252y0411 2/19/04
ECO252 QBA2
Name _______KEY_________
FIRST HOUR EXAM Hour of class registered _____
February 25 2004
Class attended if different ____
Show your work! Make Diagrams! Exam is normed on 50 points. Answers without reasons are not
usually acceptable.
I. (8 points) Do all the following.
x ~ N 1,7  Do not make diagrams of x with zero in the middle. Make up your mind! If you are
diagramming x , put the mean in the middle; if you are diagramming z put zero in the middle. Copies of
x
the diagrams are available on a separate sheet. Remember z 
and that this equation implies that if

we have z and need a value of x , as in parts j-n, we use x    z .
25 .5  1 
 1 1
z
1. P1  x  25 .5  P
  P0  z  3.50   .4998 (or .49977) Make a diagram! Your
7 
 7
diagram for z shows a Normal curve with zero in the middle and the area shaded from zero to 3.50.
Because this area starts at zero, you do not need to add or subtract.
16  1 
 0 1
z
2. P0  x  16   P
  P 0.14  z  2.14   .0557  .4838  .5395 Make a diagram!
7
7 

Your diagram for z shows a Normal curve with zero in the middle and the area shaded from -0.14 to 2.14.
Because this area is on both sides of zero, add.
25 .5  1 

3. F 25 .5 (The cumulative probability up to 25.5) F 25 .5  P z 
  Pz  3.50 
7 

 Pz  0  P0  z  3.50   .5  .4998  .9998 (or .99977) Make a diagram! Your diagram for z shows
a Normal curve with zero in the middle and the area shaded from the extreme left to 3.50. Because this area
is on both sides of zero, add.
x.015 I’m quoting from the take-home part of the exam, but you really should make a diagram
showing a Normal curve centered at zero, with 1.5% above z .015 , 48.5% between z .015 and zero and 50%
4.
below zero. “To find z .03  z.015 , recall that since z .015 is 1.5% from the top of the distribution and 50% 2
1.5% = 48.5% from zero. So P0  z  z.15   .4850 . According to the Normal table
P0  z  2.17   .4850 . So z.015  2.17. ” Now, if we use x    z , we get x.015  1  2.17 7  16.19.
252y0411 2/19/04
II. (5 points-2 point penalty for not trying part a.)
A random sample is taken of the time spent waiting in line at a bank. The following data is found.
(Recomputing what I’ve done for you is a great way to waste time.)
x
x2
1
2
3
4
5
6
7
8
9
10
11
Sum
5
3
6
2
7
1
3
4
2
10
2
45
25
9
36
4
49
1
9
16
4
100
4
257
a. Compute the sample standard deviation, s , of the waiting times. Show your work! (2)
 x  45  4.090909
Solution: x 
s
2
x

2
 nx 2

n
11
n 1
72 .90909

 7.290909 s  7.290909  2.700
10
b. Compute a 90% confidence interval for the mean ,  .(3)
Given: x  4.0909 , s  2.70 , n  11 and   .10 . So s x 
s
n

257  114.090909 2
10
2.700
 0.8141 , DF  n 1  10 and
11
10
t .05
 1.812   x  t  s x  4.0909  1.812 0.8141  4.09  1.48 or 2.61 to 5.57.
2
2
252y0411 2/19/04
III. Do all of the following Problems (18 points) Show your work except in multiple choice questions.
1.
(Lee) We want to test whether the weight of an average bag of cat food is above 50 pounds at a 5%
significance level. What are our null and alternative hypotheses?
a) H 0 :   50 and H 1 :   50
b) H 0 :   50 and H 1 :   50
H 0 :   50 and H 1 :   50
d) * H 0 :   50 and H 1 :   50
e) H 0 :   50 and H 1 :   50
f) H 0 :   50 and H 1 :   50
(Lee) We want to test whether the weight of an average bag of cat food is above 50 pounds at a 5%
significance level. We take a sample of 144 and find a sample mean of 49 and a sample standard
deviation of 5. What is the value of our t or z test ratio?
a) -2.0
b) -2.5
c) *-2.4
d) -2.8
e) -3.0
x   0 49  50
1
 12
Solution: t 



 2.4
5
sx
5
5
 
144
 12 
(Lee)A company wishes to test the volume of fuel in a 50 gallon drum. Since the company assumes
that it is being cheated it uses H 0 :   50 . Assume   .01 and that from a sample of 31 it gets a
sample mean and a sample standard deviation of 1.1 gallons. From the results of the sample it
x  50
computes the ratio
. It should do the following:
 1.1 




 31 
a) Reject H 0 if the ratio is below -2.576.
b) Reject H 0 if the ratio is below -2.327.
c) Reject H 0 if the ratio is below -2.750.
d) *Reject H 0 if the ratio is below -2.457.
e) Reject H 0 if the ratio is below -2.576 or above 2.576.
f) Reject H 0 if the ratio is above one of the numbers in a-d.
g) None of the above – supply correct value or values.
x  50
Solution: The alternate hypothesis is H 0 :   50 . We thus do a left-sided test of t 
 1.1 




 31 
30
 2.457
against  tn 1  t .01
c)
2.
3.
3
252y0411 2/19/04
4.
(Lee)The price paid by students at an WCU for a new textbook is believed to be Normally
distributed with a population standard deviation of $15.75. A random sample of 50 students has a
sample mean of $47.80. A 90% confidence interval for the population mean is:
a) *$44.14 to $51.46
b) $43.04 to $52.56
c) $46.34 to $53.66
d) $44.10 to $51.50
e) $45.06 to $50.54
15 .75

 2.22739 , and
n
50
 1.645   x  z 2 s x  47.80  1.6452.22739  47.80  3.66 or 44.04 to 51.46.
Solution: x  47 .80 ,   15 .75, n  50 and   .10 . So  x 
z .05
s
5.
In problem 4, is the price paid by students in the sample significantly different from $44.12? (1)
Solution: Yes, since 44.12 is not on the correct confidence interval. (Answer depends in the
interval that you selected in problem 4.)
6.
The head of an accounting department wants to test to see if the proportion of bad invoices is
below 0.1%. She examines 10000 invoices (, finds 4 bad ones) and tries to test the null hypothesis
with a confidence interval   .05  . The appropriate confidence interval is:
a)
p  .0004  1.960 
b)
p  .0004  1.960 
c)
* p  .0004  1.645 
d)
p  .0004  1.645 
.001 .999 
10000
.0004 .9996 
10000
.0004 .9996 
10000
.0004 .9996 
10000
e) None of the above – write in correct interval
From the formula table we have:
Interval for
Confidence
Hypotheses
Interval
Proportion
p  p  z 2 s p
H 0 : p  p0
pq
n
q  1 p
sp 
H1 : p  p0
Test Ratio
z
p  p0
p
Critical Value
pcv  p0  z 2  p
p0 q0
n
q0  1  p0
p 
 H 0 : p  .001
The Hypotheses are 
 H 1 : p  .001
4
252y0411 2/19/04
7.
We wish to test that the median is above 39. If p is the proportion above 35, the correct null
hypothesis is
a) p  .5
b) * p  .5
c) p  .5
d) None of the above.
Solution: The table on page 8 of this solution says that if p is the proportion above  0 ,
 H 0 :   0
 H : p .5
corresponds to  0
. Since our alternate hypothesis is H 1 :  39 , the

 H 1 : p  .5
H 1 :   0
null hypothesis is H 0 :  39 , and p  .5 is our null hypothesis.
8.
(Lee) A reporter wants to poll people about an upcoming election. From previous polls, the
proportion of people favoring the incumbent president is about .6. The reporter will be satisfied if
the error of estimation is less than .04 with a confidence level of 90%. How many people should be
polled?
a) 167
b) 372
c) 260
d) 347
e) *406
The outline says “The usually suggested formula is n 
pqz 2
, but since p is usually unknown,
e2
a conservative choice is to set p  0.5 . This is the formula everyone forgets that we covered.”
We don’t need to use .5, since we have an estimate of p  .6, which implies q  .4. So we try
n
pqz 2
e2

.6.4 1.645 2
.04 2
 405 .9 .
8a. (Extra Credit) In section 8.7 of the text, the authors present sample size determination using the
n0 N
finite population correction factor. Equation (8.14) says that n 
. Assume that your
n0  N  1
answer to Problem 8 is correct and use this to adjust your answer to problem 8 when the country has
only 1000 voters.
n0 N
406 1000  406000


