p-value is P p  .10   Pz  2.29   .5  .4890  .0110 . Note that this is below 5%.
b) pcv  p0  z  p becomes pcv  p0  z  p  .05  1.645 0.02179   .08584 . Make a diagram centered
2
at .05 with p on the horizontal axis. The rejection region is the area above p cv  .08584 Since
p  x  10
 .10 is in the rejection region, reject the null hypothesis.
n
100
pq
.10 .90 

 0.0009000  0.03000 Since the alternative hypothesis is H 1 : p  0.05 , and
n
100
the confidence interval must go in the same direction, p  p  z s p becomes p  p  z s p
c) s p 
2
 .10  1.645 0.0300   .05065 . Make a diagram centered at p  .10 with p on the horizontal axis.
Represent the confidence interval by shading the area above p  .05065 . Since p 0  .05 is not on the
rejection region, reject the null hypothesis. Minitab output follows Problem 6.
6. (Extra credit) Repeat problem 5 using a sample of 20 of which 2 are found defective and the binomial
distribution. (4)
Solution: H 0 : p  0.05 , H 1 : p  0.05 . p 0  0.05 , n  20 , x  2,   .05 . Note that because we will
reject the null hypothesis if the observed proportion is too far above 5%, this is a right-sided test. We use
the binomial table with p  .05 and n  20.
The p-value approach is the easiest. The p-value is Px  2  1  Px  1  1  .7358  .2642 . Note that this
is above 5%, so we cannot reject the null hypothesis.
If you want to find a critical value for x , Note that Px  3  .98410 is the first number in the p  .05
above .95. This means that Px  4  1  Px  3  1  .9841  .0159 . We can reject the null hypothesis
13
500y0712 1/8/07
only if there are 4 or more defective items. Since out observed number of defective items is 2, we cannot
reject the null hypothesis.
Annotated Minitab output follows.
————— 1/18/2007 10:36:40 PM ————————————————————
Welcome to Minitab, press F1 for help.
MTB > POne 100 10;
SUBC>
Test 0.05;
SUBC>
SUBC>
Alternative 1;
UseZ.
The command pOne sets up a confidence interval or
hypothesis test for a sample of 100 with x = 10. A
semicolon indicates that the instruction is incomplete.
This subcommand sets up a test of the proportion’s
equaling .05.
This subcommand makes the alternative hypothesis ‘>.’
This subcommand tells the computer to use the Normal
approximation to the binomial distribution.
Test and CI for One Proportion
Test of p = 0.05 vs p > 0.05
Sample
1
X
10
N
100
Sample p
0.100000
95%
Lower
Bound
0.050654
MTB > POne 20 2;
SUBC>
SUBC>
Z-Value
2.29
P-Value
0.011
The command pOne sets up a confidence interval or
hypothesis test for a sample of 200 with x = 2. A
semicolon indicates that the instruction is incomplete.
Test 0.05;
Alternative 1.
Since ‘UseZ’ does not appear, the binomial will be used.
Test and CI for One Proportion
Test of p = 0.05 vs p > 0.05
Sample
1
X
2
N
20
Sample p
0.100000
95%
Lower
Bound
0.018065
Exact
P-Value
0.264
14