Math 252 Final Exam Questions: Name and ID
Number:
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7 |
EXPACTATÝON(5P) |
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NOT:
SHOW ALL YOUR WORK. ANSWERS WITHOUT WORK WILL NOT BE
GRADED!!!
1.
[30 points] The monthly rental for a two-bed room apartment in
a)
What is the probability that a simple random sample of a 49 two bedroom apartments
will provide a sample mean monthly rental within
75YTL of the population mean?
b) What is the probability that
a simple random sample of a 49 two bedroom apartments will provide a sample
mean monthly rental within
25YTL of the population mean?
c)
Discuss the results in parts (a) and (b). Compare the results. In order to have
same probability in part (b) what should we do?
Solutions:
a)
P(
-75<
<
+75)=P(
)
P(
-75<
<
+75)=P(
) 
P(
-75<
<
+75)=P(-5,25<Z<5,25)=almost 1
b)
P(
-25<
<
+25)=P(
)
P(
-25<
<
+25)=P(
) 
P(
-25<
<
+25)=P(-1,75<X<1,75)=0,92
c) In the second case
probability is less than the first one because margin of error around the mean
is smaller in the second case. In order to have same probability in the second
case we need to increase the sampling size.
2.
[20 points] The number of hours Turks sleep each night
varies considerably with 12% of the population sleep less than six hours to 3%
sleeping more than eight hours. The following sample of 25 individuals reports
the hours of sleep per night
|
6,5 7,3 5,6
5,1 6,5 6,6
6,9 6,2 6,4
7,1 7,9 8,5
6,0 6,9 6,6
7,5 7,2 8,6
7,1 5,5 6,4
6,7 7,7 5,9
6,3 |
a)
What is the point estimate of the population mean number of hours of sleep each
night?
b)
Assume that the population has a normal distribution; develop a 95% confidence
Interval for the population mean number of hours of sleep each night
Solution:
a)

b)

3.
[20 points] A Turkish Retail Survey found that 9% of Turkish customers used
internet to buy gifts during the 2005 holiday season. If 1089 customers
participated in the survey, what is the margin of error and what is the
interval estimate of the population proportion of customers using the internet
to buy gifts? Use a 95% confidence.

4.
[30 points] Suppose a new production method will be
implemented if a hypothesis test supports the conclusion that a new method
reduces the mean operating cost per hour.
a) State the appropriate null and alternative
hypothesis if the mean cost for the current production method is 120YTL
b)
What is the TYPE 1 error in this situation? What are the consequences of making
this error?
c)
What is the TYPE 2 error in this situation? What are the consequences of making
this error?
a)
Null Hypothesis:
>120 Alternative:
<120
c)
Claiming
<120YTL when the new method does not lower costs. A
mistake could be implimenting the method when it does not help
d)
Concluding
>120YTL when the method really would lower costs. This
could lead to not implementing a method that would lower the costs.
5.
[35 points] The population mean earnings per share for
financial services corporation including AK BANK, GARANTÝ, FORTIS, ÝÞ BANKASI
and CITI BANK was 3YTL. A sample of 11 financial services corporation provided
the followings per share data:
1,95 2,66 3,79 4,01 3,17 3,05 2,39 2,2 3,14 2,98 2,66
a)
Formulate the null and alternative hypotheses that can be used to determine
whether the population means earnings per share in 2001 differ from 3YTL
reported in 2000
b)
Using
=0.05, what are critical values for the statistic, and what
is the rejection rule
c)
Compute the sample mean
d)
Compute the sample standard deviation
e)
Compute the value of test statistic.
f)
What is your conclusion?
g)
Approximate the P-Value.
Solution
a)
![]()
b)

Then The rejection
Rule is t>2,228 or t<-2,228
c)

d)

e)

f)
since -0,1435>-2,228
then we retain the null hypothesis
g)
P-Value=2P(t<-0,1435)=2P(t>0,1435)=0,888746
(EXACT)

APPROXIMATION ![]()
6. [35 points] The following shell of an ANOVA source table needs to be completed
then answer a through

a)
To what does SOURCE refer?
b)
How many group means are being compared? How do you
know this?
c)
How many total subjects were in the study? How do
you know this?
d)
How many subjects are in each group? How do you
know this?
e)
How many degrees of freedom are there?
f)
Was the null hypothesis rejected or retained? WHY?
g)
Two of the group means are 64.2 and 31.7. Can you
use these mean values to interpret this finding for independent variable X and
dependent variable Y? WHY?
Solution:
a)
Source is telling us where the total variabilities
around the overall mean is caming from.
b)
Since MS=156/(k-1) then k-1=156/52 then k-1=3 and
k=4 so there are 4 groups
c)

d)
There is not sufficent information to find number
of elements of each group
e)
3 for Between Groups and 19 for Within groups
f)
Since F=3,64 >3,133 then we can reject the Null
Hypotesis
g)
No because there is not any relation between the
means of the groups since we reject the Null Hypotesis
7. [30 points] With the growth of
internet service providers, a researcher decides to examine whether there is a
correlation between cost of internet service per month (rounded to the nearest
dollar) and degree of customer satisfaction (on a scale of 1 - 10 with a 1
being not at all satisfied and a 10 being extremely satisfied). The researcher
only includes programs with comparable types of services. A sample of the data
is provided below.
|
dollars |
satisfaction |
|
11 |
6 |
|
15 |
4 |
|
9 |
9 |
|
5 |
6 |
|
19 |
5 |
|
25 |
10 |
a)
Develop a scatter
diagram for these data with the dollar is independent variable and satisfaction
dependent variable
b)
Develop an estimated
regression equation that can be used to predict satisfaction given the amount
of dollar spent for internet users
c)
Use the estimated
regression equation to predict satisfaction score with 20 dollars spending for
the service.
Solution:
a)
b)
|
|
X |
Y |
x-meanx |
y-meany |
(x-meanx)*(y-meany) |
(x-meanx)^2 |
|
|
11 |
6 |
-3 |
-0,66667 |
2 |
9 |
|
|
15 |
4 |
1 |
-2,66667 |
-2,666666667 |
1 |
|
|
9 |
9 |
-5 |
2,333333 |
-11,66666667 |
25 |
|
|
5 |
6 |
-9 |
-0,66667 |
6 |
81 |
|
|
19 |
5 |
5 |
-1,66667 |
-8,333333333 |
25 |
|
|
25 |
10 |
11 |
3,333333 |
36,66666667 |
121 |
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TOTAL |
84 |
40 |
|
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22 |
262 |
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14 |
6,666667 |
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b1= |
0,083969466 |
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b0= |
5,491094148 |
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y=5,49+0,08X
c)
y=5,49+0,08*20
y=5,49+0,16=5,65 close to 6