MAT222 Test 2
Fall 2009
1. Out of a random sample of 50 accounts in a certain department store, 40 out of 50
customers had paid their outstanding bills on time for a given month.
(a) Construct a 95% confidence interval of the true proportion of customers who had paid
their outstanding bills on time. (10 points)
(b) Test the null hypothesis that p= .6 versus the alternative hypothesis that p >.6 at .05
level of significance. (10 points)
(a).
x
40
pˆ  
 .8
n
50
SE 
(b).
p̂(1 - p̂ )
n

.16
 .0032  .0565
50
95% confidence interval for p: p̂ z * SE = (.8)  1.96(.0565)
=.8  .1108 [.7892, .9108]
H0: p = .6l vs Ha: p > .6
z
p̂ - .6
p0 (1  p0 )
n

.8 - .6
.2
 2.89 p-value = P(Z > 2.89) < .0019 <
.6(1  .6) .0693
n
05
Reject H0 at level  = .05.
2. A test question is considered good if it differentiates between prepared and unprepared
students. The first question on a test was answered correctly by 62 of 80 prepared
students and by 26 of 50 unprepared students.
(a) Perform a significance test for the null hypothesis H 0 : p1  p2 versus H a : p1  p2
where p1 is the population proportion of correct answer for prepared students and p 2 is
the population proportion of correct answer for unprepared students. What do you
conclude? (10 points)
(b) Find a 95% confidence interval for p1 - p 2 . (10 points)
62
26
pˆ 1 
 .775. pˆ 2 
 .52
(a)
80
50
62  26 88
pˆ pooled 

 .677
80  50 130
SE Dp  p̂(1 - p̂)(
z
1
1
1
1
 )  .677  .323(  )  .0843
n 1 n2
80 50
p̂1 - p̂ 2 .775  .52

 3.02
SE Dp
.0843
p-value = P(Z > 3.02) = ..0018 < 05
Reject H0 at level  = .05; The proportion of correct answer for prepared
students is higher than the proportion of correct answer for unprepared students.
(b)
SE Dˆ 
p̂1 (1 - p̂1 ) p̂ 2 (1 - p̂ 2 )
.775(1  .775) .52(1  .52)



 .00218  00499  .0847
n1
n2
80
50
95% confidence interval for p1 - p2: D z * SE D̂ = (.775 - .52)  1.96(.0847)
=.255  .166
[.089, .4221]
3. The following table pertains to a study of the relationship between the speed of
promotion and the standard of clothing of bank employees:
Poorly dressed
Slow
10
Speed of Promotion
Average
18
Fast
14
Total
42
Well dressed
16
22
20
58
Total
26
40
34
100
(a) Give the joint distribution of “speed of promotion” and “dressed” for this table. (5
points)
Speed of Promotion
Slow
Average
Fast
Poorly dressed
.10
.18
.14
Well dressed
.16
.22
.20
(b) Computer the marginal distribution for the speed of promotion. (5 points)
P(slow) = 26/100 = .26, P(Average) = 40/100 = .40, P(fast)=34/100 = .34
(c) Compute the conditional distribution of the “speed of promotion” for “poorly
dressed” and “well dressed”. (5 points)
P(slow|poorly dressed) = 10/42 = .238 P(Average|poorly dressed) = 18/42 = .428
P(fast|poorly dressed)=14/42 = .333.
P(slow|well dressed) = 16/58 = .276 P(Average|well dressed) = 22/58 = .379
P(fast|well dressed)=20/58 = .345.
We wish to use a chi-square test at .05 level of significance to test whether the speed
of promotion and the standard of clothing are independence.
(d) State the null and alternative hypotheses for this problem. (5 points)
H0 : Speed of promotion and Clothing are independent/or unrelated.
Ha : Speed of promotion and Clothing are dependent/or related.
.
(e) Test the hypotheses in (d) at 5% level of significance and make a conclusion. (10
points) Expected counts: 10.92, 16.8, 14.28,; 15.08, 23.20, 19.72
2  
(observed  exp ected ) 2
 .0775  .0857  .0055  .0561  .0620  .0039  .2907
exp ected
degree of freedom = (2 – 1)(3 – 1) = 2
.25 < p-value Fail to reject H0
Speed of promotion and Clothing are independent/or unrelated.
4. (a) Find a 95% confidence interval for the slope in the following setting:
n = 20, yˆ  - 2.8  5.3 x , and SEb1 = 2.2
(5 points)
Use d.f. = 18, t* = 2.093
-5.3  (2.101)(2.2) = -5.3  4.62 = [-9.92, -.68]
(b) Test the null hypothesis that the slope is 0 versus the two-sided alternative. (5 points)
H0: =0 versus Ha:  0, at  =.05,
t = -5.3/2.2 = -2.41, p-value = 2P{t19 < -2.41}
d.f. = 18, 2.214 < 2.41 < 2.552  .01 < upper tail probability < .02
Hence .02 < p-value < .04
Reject H0 at level  = .05
5. Exercise 2.144 on Page 163 of our textbook gives the modulus of elasticity (MOE) and
the modulus of rupture (MOR) for 32 plywood specimens. We regress MOR on the MOE
for the 32 specimens. Here is part of the MINITAB output for the regression:
Results for: EX10_021.MTP
Regression Analysis: mor versus moe
The regression equation is
mor = 2653 + 0.00474 moe
Analysis of Variance
Source
Regression
Residual Error
Total
DF
1
30
31
SS
75573461
45987465
121560926
MS
F
75573461 49.30
1532915.5
P
0.000
(a) Complete the analysis of variance table by filling in the “Residual Error” row. (5
points)
(b) What are the values of the regression standard errors s and the squared correlation
r2? (5 points)
S = square root of MSE =1239.1; r2 = SSM/SST = 75573461/121560926 = .62
(c) The standard deviation of the moe is 325293. Find the standard error for the leastsquares slope b1. (5 points)
1
Sx 
( xi  x ) 2  325293   ( x i  x ) 2  325293 31  1781703.1

n 1
SEb1 
s
 (x
i
 x)
2

1239.1
 .000695
1781703.1
(d) Give a 95% confidence interval for the slope 1 of the regression line. (5 points)
d.f =30, t* = 2.042
95% confidence interval for the slope 1:
b1  t * SEb1  .00474  2.042(.000695)  .00474  .00142  [.00332,.00616]