STA 6166 – Exam 1 – Fall 2012
PRINT Name ___________________________
Conduct all tests at  = 0.05 significance level.
Q.1 The probability of randomly selecting the correct response on a multiple choice question with five choices is 0.20
(assuming zero knowledge). Suppose an exam consists of 6 multiple choice questions, each with five choices.
p.1.a. How many correct responses would you expect a student to pick by randomly selecting answers?
Y ~ B(n  6,   1/ 5)  E (Y )  n  6(0.2)  1.2
p.1.b. What is the probability a student gets none of the questions correct?
0
6
 6! 
6
Y ~ B(n  6,   0.2)  P Y  0   
  0.2  1  0.2   (0.8)  0.262144
0!6!


Q.2 A bridge holds up to 25 cars. It is known the weight of individual cars is normally distributed with  = 2100 lbs. and
 = 500 lbs.
p.2.a. What is the sampling distribution of the sample mean weight for n=25 cars?

500


Y ~ N    2100,   500   Y ~ N  Y    2100,  Y 

 100 
n
25


p.2.b. If the maximum load for which the bridge is designed is 80,000 lbs., what is the probability a full load of cars
(n=25) will exceed the design limits?

 Y  80000  3200   P  Z  Y  Y  3200  2100  11  0
P   Y  80, 000   P  Y 



n
25
Y
100




Q.3. Find the mean, median, standard deviation for the following data.
16 10 18 15 11 9 5
Y
S
16  10  18  15  11  9  5 84

 12 M  Y(4)  11
7
7
(16  12) 2  (10  12) 2  (18  12) 2  (15  12) 2  (11  12) 2  (9  12) 2  (5  12) 2
124

 4.55
7 1
6
Q.4 A veterinarian is interested in estimating the average amount she spends monthly on vaccines. She records the dollar
amount for 25 randomly chosen months and finds the mean to be $370 and the sample standard to be $25. Give a point
estimate and a 95% confidence interval for the population mean monthly amount spent on vaccines by this veterinarian.
S
25

5
t.025,251  2.064
n
25
95%Confidence Interval for : 370  2.064(5)  370  10.32   359.67,380.32 
Point Estimate: Y  370
sY 
Q.5. A home gardener (a statistician) plants two varieties of tomato plants. Variety B is advertised to produce higher
yields. During the harvest season he records the yields of the individual plants. Do the data show significant evidence to
support the claim that variety B produces a higher median yield? Use the Wilcoxon Rank-Sum Test with .
Variety A
Variety B
40.3 (1)
55.4 (9)
51.3 (4)
55.2 (8)
54.4 (6) 46.8
56.0 (10) 53.4
(2)
(5)
49.2 (3)
54.5 (7)
Rank Sum for Variety A= 1+4+6+2+3 = 16
Rank Sum for Variety B 9+8+10+5+7 = 39
Conclude Variety B higher if Rank Sum for A ≤ 19 (1-sided,  = 0.05)
Q.6. You want to estimate the difference between two means within + 20 of the true population mean difference with 95%
confidence (that is, E=20). Assuming the common standard deviation is 15, how many observations should you select
from each population, based on independent samples?
1 1
2
E  z /2
  z /2
n1 n1
n
2
 20  1.96(15)
n
2 1.96(15) 
 n
 4.32  5  n1  n2
202
2
Q.7. A study measured a physical characteristic of identical twins. The purpose of the study to was to determine if there
was a significant difference in the means for the younger twin versus the older twin. A summary of the data is (Difference
= Younger - Older for each pair).
Twin
Younger
Older
Difference
N
16
16
16
Mean
1.76
1.56
0.20
Std. Dev.
0.24
0.30
0.24
Is there significant evidence that younger twins differ on average from older twins? H0: D= 0 vs HA: D≠ 0
Note: This is a Paired difference test (Twins=pairs)
p.7.a. Test Statistic: tobs 
D
SD
n

0.20
0.20

 3.33
0.24 16 0.06
p.7.b. Reject H0 if the test statistic falls in the range(s) |tobs| ≥ t.025,15 = 2.132
p.7.c. The P-value is larger than or
smaller than
0.05
Q.8. A diagnostic test has 95% sensitivity (the probability a person with the condition tests positive = 0.95) and 95%
specificity (the probability a person without the condition tests negative = 0.95). In a population of people given the test,
1% of the people have the condition (probability a person has the condition = 0.01).
p.8.a. What proportion of the people will test positive?
P T  | C   0.95
P  T    P T 
 
 
C   P T
C   P  C  P T | C   P C  P T
P  C   0.01
P T  | C  0.95  P T  | C  0.05




 
P C  1  0.01  0.99
| C  0.01(0.95)  0.99(0.05)  .0095  .0495  .059
p.8.b. Given a person has tested positive, what is the probability he/she has the condition?
P C | T


P T 
P T
C



P  C  P T  | C 
P T



0.01(0.95) 0.0095

 0.161
0.059
0.059
Q.9. You will be conducting a study to compare a measure of fitness for two types of exercise programs. You will use a
significance level = 5%. If the two means differ by more than 5 units in either direction, you want to have a probability of
at least 90% of rejecting Ho. Assuming a common  = 4.0, what is the required sample size for each program, based on
independent samples? Assume for now, the sample and population means are higher for program 1
Reject H 0 if: Z obs 
Y1 Y 2
Y1 Y 2

 z.025
1 1
2



n1 n2
n
 Y 1  Y 2  z.025
2
n


2
2 
When 1  2  5 : Probabilitity we fail to reject H 0    P  Y 1  Y 2  z.025
| Y 1  Y 2 ~ N  5, 


n
n  



2

Y 1  Y 2  5 z.025 n  5
5
 PZ 

 z.025 

2
2
2




n
n
n



Goal: Want  =1-Power=1-0.90=0.10  z.025 
5


2
n
 z.025  z.10







5
2

n
 5 
n
  z.025  z.10
 2 
  z.10

n
 z.025  z.10  
2 2  z.025  z.10 
2(16)(1.96  1.282) 2
 n

 13.45  14  n1  n2
52
25
2
5
2