 288 .97 . Use 299.
Solution: n 
n 0  N  1 406  999
1405
5
252y0411 2/19/04
9.
(Lee) A sample of size 30 is taken from a Normally distributed population. The sample standard
deviation is 4 and we use the 5% significance level. Which of the following (Circle one or more!)
are (is) true? (3)
a) If the null hypothesis is H 0 :   5 , it will be rejected.
b) *If the null hypothesis is H 0 :   5 , it will not be rejected.
c) If the null hypothesis is H 0 :   5 , it will be rejected.
d) *If the null hypothesis is H 0 :   5 , it will not be rejected.
e) If the null hypothesis is H 0 :   5 , it will be rejected.
f) *If the null hypothesis is H 0 :   5 , it will not be rejected.
From the formula table (but the outline is better)
Interval for
Confidence
Hypotheses
Test Ratio
Interval
VarianceH 0 :  2   02
n  1s 2
n  1s 2
2 
2  2
Small Sample
 02
 .5 .5 2 
H1: :  2   02
s 2DF 
z 
VarianceH 0 :  2   02
 
Large Sample
 z  2DF 
2  2  2DF   1
H1 :  2   02
Critical Value
 cv2 
n  1s 2
 .25.5 2 
2
n  30, s  4 , df  n  1  29 and   .05 .  2 
n  1s 2
 02

n  14 2
52
 18 .5600 . For a
29
 42 .5570 and the rejection zone is above 42.5570 or
one-sided test either  n 1   .05
29
1n1   .95
 17 .7084 and the rejection zone is below 17.7084. For a 2-sided test,
29
29
1n1   .975
 16.0471 and  n1   .025
 42.7224 , so the refection zones are below
2
2
16.0471 and above 42.7224. Since 18.5600 is in none of these rejection zones, we would not
reject any of the null hypotheses.
A diagram will help here. Make a curve skewed to the right. Mark off a 2.5% upper ‘reject’ zone above
42.7224, a 5% upper ‘reject’ zone above 42.5570, a 2.5% lower ‘reject’ zone below 14.0471 and a 5%
lower ‘reject’ zone below 16.0471. Note that the ‘do not reject’ areas are largely the same for all three null
hypotheses because the null hypothesis is treated as if it were H 0 :   5 in all 3 cases.
6
252y0411 2/19/04
ECO252 QBA12
FIRST EXAM
February 25 2004
TAKE HOME SECTION
Name: _________KEY_______________
Student Number and class: _________________________
IV. Do at least 3 problems (at least 7 each) (or do sections adding to at least 20 points - Anything extra
you do helps, and grades wrap around) . Show your work! State H 0 and H 1 where appropriate. You
have not done a hypothesis test unless you have stated your hypotheses, run the numbers and stated
your conclusion. Use a 95% confidence level unless another level is specified.
1. You are an investment advisor and want to determine the suitability of the residents of a retirement
community as a clientele. You take a random sample of 10 incomes from the community and get the
following data (in thousands of dollars): 606 498 529 547 500 475 487 497 628 561.
Before you go any further, make your own unique data set by replacing the last digit of each of the first six
numbers with the digits of your Student number. For example, Seymour Butz’s student number is 976500,
so he will change the data to 609 497 526 545 500 470 487 497 628 561.
You want to check whether the mean income is above $500 thousand. Work in thousands. Use a
95% confidence level.
a. Find the sample mean and sample standard deviation of the incomes in your data, showing your work. (1)
b. State your null and alternative hypotheses (1)
c. Test the hypothesis using a test ratio (1)
d. Test the hypothesis using a critical value for a sample mean. (1)
e. Test the hypothesis using a confidence interval (1)
f. Find an approximate p-value for the null hypothesis. (1)
g. On the basis of your tests, should you decide to sell advising in this community? (1)
h. (Extra credit) Actually the data I used came from a skewed distribution. So assume that you expand your random
sample to 12 by adding the following data points: 501 619. (Seymour’s data would now read 609 497 526 545 500 470
487 497 628 561 501 619.) and test that the median is above 500. (3)
Solution: a) Find the sample mean and sample standard deviation of the incomes in your data, showing
your work. (1) Since everyone will get different answers, I used the original numbers.
x  5328 ,
x 2  283198 , n  10 .
So
x
x2
1
2
3
4
5
6
7
8
9
10
606
498
529
547
500
475
487
497
628
561
5328


 x  5328  532 .8
x
367236
248004
279841
299209
250000
225625
237169
247009
394384
314721
2863198
n
s

x
10
2
 nx 2
n 1
2863198  10 532 .82
 2715 .5111
9
s  2715 .5111  52 .1106
Note that most people who used the definitional formula instead of the computational formula used above
got the wrong answer.
b. State your null and alternative hypotheses (1)
b) H 0 :   500 , H 1 :   500 . Note that s x 
From the formula table:
Interval for
Confidence
Interval
Mean (
  x  t 2 s x
unknown)
DF  n 1
s

n
Hypotheses
H0 :   0
H1 :    0
2715 .5111
 271 .55111  16 .4788 .
10
Test Ratio
t
x  0
sx
Critical Value
xcv    t 2 s x
7
252y0411 2/19/04
x   0 532 .8  500

 1.990 .
sx
16 .4788
c. Test the hypothesis using a test ratio (1) The test ratio is t calc 
9
Since this is a one-sided, right-tail test, pick tn 1  t .05
 1.833 . Make a diagram. Show an
almost Normal curve centered at zero. The ‘reject’ zone is the area above 1.833. Since 1.990 is in
the ‘reject’ zone, reject the null hypothesis.
d. Test the hypothesis using a critical value for a sample mean. (1) Because the alternate
hypothesis is   500, we need a critical value for x above 500. Since this is a one-sided test,
we use t n 1  t 9   1.833 . The two-sided formula , x    t n1 s , becomes

cv
.05
0

2
x
xcv   0  tn1 s x  500  1.833 16.4788   500  30.2056  530 .2. Make a diagram showing an
almost Normal curve with a mean at 500 and a shaded 'reject' zone above 530.2. Since
x  532 .8 is above 530.2, we reject H 0 .
e. Test the hypothesis using a confidence interval (1) Because the alternate hypothesis is
  500, we need a ' ' confidence interval.   x  t  2 s x becomes   x  t s x
 532 .8  1.833 16.4788   532 .88  30.206 or   502.67 . Make a diagram showing an almost
Normal curve with a mean at x  532 .8 and the confidence interval above 502.67 shaded. Since
 0  500 is below 502.67 and thus noting the confidence interval, we reject H 0 .
f. Find an approximate p-value for the null hypothesis. (1) Because the alternate hypothesis is
9
 1.833 and
  500, the p-value is Px  532 .8  Pt  1.990  . The t table says that t .05
9
t .025
 2.262 . This means that Pt  1.833   .05 and Pt  2.262   .025 . Since 1.990 is
between these values of t, we can say .05  Pt  1.990   .025 , or .05  p  value  .025 .
g. On the basis of your tests, should you decide to sell advising in this community? (1) Since we
have shown that community mean income is above 500 thousand, we will sell in this community.
h. (Extra credit) Actually the data I used came from a skewed distribution. So assume that you
expand your random sample to 12 by adding the following data points: 501 619. (Seymour’s
data would now read 609 497 526 545 500 470 487 497 628 561 501 619.) and test
that the median is above 500. (3) The table in the outline is below.
Hypotheses about
Hypotheses about a proportion
a median
If p is the proportion If p is the proportion
above  0
below  0
 H 0 :   0

 H 1 :   0
 H 0 : p .5

 H 1 : p .5
 H 0 : p .5

 H 1 : p .5
 H 0 :   0

H 1 :   0
 H 0 :   0

H 1 :   0
 H 0 : p .5

 H 1 : p .5
 H 0 : p .5

 H 1 : p  .5
 H 0 : p .5

 H 1 : p  .5
 H 0 : p .5

 H 1 : p  .5
8
252y0411 2/19/04
We are testing to see if the median is above 500, so if p is the proportion above 500 and we use Seymour’s
 H 0 : p .5
data, our hypotheses are 
, and since seven of the eleven numbers we have if we drop 500 are
 H 1 : p  .5
above, we check the binomial table with n  11, p  .5 to find p  value  Px  7
 1  Px  6  1  .72559  .2744 . since this is below 5% we cannot reject H 0 .
2. Continue with the hypotheses of Problem 1. Instead of the sample standard deviation you found in
Problem 1 assume that the population standard deviation is 50. Use the sample mean that you found in
question 1.
a. Find a p-value for the null hypothesis. (1)
b. Create a power curve for the test. (95% confidence level) (6)
Solution: a. Find a p-value for the null hypothesis. (1) We are testing H 0 :   500 , H 1 :   500 . We
know that x  532 .8 n  10 and   .05 . We have now been told that   50 .
From the formula table we have:
Interval for
Confidence
Hypotheses
Test Ratio
Critical Value
Interval
Mean (
H0 :   0
x  0
  x  z 2  x
xcv    z  2  x
z
known)

H1 :    0
x
x 


2

x   0 532 .8  500
50 2
 250  15 .8114 . So z 

 2.07 . Since the
10
x
15 .8114
n
n
alternate hypothesis is H 1 :   500 , the p-value is (from the Normal table) Px  532 .8
 Pz  2.07   .5  .4808   .0192 .
b. Create a power curve for the test. (95% confidence level) (6) This is a one-sided test, so we are
only worried about values of the sample mean above 500 . We use  x  15.8114 . Because of the
alternate hypothesis, we wanted a critical value above 500, so we used x cv    z  x
 500  1.645 15 .8114   526 .01. You can’t do a power curve without a critical value.
Make a diagram with a Normal curve centered at 500 and a reject zone above 526.01. Half of
the distance between 500 and 526.01 is 13.00, so I add this to  0  500 four times and find the
power at 1  500 , 1  513 , 1  526 .01, 1  539 , and 1  552 . Remember that


Power  1   , and that   P Not rejecting H 0 H 0 is false . We do not reject H 0 if x is less

x  1 
than or equal to the critical value of 526.01, so   Px  526 .01   1   P z  cv

x 

526 .01  1 

 Pz 
.
15 .8114 

9
252y0411 2/19/04
1  500 :
We get the following:
526 .01  500 

  Pz 
  Pz  1.645   .5  .4500  .95 . power  1    .05 . You don’t
15 .8114 

have to do this one since you learned in class that if 1   0 ,   1   and power   .
1  513 :

  Pz 

526 .01  513 
  Pz  0.82  .5  .2939  .7989 power  1    .2011
15 .8114 

1  526 .01 :   P  z 
526 .01  526 .01 
  Pz  0  .5. You don’t have to do this one since you learned
15 .8114


in class that if 1  xcv ,   .5 and power  .5.

1  539 :   P  z 

526 .01  539 
  Pz  0.82  .5  .2939  .2061 . power  1    .7939
15 .8114 
526 .01  552 
  Pz  1.64  .5  .4495  .0505 . power  1    .9495.
15 .8114 

If we are really picky, we could try a point 13 higher than this.
526 .01  565 

1  565 :   P  z 
  Pz  2.47  .5  .4932  .0068 . power  1    .9932.
15 .8114 

You now have the power for 5 points above or equal to 500. Make a diagram. Put zero through
one on the y-axis and 500 to 570 on the y-axis.

1  552 :   P  z 
3.
a. Using the sample standard deviation that you found in question 1, test the hypothesis that the standard deviation is
48.(1)
b. Test that the standard deviation equals 48. (1)
c. Do a 2-sided confidence interval for the standard deviation (1)
d. (Extra Credit) Test the hypothesis that the standard deviation is above 48 using a confidence interval. (2)
e. Create a two sided confidence interval for the mean using a the sample mean you found, a
population standard
deviation of
50 and a confidence level of 97%. (2)
f. Assume that you wanted to create a 99% confidence interval for the mean income with an error of 2 (thousand) ,
above
how
level
large a sample would you need? Assume that the population standard deviation is 50. (1)
g. With some illustrative calculations show the effect on the size of the sample you need in f) of (i) a higher confidence
and (ii) a larger population variance. (2)
Solution: a. Using the sample standard deviation that you found in question 1, test the hypothesis that the
standard deviation is above 48.(1) : From the formula table (but the outline is better)
Interval for
Confidence
Hypotheses
Test Ratio
Critical Value
Interval
VarianceH 0 :  2   02
n  1s 2
n  1s 2
n  1s 2
2


