Problem #1
A total of 200 people in each of three grocery stores were asked if they favored, opposed,
or were indifferent to the sale of lottery tickets in grocery stores. The results of the survey
are summarized below. Test the hypothesis at α= 0.05.
Opinion
Favor
Oppose
Indifferent
Store A
21
101
78
Store B
78
63
59
Store C
94
27
79
Answer
The null hypothesis tested is
H0: There is no significant difference in the proportion of opinion in the three stores.
The alternative hypothesis is
H1: There is significant difference in the proportion of opinion in the three stores.
The test statistic used is
The test statistic used is
2  
(O  E )2
E
where
O–
Observed frequency
EExpected frequency
The expected frequencies are calculated assuming that the null hypothesis is
true.The Expected frequencies are given below. They are calculated using the
formula ,
Eij 
Ri C j
G
,where Ri , ith row total, Cj jth column total and G is the
grand Total.
Rejection criteria: Reject the null hypothesis if the calculated value of Chi
square is greater than the critical value of chi square with 4 d.f
Details
The expected frequencies are given below
Opinion
Favor
Oppose
Indifferent
Store A
Store B
Store C
64.33333
63.66667
72
64.33333
63.66667
72
64.33333
63.66667
72
The Chi square contribution from each cell is
Opinion
Store A
Store B
Favor
29.18826
2.903282
Oppose
21.8918
0.006981
Indifferent
0.5
2.347222
Store C
13.68048
21.11693
0.680556
Results
Critical Value
9.487729
Chi-Square Test Statistic 92.31551
p-Value
4.24E-19
Reject the null hypothesis
Conclusion: Reject the null hypothesis as the calculated value of test statistic is greater
than the critical value of Chi square at 0.05 significance level. The sample provide
enough evidence to support the claim that there is significant difference in the
proportion of opinion in the three stores.
Problem #2
A professor of economics wants to study the relationship between income (y in $1000s)
and education (x in years). A random sample eight individuals is taken and the results are
shown below.
Education
Income
16
58
11
40
15
55
8
35
12
43
10
41
13
52
14
49
Test to see if there is any significant relationship between income and education. Use α =
.05.
Answer
The correlation between X,Y is given by the formula
n
r
n
n
i 1
i 1
n X iYi   X i  Yi
i 1
2
n
 n

 n 
n X    X i  .n Yi 2    Yi 
i 1
 i 1  i 1
 i 1 
n
2
2
i
Education: X
16
11
15
8
12
10
13
14
99
Income: Y
58
40
55
35
43
41
52
49
373
X
256
121
225
64
144
100
169
196
1275
Y
3364
1600
3025
1225
1849
1681
2704
2401
17849
XY
928
440
825
280
516
410
676
686
4761
Thus the calculated value of correlation based on the above formula is
r
n
n
n
i 1
i 1
i 1
n X iYi   X i  Yi
=

8*4761  99*373

n
n
8*1275   99  .8*17849   373
 n

 n 
n X i2    X i  .n Yi 2    Yi 
i 1
 i 1  i 1
 i 1 
= 0.960345
2
2
2
2
The null hypothesis tested is
H0:  =0
H1:  ≠ 0
(- population correlation coefficient)
Test Statistic used is
t
r
1 r2
n  2 ~ tn  2
Rejection criteria: Reject the null hypothesis , if the calculated value of t is greater than
critical value of t with (n-2) d.f at 0.05 significance level
The calculated value of t = 8.4370
d.f
=6
Critical value = 2.45
Conclusion: Reject the null hypothesis that the population correlation coefficient is zero.
The sample provide enough evidence to support the claim that H1:  ≠ 0
Problem #3
Thirty-five employees who completed two years of college were asked to take a basic
mathematics test. The mean and standard deviation of their scores were 75.1 and 12.8,
respectively. In a random sample of 50 employees who only completed high school, the
mean and standard deviation of the test scores were 72.1 and 14.6, respectively.
a. Can we infer at the 10% significance level that a difference exists between the two
groups?
b. Estimate with 90% confidence the difference in mean scores between the two groups
of employees. Explain how to use this interval estimate to test the hypotheses.
Answer
The null hypothesis tested is
H0: There is no significant difference in the mean score of two groups of employees
The alternative hypothesis is
H1: There is significant difference in the mean score of two groups of employees.
The test statistic used is
t
X1  X 2
S
(n1  1) S12  (n2  1) S22
n1n2
~ tn1  n1 2 where S 
n1  n2  2
n1  n2
Rejection criteria: Reject the null hypothesis, if the calculated value of t is
greater than the critical value of t at 0.1 significance level.
Details
Significance Level : 0.1
t Test for Differences in Two Means
Data
Hypothesized Difference
Level of Significance
Population 1 Sample
Sample Size
Sample Mean
Sample Standard Deviation
Population 2 Sample
Sample Size
Sample Mean
Sample Standard Deviation
0
0.1
35
75.1
12.8
50
72.1
14.6
Intermediate Calculations
Population 1 Sample Degrees of Freedom
34
Population 2 Sample Degrees of Freedom
49
Total Degrees of Freedom
83
Pooled Variance
192.9566
Difference in Sample Means
3
t Test Statistic
0.979943
Two-Tail Test
Lower Critical Value
Upper Critical Value
p-Value
-1.66342
1.66342
0.329962
Conclusion: Fails to reject the null hypothesis. Here the calculated value of test statistic
is less than the critical value.
Confidence Interval
The 90% Confidence interval for the difference of mean scores between the two groups
of employees is given by

n1n2 
(n1  1) S12  (n2  1) S22
X

X

t
S
/
where
S

 1

2
 / 2, n1  n2  2
n1  n2 
n1  n2  2

Details
Population 1 Sample
Sample Size
Sample Mean
Sample Standard Deviation
Population 2 Sample
Sample Size
Sample Mean
Sample Standard Deviation
35
75.1
12.8
50
72.1
14.6
Intermediate Calculations
Population 1 Sample Degrees of Freedom
34
Population 2 Sample Degrees of Freedom
49
Total Degrees of Freedom
83
Pooled Variance
192.9566
Difference in Sample Means
3
t Test Statistic
0.979943
Upper confidence Limit = 8.092392
Lower Confidence limit = -2.092392
Since this confidence interval contain the value ‘0’ we can conclude that there is no
significant difference in the difference of mean scores of two groups.