STAT 200 Final Examination

This is an open-book exam for STAT 200 Introduction to Statistics course. Answer all 20 questions and show all work and reasoning. Complete the exam individually and sign the honor statement. Record your answers and work in this document.

STAT 200 Final Examination Fall 2018 OL4_US2 Page 1 of 14
STAT 200 Introduction to Statistics Name______________________________
Final Examination: Fall 2018 OL4/US2 Instructor __________________________
Answer Sheet
Instructions:
This is an open-book exam. You may refer to your text and other course materials as you work
on the exam, and you may use a calculator.
Record your answers and work in this document.
Answer all 20 questions. Make sure your answers are as complete as possible. Show all of
your work and reasoning. In particular, when there are calculations involved, you must
show how you come up with your answers with critical work and/or necessary tables.
Answers that come straight from calculators, programs or software packages without
explanation will not be accepted. If you need to use technology to aid in your calculation,
you have to cite the source and explain how you get the results. For example, state the
Excel function along with the required parameters when using Excel; describe the detailed
steps when using a hand-held calculator; or provide the URL and detailed steps when using
an online calculator, and so on.
Show all supporting work and write all answers in the spaces allotted on the following pages.
You may type your work using plain-text formatting or an equation editor, or you may hand-
write your work and scan it. In either case, show work neatly and correctly, following standard
mathematical conventions. Each step should follow clearly and completely from the previous
step. If necessary, you may attach extra pages.
You must complete the exam individually. Neither collaboration nor consultation with
others is allowed. It is a violation of the UMUC Academic Dishonesty and Plagiarism
policy to use unauthorized materials or work from others. Your exam will receive a zero
grade unless you complete the following honor statement.
Please sign (or type) your name below the following honor statement:
I understand that it is a violation of the UMUC Academic Dishonesty and Plagiarism policy to
use unauthorized materials or work from others. I promise that I did not discuss any aspect of
this exam with anyone other than my instructor. I further promise that I neither gave nor
received any unauthorized assistance on this exam, and that the work presented herein is
entirely my own.
Name _____________________ Date___________________

STAT 200 Final Examination Fall 2018 OL4_US2 Page 2 of 14
Record your answers and work.
Proble
m
Number
Solution
1
Answer:
(a) (iii)
(b) (i)
Justification for (a) and (b):
(a) Set of income responses from al female in the US is considered to be a
parameter because is shows the median of the given population.
(b) Set of income responses from 3500 females in the US would be a sample
because it represents the population.
2
Answer:
(a) (iii)
(b) (iii)
Justification for (a) and (b):
(a) Student id can easily be represented in a specified manner and also a
numerical variable and hence it would be ordinal.
(b) Systematic sampling is suitable procedure to select random sample. This is
because each and every nth item selected would take a random number.

STAT 200 Final Examination Fall 2018 OL4_US2 Page 3 of 14
3
Answer:
(a)
(b) % of midterm scores be at least 80 would be 55%.
Work for (a) and (b):
% of midterm scores be at least 80 would be = (8+3)/20} = 0.55 or 55%
4 Answer:
(a) Range value = 53
(b) Median value = 83.50
(c) Mode value = 84
Work for (a), (b) and (c):
First arranging the data from smallest to largest order.
Range value = Highest value - Lowest value = (98-45) = 53
Median value = {(20/2)th value +((20+1)/2)th value}/2 = (83+84)/2 = 83.5
Highest number of frequencies has been observed for 84 and hence, mode would be
84.

STAT 200 Final Examination Fall 2018 OL4_US2 Page 4 of 14
5
Answer:
(a) Mean = 75.70
(b) Standard deviation = 12.89
(c) Average() and Stdev.s() are the two function which has been used to find the
sample average and sample standard deviation.
Work for (a) and (b):
6
Answer:
(a) P (diamond or clover) = ½
(b) P (not a spade) =3/4
Work for (a) and (b):
(a) Probability of drawing a diamond = 13/52
Probability of drawing a clover =13/52
Hence,
P (diamond or clover) = 13/52 + 13/52 = ½
(b) Probability that car is a spade = 13/52
P (not a spade) = 1 – 13/52 = ¾
7 Answer:
(a) P(white ball and white ball) with replacement = 0.36
(b) P(white ball and white ball) without replacement = 0.33
Work for (a) and (b):
Probability that 1st ball would be white = 6/10
Probability that 2nd ball would also be white = 6/10
It is because there are 6 white balls and with replacement there would be again 6
white balls.
P(white ball and white ball) with replacement ¿
6
10∗6
10 = 36
100

STAT 200 Final Examination Fall 2018 OL4_US2 Page 5 of 14
Probability that 1st ball would be white = 6/10
Probability that 2nd ball would also be white = 5/10
It is because there are 6 white balls and without replacement there would be 5 white
balls.
P(white ball and white ball) with replacement ¿
6
10∗5
10 = 30
100
8
Answer:
(a) Yes!! It is because order for which the store would be visited would also be
different.
(b) Permutation. It is because order matter in the scheduling and assistant would be
able to choose any of the five stores from twenty stores and hence, permutation is
take into account for different schedules.
(c) Total number of different schedule that assistant would recommend = 1860480
Work for (c):
9
Answer:
(a) No!! This is because Mimi is taking two books from the same type of book series
which is Fun series and hence, order does not make any difference in the book
selection.
(b) Combination
(d) Different ways of choosing 3 books from 8 = 56
Work for (c):
10 Answer:
(a)

STAT 200 Final Examination Fall 2018 OL4_US2 Page 6 of 14
(b) mean = 1.5 , standard deviation = 0.87 .
Work for (a) and (b):
There are eight possible outcomes when a fair coin has been tossed three times.
Mean E ( X ) =( 0∗1
8 ) +
( 1∗( 3
8 )) + ( 2∗3
8 )+ ( 3∗1
8 )=1.5
E ( X2 ) =0+1∗( 3
8 )+ 22∗( 3
8 )+ 32∗( 1
8 )=3

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11
Answer:
(a) n = 15 , p = 0.30 , and q = 0.70 .
(b) P = 0.485
(d) BINOMDIST () would be used to find the requisite probability.
Work for (b):
Probability that Mimi scores at east 5 from the 15 field goals =
12
Answer:
(a) P = 0.6678
(b) 90th percentile value = 12.56
(c) NORMSINV function and standard normal table has been used to find the
probability and percentile.
Work for (a) and (b):
P ( 8.5<X <12.5 )=P ( 8.5−10
2 <z < 12.5−10
2 )=P(−0.75<Z <1.25)
P ( 8.5< X <12.5 ) =0.6678 (standard normal table)
The z value in case of 90th percentile = NORMSINV (0.90) = 1.281

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13
Answer:
(a) Standard deviation (for sample mean value) = 1.522
(b) P =0.8936
(c) Standard normal table has been used to find the probability.
Work for (a) and (b):
Standard deviation (for sample mean value) ¿ 9.13
√36 =1.522
P ( 150<X <155 ) =P ( 150−152.8
1.522 < z< 155−152.8
1.522 )=P (−1.84< Z <1.45)
P ( 150<X <155 ) =0.8936 (standard normal table)
14 Answer:
(a) 95% confidence interval is defined as [0.5068 0.5557]
(b) From the above, it is apparent that with 95% confidence, one can conclude that
mean proportion of adult population living in a house where there is no landline
would lie between 0.5068 and 0.5557.
Work for (a):
The confidence interval has been determined using the KADD add in to Excel.
Relevant output is exhibited as follows.

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15
Answer:
(a) 95% confidence interval is defined as [1080.40 1119.60]
(b) From the above, it is apparent that with 95% confidence, one can conclude that
mean SAT scores would occupy the range of 1080.40 and 1119.60.marks.
Work for (a):
The confidence interval has been determined using the KADD add in to Excel.
Relevant output is exhibited as follows. Z is used as population standard deviation is
known.

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16
Answer:
(a) One sample z test for population proportion.
(b) Null hypothesis (H0): p <= 0.50
Alternative hypothesis (H1): p>0.50
(c) Z statistics = 2.50
(d) Corresponding p value = 0.0124
(e) As p value< significance level, hence H0 would be rejected and H1 would be
accepted.
(f) Based on the above, it may be concluded that adult proportion living in houses
where landline phones is absent is expected to be greater than 50%.
Work for (c) and (d):
The z and p values have been determined on the basis of an online calculator whose
URL is shown below.
https://www.medcalc.org/calc/test_one_proportion.php

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17
Answer:
(a) The t test for two dependent samples (matched pairs) is the suitable choice
considering that the subjects are the same and essentially the observations are taken
at two different times.
(b) (ii) μ1−μ2=0( μd=0)
(c) (i) μ1−μ2>0 ( μd >0)
(d) test statistic is 3.138
(e) Corresponding one tail p value has come out as 0.017
(f) As p value< significance level, hence H0 would be rejected and H1 would be
accepted.
(g) Based on the above, it may be concluded that average number of words that are
recalled after 1 hour tend to be higher than the corresponding average after 24 hours.
Work for (d) and (e):
The relevant test has been performed in excel with the following output.

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18
Answer:
(a) (a) Chi-square goodness of test fit seems suitable considering that both the variables
are categorical and hence none of the other options seem to be suitable.
(b)
(c) (b) Null hypothesis H0: The published value does not significantly deviate from the
true proportion.
(d)
Alternative hypothesis H1: The published value does significantly deviate from the
true proportion.
(c)Test statistic i.e. chi-square has come out as 364.9
(e)
(f)
(g) (d) The corresponding p value has come out as 0.00
(e) As p value< significance level, hence H0 would be rejected and H1 would be
accepted.
(h)
(f) Based on the above, it may be concluded that there is significant deviation
between the published values and true proportion.
Work for (c) and (d):
Expected value = Expected proportion * Observed value
X2 =364.9
Degree of freedom = Number of obseration-1 =5-1= 4
The p value has been computed through online calculator and the relevant URL is
shown below.
https://www.socscistatistics.com/pvalues/chidistribution.aspx
19 Answers:
(a) Final exam score( y)=10.61+(0.87∗Average quiz score ( x))

STAT 200 Final Examination Fall 2018 OL4_US2 Page 13 of 14
(b) Final exam score (@ average quiz score = 80) = 79.81
(c) Final exam score (@ average quiz score = 40) = 45.21
(d) 79.81 final exam score is termed as more, closer to the true final exam score.
Work for (a), (b), and (c):
Regression output from excel
Y = 10.61 +0.87X
Y for X = 80, Y = 10.61 +0.87*80 = 79.81
Y for X = 40, Y = 10.61 +0.87*40 = 45.21
20 Answer:
(a) t test for two independent sample
(b) ANOVA
Justification for (a) and (b):
(a) Since there are only two samples with numerical values which require comparing
of means and have unknown population standard deviations, hence t test for two
independent sample seem appropriate.

STAT 200 Final Examination Fall 2018 OL4_US2 Page 14 of 14
(b) In this case, since the mean of more than 2 samples ought to be compared, hence
t test is not suitable. Also, considering the non-categorical nature of data, Chi square
is also not suitable. Hence, ANOVA is the suitable choice here.