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UNIVERSITY OF REGINA
Department of Mathematics and Statistics
Elementary Statistics For Applications
Stat 100
Final Exam, Fall 2013
Time: 3 hours Full Name:
Pages: 10 Student Number:
Instructor: (check one)
◻ Peter Douglas (100-001)
◻ Robert Petry (100-C01)
◻ Vijayaparvathy Agasthian (100-L01/L02)
◻ Patrick Maidorn (100-991)
INSTRUCTIONS
- All work and answers are to be placed in this booklet. If you require more space for an answer, work on the back of the sheet and indicate where the work is to be found.
- Place final answers in blanks at right when provided.
- Each question will be graded for both the correct answer and supporting work.
- The marks are placed beside each question.
- Only calculator, formula sheet, and the provided probability tables and scrap paper are permitted in the exam.
For instructor use only:
Page Marks Score
2 15
3 10
4 15
5 10
6 11
7 11
8 10
9 10
10 8
Total: 100
Marks
(15) 1. Joe’s Nut Bar sells snacks to office workers. A random selection of Nut Bar products had the following prices:
(a) Is the variable discrete or continuous? (Circle your answer at right.)
discrete/continuous
Solution: Discrete as prices do not come in fractions of a cent.
(b) Determine the mode. Answer:
Solution: mode = $ 7. 50
(c) Determine the mean. Answer:
Solution: ¯x =
(d) Determine the standard deviation. Answer:
Solution:
s =
ÁÁ
À^968.^1875 −^
( 89. 75 )^2 9 9 − 1
(e) Determine the first and third quartiles. Answer: Q 1 =
Q 3 =
Solution: Q 1 = x^14 ( 9 + 1 ) = x 2. 5 = $ 7. 50
since x 2 = x 3 = $ 7. 50
Q 3 = x (^34) ( 9 + 1 ) = x 7. 5 =
x 7 + x 8 2
(f) Is $15.25 an outlier in the above dataset? Circle your answer at right and show the relevant calculation used to determine your result below.
yes/no
Solution: The interquartile range is IQR = Q 3 − Q 1 = $ 12. 75 − $ 7. 50 = $ 5. 25 The lower fence and upper fence are
Lower Fence = Q 1 − 1. 5 (IQR) = $ 7. 50 − 1. 5 ($ 5. 25 ) = −$ 0. 375
Upper Fence = Q 3 + 1. 5 (IQR) = $ 12. 75 + 1. 5 ($ 5. 25 ) = $ 20. 625 Since $15.25 lies within this range it is not an outlier.
(10) 3. Katniss Everdeen is an expert archer who can hit the bulls-eye on a target in 90% of her shots. (a) In a particular training session Katniss takes 8 shots at the target. What is the probability that she hits the bulls-eye at least 6 times?
Answer:
Solution:
P (x = 6 ) = 8 C 6 ( 0. 9 )^6 ( 0. 1 )^2 = 0. 14880348 ≈ 0. 1488 P (x = 7 ) = 8 C 7 ( 0. 9 )^7 ( 0. 1 )^1 = 0. 38263752 ≈ 0. 3826 P (x = 8 ) = 8 C 8 ( 0. 9 )^8 ( 0. 1 )^0 = 0. 43046721 ≈ 0. 4305
So P ( 6 ≤ x) = P (x = 6 ) + P (x = 7 ) + P (x = 8 ) = 0. 96190821 ≈ 0. 9619
(b) Over a two-week period Katniss takes 240 shots at the target. Use the normal approximation to the binomial to estimate the probability that she hits the target at least 220 times.
Answer:
Solution: The mean for the binomial is μ = ( 240 )( 0. 9 ) = 216 with standard deviation σ =
- 6 = 4 .647580015. The probability is
P ( 220 ≤ xbinomial) = P ( 219. 5 < xnormal) = P
< z
= P ( 0. 753080095 < z) = P (z < − 0. 75 ) = 0. 2266
Note if no continuity correction is used the (incorrect) answer is 0..
(5) 4. A Toronto newspaper wishes to conduct a poll to estimate the current approval rating of the mayor. The question asked will be “Do you approve of the actions of the mayor?” with the only choice of answers being Yes or No. How large a sample should be selected if the newspaper wants its poll results to be accurate within 4% of the true population proportion, 19 out of 20 times?
Answer:
Solution: E = 0 .04, z = 1 .960 for 1−α = 19 ~ 20 = 0 .95, and proportion estimate (since none given) is 0.50. Sample size is:
n = ( 0. 50 )( 0. 50 )
2 = 600. 25 = 601 residents
(10) 5. The weight of male reindeer of the R. Santaclausus subspecies is normally distributed with mean 102.4 kg and standard deviation 13.9 kg. (a) What proportion of these reindeer would weigh more than 118.0 kg?
Answer:
Solution:
P (118 kg < x) = P
118 .0 kg − 102 .4 kg 13 .9 kg
< z
= P ( 1. 122302158 < z) = 1 − P (z < 1. 12 ) = 1 − 0. 8686 = 0. 1314
(b) Rudolph is a fairly small reindeer. If exactly 10% of the reindeer weigh less than Rudolph, how much does Rudolph weigh?
Answer:
Solution: 10% of the probability in a normal distribution is less than z = − 1 .28. Hence x = μ + zσ = 102. 4 + (− 1. 28 )( 13. 9 ) = 84 .608 kg ≈ 84 .6 kg
(c) If 36 reindeer are randomly selected and their average weight calculated, what is the probability that the mean weight is less than 100.0 kg?
Answer:
Solution: Standard error of the mean is σx¯ = σ~
n = 13. 9 ~
36 = 2 .31666666 kg and so
P (x¯ < 100 kg) = P z <
100 .0 kg − 102 .4 kg
- 316... kg
= P (z < − 1. 035971223 ) = P (z < − 1. 04 ) = 0. 1492
(11) 7. Statistics Canada reports that half of all households have at least one pet, with pet ownership higher in Alberta, Manitoba, and Saskatchewan and lower in the Maritimes, British Columbia, Ontario, and Quebec. (a) In a random sample of 300 households across Canada, 126 households said that they owned at least one pet. Does this data provide sufficient evidence to indicate that the proportion of households with at least one pet is different from that reported by the Statistics Canada? Test using a level of significance of α = 0..
Solution: Critical z-values for the two-tailed test (“different”) are ± 1.. Sample proportion p = 126 ~ 300 = 0 .42. Standard error of proportion is ¾ ( 0. 5 )( 0. 5 ) 300
Calculated z-value for the data is therefore
z =
Therefore we reject H 0 at α = 0 .05 and accept Ha. The evidence supports that the proportion of households with at least one pet differs from the 50% reported.
(b) What is the P -value for the hypothesis test?
Solution: The area from the normal curve table for z = − 2 .77 is 0.0028. Dou- bling for the two tails gives
P -value = 2 ( 0. 0028 ) = 0. 0056
which is less than α = 0 .05 as expected for a test in which the null hypothesis was rejected.
Answer:
(10) 8. Boston Bobby is a die-hard baseball fan and supports his hometown team the Boston Red Sox. Bobby has an ongoing argument with New York Nat over the defensive capabilities of the Boston Red Sox and the New York Yankees team. Bobby claims that the Red Sox have a superior defense and uses the following statistics to back up his claim.
Runs Allowed per Game
Games Mean Standard Dev.
Team Played (runs) (runs)
Boston Red Sox 35 3.96 1.
New York Yankees 32 4.25 1.
Bobby claims that Boston will allow fewer runs on average than New York. Test this hypothesis using a 5% level of significance.
Solution: Assuming a left-tailed test (μB − μN < 0) the critical z-value is − 1 .645. The standard error of the difference of means is
( 1. 07 )^2 35 +^
( 1. 15 )^2 32 =^0 .272102101 runs. So the calculated z-value for the data is
z =
Therefore we fail to reject H 0 at α=0.05. The evidence does not support Bobby’s claim. Alternatively the P -value for z = − 1 .07 is 0.1423 which is greater than α = 0. 05 resulting in the same conclusion. Note that a z-test was valid here due to the large sample sizes (n ≥ 30).
(8) 10. Data from a study on the number of absences and the final grades of seven randomly selected students from a statistics class is given in the following table.
Number of absences (x) Final Grade (y) 6 82 2 86 15 43 9 74 12 58 5 90 8 78
The following summations for the data have already been calculated: Σx = 57, Σy = 511, Σx^2 = 579, Σy^2 = 38993, Σxy = 3745
(a) The correlation coefficient for the relationship between number of absences and final grade is calculated to be r = − 0 .9442. Interpret the meaning of this value in plain English.
Solution: The negative sign indicates a negative correlation between the vari- ables. (As absences x increase the grade y decreases.) The magnitude of r indicates that the correlation is extremely high. Therefore, in summary, there is an extremely high negative correlation.
(b) Find the equation of the best fit (least-squares) regression line.
Solution: Using the following (or other) standard formulas gives:
a =
(∑ x^2 ) ⋅ (∑ y) − (∑ x) ⋅ (∑ xy) n ⋅ (∑ x^2 ) − (∑ x)^2
=
( 7 ) ⋅ ( 579 ) − ( 57 )^2
b =
n ⋅ (∑ xy) − (∑ x) ⋅ (∑ y) n ⋅ (∑ x^2 ) − (∑ x)^2
=
( 7 ) ⋅ ( 579 ) − ( 57 )^2
= − 3. 621890547 ≈ − 3 .62 %~absence
The linear regression line ˆy = a + bx is therefore
ˆy = 102. 5 − 3. 62 x [%]
Answer:
