trdhgjhgfgcvbvhhgjghbntewghbnmbnbnbjhguyghjbnuyyuyfgbnbjhbhgbbhuiyti7rjfhgvhgrrtfgfugiuyuyjuiyyjhyt76yttyyupiuiojoioii0909ooeoreferfergergferfewfwefwef2d33rf34

1. Hypothesis
testing - II
DR ROSHNY UNNIKRISHNAN
IBS BANGALORE

2. Session objectives
Difference Between Two Population Means: 𝜎1 and 𝜎 2 Known
Difference Between Two Population Means: 𝜎 1 and 𝜎 2
Unknown
Difference Between Two Population Means: Matched Samples
Difference Between Two Population Proportions

3. Difference Between Two
Population Means: 𝜎1
and 𝜎 2 Known

4.  Expected Value
Sampling Distribution of
E x x
( )
1 2 1 2
  
 
 Standard Deviation (Standard Error)

 
x x
n n
1 2
1
2
1
2
2
2
  
where: 𝜇1 = standard deviation of population 1
𝜇2 = standard deviation of population 2
n1 = sample size from population 1
n2 = sample size from population 2

5.  Interval Estimate
Interval Estimation of 1 - 2:
 1 and  2 Known
2 2
1 2
1 2 /2
1 2
x x z
n n

 
  
where:
1 -  is the confidence coefficient
 The point estimator of the difference between the
means of the populations 1 and 2 is .
x x
1 2


6. Hypothesis Tests About  1   2:
 1 and  2 Known
 Hypotheses
1 2 0
2 2
1 2
1 2
( )
x x D
z
n n
 
 


 
 
1 2 0
:
a
H D
 
 
0 1 2 0
:
H D
 
 
0 1 2 0
:
H D
 
 
1 2 0
:
a
H D
 
 
0 1 2 0
:
H D
 
 
1 2 0
:
a
H D
Left-tailed Right-tailed Two-tailed
 Test Statistic
For table value or critical value df = N₁ + N₂ - 2
When 𝐷0 =
0 𝑡ℎ𝑒 𝑛𝑢𝑙𝑙 ℎ𝑦𝑝𝑜𝑡ℎ𝑒𝑠𝑖𝑠
𝑖𝑠 𝐻0: 𝜇1 − 𝜇2 = 0

7. Pg 612
b. With confidence level of 95% what is your conclusion?
C. Find the point estimate and Construct a 90% confidence interval for difference between
two population means

8.  
1 2 0
2 2 2 2
1 2
1 2
(25.2 22.8) 0
2.03
(5.2) 6
40 50
x x D
z
n n
 
   
  


a)
b) At 95% z= 1.645
Hence 2.03 falls in rejection region reject null
c) Point estimate = 25.2-22.8 = 2.4
Interval estimate is 2.4±1.645(
5.2 2
40
+
6 2
50
(0.456 to 4.344)

9. Q 7 Pg 613
Consumer Reports uses a survey of readers to obtain customer satisfaction
ratings for the nation’s largest retailers (March 2012). