2565 Midterm2 2016 Solution

York University MATH MATH 2565 (Winter (Winter 2016): Introduction to Applied Statistics Midterm II Examination – Thursday March 17, 2016, 10 am - 11:10 am Last Name:

Given Names:

Student Number: Signature Signature : DO NOT WRITE IN THIS AREA Read the following instructions carefully

1. Stud Studen ents ts are allow allowed ed to brin bring: g: a.

One One oneone-si side ded d

8.5”x11” 8.5”x11” formula sheet b. Calculator(s Calculator(s)) (cannot use Ques Questi tion on

Max Ma x Scor Scoree

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10

2. All work must must be done in the free space provided provided be-

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10

low for each problem or on the back of the  preceding

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10

page.

4

10

5

10

TOTAL

50

laptop, cell phone or ipad as calculator)

3. No other sheets sheets of paper will be accepted accepted or marked. marked. 4. This This exam consist consistss of 9 pages in total. total. Last Last 3 pages pages contain three tables.

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Your our Scor Scoree

Question 1  (10 marks) Research concerning purchasing habits at the grocery store suggests

that impulse buys make up 60% of all purchases. Suppose 100 customers are selected at random. (a) Let X   be the number of customers who make an impulse purchase out of the 100 randomly selected customers. Find the mean and variance of  X .

Solution µ = np = 100(.6) = 60, σ2 = np(1

− p) = 100(.4)(.6) = 24

(b) Find the variance of the sample proportion  p = ˆ

Solution var p = ˆ

p(1− p) n

=

.6(.4)

100

X   . n

= .0024

(c) find the probability that at most one customer makes an impulse purchase.

100(.4)100 + +100(.6)(.4)99 ∼ 0

Solution

0

1

2

Question 2  (10 marks) Let X 1 , . . . , X16    be a random sample from the N (µ, σ 2 ) distribution

with µ  = 5. The sample mean is 4.8 and the sample standard deviation is s = .5967 ¯= (a) Give the sampling distribution of  X 

¯ Solution X 

X 1 +...+X 16

16

σ2

 ∼ N (5, 16 )

 ¯ (b) Assume σ 2 = 4 is known. Find P (X 

 ≤ 5.2)

Solution

 ¯ P (X 

 ≤ 5.2)

= P (

√ n(X   ¯ − µ) σ

≤ 4(5.22− 5) ) = P (Z  ≤ .4) = .6554

 ¯ (c) Assume σ 2 is unknown. Find P (X 

 ≤ 5.2)

Solution

 ¯ P (X 

 ≤ 5.2)

= P (

√ n(X   ¯ − µ) s

− 5) ) = P (T 15 ≤ 1.3406) = .90. ≤ 4(5.2 .5967 3

Question 3   (10 marks) Suppose a random sample of size n   = 80 is obtained from a

ˆ  be the sample proportion of successes. population with probability of success p = .35. Let P   ˆ (a) Find the value a  such that P (P 

 ≤ a) = 0.1093.

Solution a = p

  (1− )

− 1.23

 p

 p

n

= 0.284408

 ˆ (b) Find the value b  such that P (P 

 ≥ b) = 0.0102.

Solution b = .35 + (2.32) sqrt((.35) (.65)/80) = 0.4737182

∗

∗  ˆ ≥ 0.4607) = 0.0102.(Round up to the nearest (c) What should n   be if you want P (P  integer).

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Question 4   (10 marks) Parrot Jungle Island is a roadside attraction in Miami, Florida,

featuring tropical birds, crocodiles, and over 2000 varieties of plants and flowers. A new advertising campaign was developed to attract more out-of-state visitors. In a random sample of 270 visitors, 189 were area residents. (a) Find the sample proportion of visitors who were area residents.

Solution:  p=189/270=0.7; n ˆ  p = ˆ 189

≥ 5, n(1 − p)ˆ = 81 ≥ 5

(b) Find a 99% confidence interval for the true proportion of visitors who were area residents.

Solution:   0.7+2.5758*sqrt(0.3*0.7/270)=0.7718356; 0.7-2.5758*sqrt(0.3*0.7/270)=0.6281644

(c) Suppose that you have no prior knowledge of the proportion of visitors who were area residents. Find the sample size necessary for a 99% confidence interval with a bound on the error of estimation of 0.05.

∗ ∗ {z0 005 /B}2=663.4746, therefore n = 664.

Solution: n = 0.5 0.5

.

5

Question 5  (10 marks) According to the U.S. Fire Administration, approximately 25,000

fires are caused by fireworks each year in the United States. Despite numerous public warnings against the use of fireworks, the home property damage due to these fires is enormous. In a random sample of 25 fires due to fireworks, the resulting mean property damage (in dollars) was 860.75 with a standard deviation 350.50. Assume the underlying distribution of property damage due to these fires is normal. (a) Find a 99% confidence interval for the true mean property damage due to a fire caused by fireworks.

Solution:   860.75+2.7969*350.5/5=1056.813

860.75-2.7969*350.5/5= 664.6873 (b) What is the bound on the error of estimation in part (a)?

Solution:   B=2.7969*350.5/5=196.0627

(c) What is the width of a 95%  confidence interval for the true mean property damage due to a fire caused by fireworks?

Solution: 2 t0.025,24

∗

∗ 350.5/5 = 2 ∗ 2.0639 ∗ 350.5/5 = 289.3588 6