2 
2  2
Small Sample
cv
 02
 .25.5 2 
 .5 .5 2 
H1: :  2   02
VarianceLarge Sample
 
s 2DF 
 z 2  2DF 
H 0 :  2   02
z 
2  2  2DF   1
H1 :  2   02
Only the test ratio method is normally used. We know that x  532 .8, s  52 .1106 , n  10 and
 H 0 :   48 2 n  1s 2 952 .1106 2
 

 10 .6075 . The upper 5% value of chi 02
48 2
 H 1 :   48
  .05 . 
2 n 1
  .2059   16 .9190 . Make a diagram showing a chi-squared distribution
squared is  .05
centered at 9 and a shaded ‘reject’ zone above 16.9190. Since  2  10.6075 is not in the ‘reject’
zone, do not reject H 0 .
10
252y0411 2/19/04
b. Test that the standard deviation equals 48. (1) Use the same facts as in a) to test
 H 0 :   48
. This is a 2-sided test. The upper 2.5% value of chi-squared is

 H 1 :   48
n 1
9 
n 1
9 
 .2025
  .2025
 19 .0228 . The lower 2.5% value of chi-squared is  .2975
  .2975
 2.7004 .
Make a diagram showing a chi-squared distribution centered at 9 and shaded ‘reject’ zones
below 2.7004 and above 19.0228. Since  2  10.6075 is not in the ‘reject’ zones, do not reject
H0 .
c. Do a 2-sided confidence interval for the standard deviation (1) The supplement gives the
formula
n  1s 2
 2
 2 
2
n  1s 2
2
1
, which is the same as  2 
2
chi_squared were found in b), so we get
n  1s 2
 .25 .5 2 
above. The values of
9 52 .1106 2
9 52 .1106 2
, which becomes
 2 
19 .0228
2.7004
1284 .75   2  9050 .37. . If we take square roots we get 35.843    95.133 .
d. (Extra Credit) Test the hypothesis that the standard deviation is above 48 using a confidence
 H :   48
interval. (2)  0
. Since the alternate hypothesis is ‘greater than,’ we want an interval
 H 1 :   48
with ' '. For a one-sided interval, we want  2 
n  1s 2 , where the appropriate chi-squared is
2
either the lower or upper 5% value, whichever will give us the smaller value. We already know
 .2059   16 .9190 . We find  .2959   3.3251 . The first of these two value will give us a lower limit,
952 .1106 2
 27 .72 . If we take square roots, we get   5.265 . Make a diagram
16 .9190
showing a confidence interval in the entire area above 5.265. Since 48 is in this area, we do not
reject H 0 .
so  2 
e. Create a two sided confidence interval for the mean using a the sample mean you found, a
population standard deviation of 50 and a confidence level of 97%. (2) This is a two-sided
interval, so we need values on both sides of the sample mean. x  532 .8 n  10 and
  .03 . We use   xcv  z 2  x , where  x 
z .032  z.015 , recall that since z .015


