252solngr2-071 2/19/07
ECO 252 Second Graded Assignment
R. E. Bove
Name:
Class days and time:
Student Number:
Please include class and student number on what you hand in! Papers should be stapled. Your
writeup should state clearly what you did and concluded.
Problem 1: Which of the following could be a null hypothesis? Which could be an alternative hypothesis?
Which could be neither? Why?
(i) p  3 , (ii) p  .3 , (iii) p  .3 , (iv)   3 , (v)   3 , (vi)   3 , (vii) s  3 , (viii)  3 , (ix)   5 ,
(x) p  .3 , (xi)   3 , (xii)   0 ,(xiii) p  .5.
Problem 2: A rental firm believes that that the average time that a backhoe is rented is 4.2 days. In order to
verify this statement, a sample of rental records is taken, with the following results.
4, 2, 4, 3, 6, 2, 3, 2, 5, 3, 2, 4, 2, 2
You presumably know that n  14 , x  3.1429 , and  x  1 .1x . ( x is the last digit of your student
number.) Test the hypothesis that the mean is 4.2. Assume that the confidence level is 95%.
a) State your null and alternative hypotheses.
b) Find critical values for the sample mean and test the hypothesis. Show your reject region on a
diagram
c) Find a confidence interval for the sample mean and test the hypothesis. Show your results on a
diagram.
d) Use a test ratio for a test of the sample mean. Show your reject region on a diagram
e) Find a p-value for the null hypothesis using the Normal table and use the p-value to test the
hypothesis.
Problem 3: (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
30
40
60
60
60
50
Personalize the data as follows: add the digits of your student number to the last six numbers. Example: Ima
Badrisk has the student number 123456; so the last six numbers become {31, 42, 63, 64, 65, 56}.
Compute the sample standard deviation using the computational formula (if you don’t know what
that means, find out!). (If you did this correctly on the last assignment, just copy your work.) Test the
hypothesis that average time watching sports is below 70 hours.
a) Test the validity of the hypothesis using a confidence level of 95% and a critical value for the
sample mean. (You cannot test the validity of a hypothesis that you haven’t stated!)
b) Find an approximate p-value for the statement.
c) Will we reject the hypothesis at a significance level of (i) .001? (ii) .01? (iii) .10? Using
the p- value explain why.
Note that none of the problems beyond this point involve sample means.
252solngr2-071 2/19/07
Problem 4: (Ken Black) It is generally believed that 79% of US companies offer flexible scheduling . A
survey of 415 accounting firms finds that 300  x have flexible scheduling. ( x is the 2nd to last digit of your
student number.) Test the hypothesis that the proportion of accounting firms that offer flexible hours is
below that for other US firms.
a) State your null and alternative hypotheses.
b) Find a test ratio for a test of the proportion.
c) Find a p-value for this ratio and use it to test the hypothesis at a 5% significance level.
Extra Credit Problem 5:
a) Finish problem 4 by finding an appropriate confidence interval for the proportion and showing
whether it contradicts the null hypothesis.
b) Use the data in problem 3 to test the hypothesis   10 .
c) Use Minitab to check your answer to problem 4. Do this three ways
First: Enter Minitab. Use the Editor pull-down menu to enable commands. Then enter the
commands below.
Pone 415 300+ x ;
(Replace 300+x with the number you used.)
Test 0.79;
Alter -1;
(Makes H1 ‘less than.’)
useZ.
(Uses normal approx. to binomial)
Second: Use the Stat pull-down menu. Choose ‘Basic Stat’ and then ‘1 proportion.’
Check ‘summarized data’ and enter your n and 300+ x . Press Options. Set ‘test proportion’ as
0.79, alternative hypothesis as ‘less than’ and check ‘Normal distribution.’ Go.
Third: Use the pull-down menu again. But before you start put 10  x yesses and 50  x noes in
column 1. Uncheck ‘summarized data’ and let Minitab know that the data are in column 1 (C1).
Other options are unchanged.
252solngr2-071 2/19/07
Problem 1: Which of the following could be a null hypothesis? Which could be an alternative hypothesis?
Which could be neither? Why?
(i) p  3 , (ii) p  .3 , (iii) p  .3 , (iv)   3 , (v)   3 , (vi)   3 , (vii) s  3 , (viii)   3 , (ix)   5 ,
(x) p  .3 , (xi)   3 , (xii)   0 ,(xiii) p  .5.
Solution: Remember the following:
α) Only numbers like , p,  2 ,  and (the population mean, proportion, variance,
standard deviation and median) that are parameters of the population can be in a
hypothesis; x , p, s 2 , s and x.50 (the sample mean, proportion, variance, standard deviation
and median) are statistics computed from sample data and cannot be in a hypothesis because
a hypothesis is a statement about a population;
β) The null hypothesis must contain an equality;
γ) p must be between zero and one;
δ) A variance or standard deviation cannot be negative.
(i) p  3 could not be H 0 or H 1 since it contains a unreasonable value for a parameter.
(ii) p  .3 could not be H 0 or H 1 because p is a sample statistic, not a parameter.
(iii) p  .3 could be H 0 since it contains a parameter and an equality. H 1 would be p  .3 .
(iv)   3 , could be H 0 since it contains a parameter and an equality. H 1 would be   3 .
(v)   3 can’t be either H 0 or H 1 since a population standard deviation cannot be negative.
(vi)   3 can be H 0 since it contains a parameter and an equality. H 1 would be   3 .
(vii) s  3 can’t be either H 0 or H 1 since the sample standard deviation is not a parameter.
(viii)   3 can be H 0 since it contains a parameter and an equality. H 1 would be   3
(ix)   5 can be H 1 since it contains a parameter and a strict inequality. H 0 would be   5 .
(x) p  .3 could not be H 0 or H 1 since p, a parameter, cannot take values below zero or above 1.
(xi)   3 could not be H 0 or H 1 since  , a parameter, cannot take values below zero.
(xii)   0 could be H 0 since it contains a parameter and an equality. H 1 would be   0 . However, note
that this H 0 says that x is a constant.
(xiii) p  .5 can’t be either H 0 or H 1 since the sample proportion is not a parameter.
Learn to make  and call it ‘mu.’ It’s not a ‘u’ and you are too young to be unable
to adjust to using a Greek letter!
252solngr2-071 2/19/07
Problem 2: A rental firm believes that that the average time that a backhoe is rented is 4.2 days. In order to
verify this statement, a sample of rental records is taken, with the following results.
4, 2, 4, 3, 6, 2, 3, 2, 5, 3, 2, 4, 2, 2
You presumably know that n  14 , x  3.1429 , and  x  1 .1x . ( x is the last digit of your student
number.) Test the hypothesis that the mean is 4.2. Assume that the confidence level is 95%.
a) State your null and alternative hypotheses.
b) Find critical values for the sample mean and test the hypothesis. Show your reject region on a
diagram
c) Find a confidence interval for the sample mean and test the hypothesis. Show your results on a
diagram.
d) Use a test ratio for a test of the sample mean. Show your reject region on a diagram
e) Find a p-value for the null hypothesis using the Normal table and use the p-value to test the
hypothesis.
H :   4.2
Solution: a)  0
From the problem statement  0  4.2, x  3.1429 ,   1to 1.9, n  14 and
H 1 :   4.2
H 0 :    0
H 1 :    0
  .05 . From the problem statement According to Table 3 or the outline, if we wish to test 
and  is known, use Test Ratio z 
  x  z 2  x , where  x 
(ii)
1.9

n
x  0
x
, Critical Value xcv   0  z  x or Confidence Interval
2
and z  z.025  1.960 . The standard error is (i)
2
1
14

1
 0.2673 to
14
3.61
 0.5078
14
14
b) Critical Values: (i) xcv   0  z  x  4.2  1.9600.2673  4.2  0.5239 or 3.6761 to 4.7239.
2

Make a diagram! Show a normal curve with a mean at 4.2 and shaded rejection regions below 3.6761 and
above 4.7239.
(ii) xcv   0  z  x  4.2  1.9600.5078  4.2  0.9953 or 3.2047 to 5.1953
2
Make a diagram! Show a normal curve with a mean at 4.2 and shaded rejection regions below 3.2047 and
above 5.1953.
Since x  3.1429 , is in the lower rejection region , reject H 0 .
c) Confidence intervals: (i)   x  z  x  3.1429  1.9600.2673  3.1429  0.5239 or 2.6190
2
to 3.6668. Make a diagram! Show a normal curve with a mean at 4.2 and shade the confidence interval
between 2.6190 and 3.6668.
(ii)   x  z  x  3.1429  1.9600.0.5078  3.1429  0.9953 or 2.1476 to 4.1382. Make a diagram!
2
Show a normal curve with a mean at 4.2 and shade the confidence interval between 2.1476 and 4.1362.
Since  0  4.2 is not on these intervals, reject H 0 .
d) (i) z 
x  0
x

x   0 3.1429  4.2
3.1429  4.2
 3.95 . (ii) z 

 2.08 Make a
0.2673
x
0.5078
diagram! Show a normal curve with a mean at 0 and shaded rejection regions below -1.960 and above
1.960. Since our value of z is in the lower rejection region
e) Since this is a two-sided test (i) p  value  2Pz  3.95 
 2.5  P 3.95  z  0  2.5  .5000   .0000 . (ii) p  value  2Pz  2.08 
 2.5  P 2.08  z  0  2.5  .4812   .0376 . Since p  value    .05 ,
reject H 0 .
252solngr2-071 2/19/07
Problem 3: (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
30
40
60
60
60
50
Personalize the data as follows: add the digits of your student number to the last six numbers. Example: Ima
Badrisk has the student number 123456; so the last six numbers become {31, 42, 63, 64, 65, 56}.
Compute the sample standard deviation using the computational formula (if you don’t know what
that means, find out!). (If you did this correctly on the last assignment, just copy your work.) Test the
hypothesis that average time watching sports is below 70 hours.
a) Test the validity of the hypothesis using a confidence level of 95% and a critical value for the
sample mean. (You cannot test the validity of a hypothesis that you haven’t stated!)
b) Find an approximate p-value for the statement.
c) Will we reject the hypothesis at a significance level of (i) .001? (ii) .01? (iii) .10? Using
the p-value explain why.
Note that none of the problems beyond this point involve sample means.
Solution: Two data sets for computations of the variance are shown here. They will be referred to as
solution 1 and solution 2. The first represents a student number of 000000 and the second 999999.
x12
x1
index
1
2
3
4
5
6
7
8
9
10
11
50
48
65
74
66
30
40
60
60
60
50
603
2500
2304
4225
5476
4356
900
1600
3600
3600
3600
2500
34661
x
n1  n 2  11 ,
1
The means are x1 
s12 
s x1 
s 22
x
2
1
 nx12
n 1
s1

n
x

2
2

x 22
x2
50
48
65
74
66
39
49
69
69
69
59
657
x
 603 ,
x
1
n
2500
2304
4225
5476
4356
1521
2401
4761
4761
4761
3481
40547
2
1

 34661 ,
x
2
x
 657 and
603
 54 .8182 and x 2 
11
x
n
2

2
2
 40547 .
657
 59 .7273 .
11
34661  1154 .8182 2 1605 .61

 160 .56 s1  160 .56  12 .671
10
10
s12
160 .56

 14 .5965  3.821
n
11
 nx 22
n 1

40547  1159 .7273 2 1306 .15

 130 .61 s 2  130 .61  11 .429
10
10
s 22
130 .61

 11 .874  3.446
n
11
n
H 0 :   70 , H 1 :   70
s x2 
s2

a) From Table 3 x cv    tn 1 s x is the formula for a two sided critical value when the population
2
standard deviation is unknown. 1    .95   .05
n 1
10
 t .05
 1.812
critical value below 70 so use t .05

2
 .025
10
t n1  t.025
 2.228 , but we want a
2
252solngr2-071 2/19/07
Solution 1: xcv    tn1 s x  70 1.8123.821  70  6.9237  63.0763 Make a diagram. Show an
almost-Normal curve with a mean at 70, and ‘reject’ region below 63.0763. Since x1  54 .8182 falls in the
‘reject’ region, reject the null hypothesis.
Solution 2: x cv    tn 1 s x  70  1.812 3.446   70  6.2442  63 .7558 Make a diagram. Show an
2
almost-Normal curve with a mean at 70 and a ‘reject’ region below 63.7558. Since x 2  59 .7273 falls in
the ‘reject’ region, reject the null hypothesis.
x  0
b) From Table 3, t calc 
. Since this is a left-sided test we want Pt  t calc  . Solution 1:
sx
54 .8182  70
 2.193 . According to the t-table Pt  2.764   P(t  2.764 )  .01 ,
6.9237
Pt  2.228   P(t  2.228 )  .025 and Pt  1.812   P(t  1.812 )  .05 .
t calc 
c) Since our calculated t is between the last two values, we can say .01  pvalue  .05 . Since this is below
10%, we reject the null hypothesis for a significance level of 10%, but not for 1% or 0.1%.
Problem 4: (Ken Black) It is generally believed that 79% of US companies offer flexible scheduling . A
survey of 415 accounting firms finds that 300  x have flexible scheduling. ( x is the 2nd to last digit of your
student number.) Test the hypothesis that the proportion of accounting firms that offer flexible hours is
below that for other US firms.
a) State your null and alternative hypotheses.
b) Find a test ratio for a test of the proportion.
c) Find a p-value for this ratio and use it to test the hypothesis at a 5% significance level.
Solution: a) H 0 : p  .79 , H 1 : p  .79
b) z 
p  p0
p
. p 
p0 q0
300
309
.79 .21
 .7229 to p 
 .7446 .

 .00039976  .01999 p 
415
415
n
415
.7229  .79
.7446  .79
 3.35 to
 2.27
.01999
.01999
c) Pz  3.35   .5  .4996  .0004 and Pz  2.27   .5  .4884  .0116 Since these are both below .05,
they would result in a rejection of the null hypothesis.
Thus z 
Extra Credit Problem 5:
a) Finish problem 4 by finding an appropriate confidence interval for the proportion and showing
whether it contradicts the null hypothesis.
pq
Solution: The two sided interval for a proportion is p  p  z s p where s p 
and q  1  p . We
2
n
know that p  .7229 to p  .7446 . For a one-sided test, p  p  z s p , where z  z .05  1.645 . For the
.7229 .2771 
 .0004827  .02197 . The interval is
415
p  .7229  1.645 .02197   .7590 . For the second value q  1  p  1  .7446  .2554 , so that
first value q  1  p  1  .7229  .2771 , so that s p 
thus
.7446 .2554 
 .0004582  .02141 . The interval is thus p  .7446  1.645 .02141   .7798 . No
415
matter which of these is true, saying p  .7590 or p  .7798 contradicts the null hypothesis H 0 : p  .79 .
sp 
252solngr2-071 2/19/07
b) Use the data in problem 3 to test the hypothesis   10 .
Solution: The test ratio for this problem is  2 
n  1s 2
 02
and for 10 degrees of freedom and a two sided
10  20 .4832 and  210  3.2470 . The two rejection zones are below 3.2470 and above
5% test,  .2025
.975
20.4832.
We had two solutions.
s12
x

2
1
 nx12
n 1

34661  1154 .8182 2 1605 .61

 160 .56 or
10
10
2
s1  160 .56  12 .671 . This means that our calculated test ratio is  calc

For s 22 
2
 calc

x
2
2
 nx 22
n 1
n  1s 2
 02


n  1s 2
 02

10 160 .56 
 16 .06 .
100
40547  1159 .7273 2 1306 .15

 130 .61 or s 2  130 .61  11 .429 ,
10
10
10 130 .61
 13 .06 . Since neither of these is in the ‘reject zone described above, do not
100
reject the null hypothesis. We, of course can do this as a confidence interval or a critical value. Since this is
a 2-sided test, the confidence interval has the form  2 
us
n  1s 2
20 .4832
 2 
n  1s 2
3.2470
or
n  1s 2
.25 .5 2 
10 160 .56   2  10 160 .56
20 .4832
2
includes 10. If we want critical values, use s cv

3.2470
 .25 .5 2  02
n 1
. For the first solution, this would give
or 8.85    22.24 , an interval that
. The two critical values would be
20 .4832 10 2
3.2470 10 2
2
 204 .83 and s cv

 32 .47 . Since both values of s 2 fall within these
10
10
critical values, we cannot reject the null hypothesis.
2
s cv

c) Use Minitab to check your answer to problem 4. Do this three ways
First: Enter Minitab. Use the Editor pull-down menu to enable commands. Then enter the
commands below.
Pone 415 300+ x ;
(Replace 300+x with the number you used.)
Test 0.79;
Alter -1;
(Makes H1 ‘less than.’)
useZ.
(Uses normal approx. to binomial)
Second: Use the Stat pull-down menu. Choose ‘Basic Stat’ and then ‘1 proportion.’
Check ‘summarized data’ and enter your n and 300+ x . Press Options. Set ‘test proportion’ as
0.79, alternative hypothesis as ‘less than’ and check ‘Normal distribution.’ Go.
Third: Use the pull-down menu again. But before you start put 10  x yesses and 50  x noes in
column 1. Uncheck ‘summarized data’ and let Minitab know that the data are in column 1 (C1).
Other options are unchanged.
252solngr2-071 2/19/07
————— 2/16/2007 6:30:04 PM ————————————————————
Welcome to Minitab, press F1 for help.
MTB >
SUBC>
SUBC>
SUBC>
pone 415 300;
test 0.79;
alter -1;
usez.
#Done using commands
Test and CI for One Proportion
Test of p = 0.79 vs p < 0.79
95%
Upper
Sample
X
N Sample p
Bound
1
300 415 0.722892 0.759030
#Same pvalue as I got for x=300.
MTB > POne 415 309;
SUBC>
Test 0.79;
SUBC>
Alternative -1;
SUBC>
UseZ.
Z-Value
-3.36
P-Value
0.000
#Done using menu
Test and CI for One Proportion
Test of p = 0.79 vs p < 0.79
95%
Upper
Sample
X
N Sample p
Bound
1
309 415 0.744578 0.779790
#Same pvalue as I got for x=309.
Z-Value
-2.27
P-Value
0.012
MTB > print c1
Data Display
C1
y
n
n
n
y
n
n
n
y
n
n
n
y
n
n
y
n
n
y
n
n
y
n
n
y
n
n
y
n
n
y
n
n
y
n
n
n
n
n
n
n
n
MTB > POne C1;
SUBC>
Test 0.79;
SUBC>
Alternative -1;
SUBC>
UseZ.
Test and CI for One Proportion: C1
Test of p = 0.79 vs p < 0.79
Event = y
Variable
C1
X
11
N
60
Sample p
0.183333
95%
Upper
Bound
0.265500
Z-Value
-11.54
P-Value
0.000
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n