Spring 12 QMB3250 Exam 1 Applicable Fall 10 Exam 1 Solutions

7/31/2019 Spring 12 QMB3250 Exam 1 Applicable Fall 10 Exam 1 Solutions

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Applicable Fall 10 Exam 1 Solutions

Notes:

The Fall 2010 semester was the first time that the true/false problems wereincorporated on this exam.

Dr. Thompson provided students with a formula sheet for this exam (it issupposedly posted on e-learning)

On this exam, Dr. Thompson was a bit of a stickler about rounding, so try tomake your estimates to about 4 or 5 decimal places where you can. This is, without a doubt, the most difficult QMB exam I have ever seen (so dont

feel bad if you struggle with it).

1. Answer: (C)

When constructing an interval for the median, we use the given formula .4n 2 i.e. .4(18)

2 = 5.2 rounds to 5.

Therefore, we will select the 5th smallest value from the sample (since the question only

wants the lower bound).

Unfortunately, the numbers are not in order but if you go searching, you will find that

the 5th lowest value overall is 13.3

2. Answer: (B)

If we are testing for the idea that the firs mean moisture content is lower than pines wemust be testing the alternative

H1: Pine < Fir H1: Pine Fir < 0 (and thus H0: Pine Fir > 0)


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3. Answer: (A)

Since we can assume equal variance, we will use the pooled procedure, but first we have

to get the pooled variance

+

=

+

2 2

2 1 1 2 2

1 2

( 1) ( 1)

2p

n s n ss

n n=

2 2(10 1)3.885 (8 1)3.366

10 8 2

+

+

= 13.4468

Now that we have the value of the pooled variance, lets use it in the SE for independent

means

SE = 2

1 2

1 1

ps

n n

+

=

1 113.4468

8 10+

= 1.7394

4. Answer: (B)

As per the formula sheet, we know that to find the t statistic for two independent means,

we find the difference in sample means and divide by the SE

1 2

2

1 2

1 1

p

x xt

sn n

=

+

1 2x x

tSE

= 14.13 19.0375

1.91t

= = 2.5694

Note: Remember that this question deals with two independent samples of observations.

Therefore, we should NOT use the given mean of the differences. We only look at the

differences when we are dealing withpaired samples. Those differences are completely

unnecessary and are meant to confuse/distract you.


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5. Answer: (B)

Since we are conducting a one-sided confidence interval for a proportion, we will use

the z distribution i.e. df= and we will be careful as to how we get the critical value.

Since we are doing a one-sided interval, we do NOT use the confidence levels that welisted at the top of the t table. Instead, we use our own (written in) onesided confidence

levels.

df90% 95% 97.5% 99% 99.5%

1sided

CI

: : : : :

100-up 1.282 1.645 1.960 2.326 2.576

From the problem, we know thatp = 67/157 = .426752

Using the formula for a one-sided CI for a proportion with the FPCF

( )

*1

1

p p N np z

n N

( ).426752 1 .426752 1350 157.426752 1.645

157 1350 1

+

= .4878

6. Answer: (A)

Since there are N= 1,350 students, we estimate that at most (.4238)(1,350) = 572.13 572of them will take the exam.

Note: The actual value of 572.13 is off solely due to rounding error i.e. we cannot have

.13 people take the exam so we round to the closest whole number.


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7. Answer: (B)

Considering Kohler as group 1 and Russell as group 2, we will use the information

regarding two independent proportions with the following sample values

p1 = 27/150 = .18 andp2 = 25/200 = .125

As per the formula sheet, we know that the standard error for two independent

proportions is given as

SE =( ) ( )

1 1 2 2

1 2

1 1p p p p

n n

+ =( ) ( ).18 1 .18 .125 1 .125

150 200

+ = .03913

8. Answer: (D)

Since the interval goes from negative to positive values (i.e. it crosses zero) we do NOT

believe that there is a significant difference in the proportion of defects for these two

methods.

9. Answer: (B)

As per the review, we know that the degrees of freedom used when we can assume

EQUAL variances is df= n1 + n2 2 = 14 + 9 2 = 21

Note: The dfthat we use when we cannot assume equal variances is generally less than

the dfthat we use when we can assume equal variances (or, at least, it cannot be larger).

10. Answer: (A)

IQR = Q3 Q1 = 55 25 = 30

Note: We can get the values of Q3 and Q1 by looking at the right and left edges of the

boxplot, respectively.


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11. Answer: (B)

In order to determine if outliers exist, we can do an outlier check by computing the

upper and lower fences

lower fence = Q1 1.5(IQR) = 25 1.5(55 25) = 20 upper fence = Q3 + 1.5(IQR) = 55 + 1.5(55 25) = 100

Therefore, if we have any observations below 20 or above 100, these values will be

considered outliers. Since we have no observations this large, it looks as though we do

not have any outliers.

12. Answer: (A)

Recall that the Wilcoxon Rank Sum test is testing the medians of these two groups i.e. this

test acquires its test statistic by ranking all values.

As per the output, we can determine that the mean of the ranks for one group is 217.5/14

= 15.54 and the mean of the ranks for the other group is 58.5/9 = 6.5

However, how do we know which one is which???

Well, by looking at the box plots, we can see that the values for the West are generally

lower than those for Central.

Therefore, the lower mean rank should be associated to the lower group i.e. the West.

13. Answer: (A)

Though it is not explicitly stated in the question, we can reasonably assume that the

hypotheses being tested are as follows

H0: The median for West and Central are the same vs. H1: The median for West is lower

Since the pvalue = .00182 < = .05, we reject H0 and conclude that service times are

generally lower in the West i.e. the median for the West is lower.

Note: Since no level was given in the problem, we can assume = .05


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14. Answer: (D)

Since each golfer is driving the ball with both cubs, the data values must be paired. As

per the formula sheet, we know that the formula for a confidence interval for paired

means is

* d

d

sd t

n

ME =* d

d

st

n

If we look on the t table with 95% confidence and df= nd 1 = 9, we get a critical t value

of 2.262

80% 90% 95% 98% 99% Clevel

df

: : : : : :9 1.383 1.833 2.262 2.821 3.250

ME = * d

d

st

n

=3.26

2.262

10

= 2.332

Note: Recall that when the data values are paired, we are ONLY concerned with the

differences.

15. Answer: (B or D)

Since the company wishes toprove that the new club can drive a gold ball longer than the

leader, this must be the alternative hypothesis

H1: NewClub > LeaderNewClub Leader > 0 d > 0

thus, H0: NewClub < LeaderNewClub Leader < 0 d < 0

Therefore, both (B) and (D) are equally correct answers.


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16. Answer: (A)

Note: This question is asking us to effectively test the hypotheses laid out in the

previous question.

In testing the previous hypotheses, our t test statistic would be

0

STAT

H value

d

d

dt

s

n

= =2.810 0

3.26

10

= 2.7258

So, we know that the second half of this statement is true. Now, lets determine the

critical t value using the one-tailed = .05

.10 .05 .025 .01 .005 1Tail

df

: : : : : :

9 1.383 1.833 2.262 2.821 3.250

Therefore, both of these statements are correct.

Notes:

We know that this is a one-tailed test because we are testing H1: NewClub > Leader(that is, our alternative hypothesis has a greater than or less than symbol in it)

This question is a bit tricky because Dr. T claims that the critical value is 1.8331(which he most likely got from a computer) even though the table rounds this

value to 1.833


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Notes on Questions #1720

In my opinion, this question is written in a somewhat vague/poor fashion in thatits not entirely clear what is going on.

What we are supposed to take away from this question is that this company sellsmerchandise to most everyone at full price but occasionally people qualify for a

discount.

However, if someone qualifies for a discount and simply doesnt ask for it, theywill NOT receive the discount.

Therefore, all invoices that have a difference of 0 were those in which themerchandise was sold at the correct price. This occurs if either (1) the customer

simply does not qualify for a discount (and thus paid full price) or (2) the

customer qualifies for a discount and asks for/gets it. Thus, the only differencesthat are in the data set result from those customers who do qualify for a discount

but subsequently do not ask for/acquire it.

17. Answer: (B)

As per the output, there are 158 data points that have a difference of 0. However, as per

what is discussed above, this occurs if the customer either does not qualify for a

discount or does qualify for a discount and asks for it.

Therefore, to claim that there were 158 invoices that had no discount available is not

entirely correct.

18. Answer: (A)

As per the output, we see there were 17 differences that did not = 0 i.e. these are

invoices in which the customer DID qualify for a discount but did not get it.

Therefore, we know that 17/175 = .0971 = 9.71% of customers qualified for discount but

did no get it.


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19. Answer: (A)

As per the output (under the intermediate calculations), we see that the sum of the

differences is $1,265.94 i.e. there were $1,265.94 worth of discounts that were NOT taken.

Therefore, the mean discount not taken per these invoices is $1,265.94/17 = $74.47

Note:

We CANNOT use the value that is the average discount in the sample of 7.233because this includes all the differences of zero in their calculations.

That is, this value is computed from 1265.94/175 = 7.233

20. Answer: (B)

As per the output, we know that the upper limit for the TOTAL difference (i.e. the total

monetary value of discounts foregone by customers) is $39.960.33

21. Answer: (E)

As per the formula sheet, we know that the formula for a confidence interval for one

proportion is given as

( )

*1p p

p zn

p ME ME =( )

*1p p

zn

Therefore, we will compute the margin of error withp = 38/188 = .2021 and z* = 1.96 (i.e.

match 95% confidence and df= on the t table)

ME =( )

*1p p

zn

=

!

"#$%

#&'&" "! #&'&"( )"((

= .0574


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22. Answer: (B)

Lets use the confidence interval formula from the previous question

( )*1p p

p zn

p ME .2021 + .0574 .1447 to .2595

That is, we are 95% confident that between 14.47% and 25.95% of the homes in this

community have a computer but no internet access.

That is, we do NOT believe that 28% of homes have a computer but no internet access

because the interval does NOT include .28 or 28%

23. Answer: (B)

As per the formula sheet, the standard error (for one mean w/ FPCF) is given as

SE =1

s N n

Nn

=38.47 663 27

663 127

= 7.2567

24. Answer: (D)

Recall that the shortcut to create a confidence interval for a TOTAL is to simply createthe confidence interval for the mean and then multiple by the population size.

In the formula, we need to acquire the critical t value with 95% confidence and df= n 1

= 27 1 = 26 and we get a critical t of t* = 2.056

*

1

s N nx t

Nn

*x t SE 172.45 + (2.056)(7.2567) 157.53 to 187.37

Since this is the confidence interval for the mean, we can get the confidence interval for

the total by multiplying by the population size i.e.

157.53(663) to 187.37(663) $104,442.39 to $124,226.31


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25. Answer: (B)

As per the review, we generally do not incorporate the FPCF unless the sample

represents more than 5% of the population.

However, this only holds true for a confidence interval for one mean (and oneproportion).

In #24, however, we NEED the population size in order to create the confidence interval

for the population total.

Therefore, it does make an impact.

Note: In my opinion, this is a poor/deceptive question in that it tricks you into

referencing what you know about the FPCF but then turns it on its end because the

population size is a necessity in determining the population total.

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Spring 12 QMB3250 Exam 1 Applicable Fall 10 Exam 1 Solutions