2

50 2
 250  15 .8114 . To find
10
n
n
is 1.5% from the top of the distribution and 50% - 1.5% =
48.5% from zero. So P0  z  z.15   .4850 . According to the Normal table
P0  z  2.17   .4850 . So z.015  2.17. If we use this in the formula for a confidence interval,
  x  z 2  x  532.8  2.1715.81114  532.8  34.31 or 498.41 to 567.11.
11
252y0411 2/19/04
f. Assume that you wanted to create a 99% confidence interval for the mean income with an error
of 2 (thousand) , how large a sample would you need? Assume that the population standard
deviation is 50. (1) n 
n
2.576 2 50 2
2
2
z 2 2
e2
. Use z.005  2.576 ,   50 and e  2.
 64 .4 2  4147 .36 . So use at least 4148.
g. With some illustrative calculations show the effect on the size of the sample you need in f) of
(i) a higher confidence level and (ii) a larger population variance. (2)
1.96 2 50 2
 49 .0 2  2401 . So,
(i) Try a 95% confidence interval using z .025  1.960 . n 
22
since a lower confidence level gives us a smaller sample, a higher confidence level must give us a
larger sample.
2.576 2 100 2
 128 .8 2  16589 .44 , a
(ii) Double the standard deviation to 100 and get n 
22
sample size four times as large.,
4.
a. Assume that a travel agent claims that the average number of clients coming into the office in an hour has a Poisson
distribution with a mean of 5. Test this hypothesis with a 5% significance level if the following occur:
(i) 2 clients in an hour. (1)
(ii) 4 clients in 2 hours.(1)
(iii) 24 clients in 12 hours.(2)
b. A human resources director states that 80% of our employees are capable of being trained for a higher level job. A test
is
given to qualify a randomly picked sample of 200. According to the test 149 of the employees are ready for
additional
training.
(i) Is the human resources director wrong? (A yes or no without supporting calculations gets you nowhere. Do not use a
test
ratio in this section.) (2)
(ii) Find a p-value for the null hypothesis. (2)
(iii) (Extra Credit – hard) Find a power curve for this test or, at least find the power for a few values of the population
proportion. (2+++)
Solution: a. Assume that a travel agent claims that the average number of clients coming into
H 0 : Poisson5
the office in an hour has a Poisson distribution with a mean of 5. 
. Test
H 1 : Not Poisson5
this hypothesis with a 5% significance level   .05  if the following occur:
(i) 2 clients in an hour. (1) Using a Poisson table with a parameter of 5 we find
Px  2  .26503 . If we double this we get a p-value of .53006   . So do not reject
H0.
(ii) 4 clients in 2 hours.(1) In 2 hours the mean (parameter) becomes 10. So using a
Poisson table with a parameter of 10 we find Px  4  .02925 . If we double this we
get a p-value of .05850   . So do not reject H 0 .
(iii) 24 clients in 12 hours.(2) In 24 hours the mean (parameter) becomes 60. We do not
have a Poisson table with a parameter of 60, so we use a Normal distribution with a

24  60 
  Pz  4.65 
mean and variance of 60. We find Px  24   P z 

60 

 .5  .5  0 . If we double this we still get a p-value of 0   . So reject H 0 . This
illustrates the fact that the test gets stronger as the time gets longer.
12
252y0411 2/19/04
b. A human resources director states that 80% of our employees are capable of being trained for a
higher level job. A test is given to qualify a randomly picked sample of 200. According to the
test 149 of the employees are ready for additional training.
(i) Is the human resources director wrong? (A yes or no without supporting calculations gets you
 H : p  .8
nowhere. Do not use a test ratio in this section.) (2) Assume   .05 . Use either  0
 H 1 : p  .8
 H 0 : p  .8
. The second really makes more sense, I think. From the formula table we have:

 H 1 : p  .8
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
p
H1 : p  p0
pq
p0 q0
sp 
p 
n
n
q  1 p
q0  1  p0
  .05 , n  200 and p 0  .8.  p 
p
x 149

 .745 . So s p 
n 200
p0 q0
.8.2

 .000804  .02828 .
n
200
p0 q0
.745 .255 

 .000950  .03082 .
n
200
 H : p  .8
Two sided version -  0
Critical Value - pcv  p0  z  p  .8  1.960.02828
2
 H 1 : p  .8
 .8  .055 or .745 to .855. Since the reject zone is below .745 and above .855 and p  .745 is
not in it, do not reject H 0 .
Confidence Interval p  p  z s p  .745  1.960.03082
2
 .745  .060 or .685 to .805. Since this interval includes .8, do not reject H 0 .
H 0 : p  .8
One sided version - 
Critical Value - pcv  p0  z  p  .8 1.645.02828
H 1 : p  .8
 .8  .047  .753 . Since the reject zone is below .753 and p  .745 is in it, reject H 0 .
Confidence Interval p  p  z s p  .745  1.645.03082  .745  .051  .796 . Since this interval
( p  .796 ) does not include .8, reject H 0 . So I think the director was wrong.
(ii) Find a p-value for the null hypothesis. (2) z 
p  p0
p

.745  .8
 1.94 . Pz  1.94 
.02828
Pz  0  P1.94  z  0  .5  .4738  .0262 . So if the problem is 1-sided, we have
p  value  .0262 and if it’s 2-sided p  value  .0524 .
(iii) (Extra Credit – hard) Find a power curve for this test or, at least find the power for a few
values of the population proportion. (2+++)
Hey! I’m not going to do this unless someone tries it! There is an example of a single solution in
the solution to an earlier exam. Use the critical value above, pick values of p1 below (and maybe
above) .8, like .775, .753, .725 and .7. recompute  p 
p1 q1
for each one and find the
n
probability of being above the critical value.
13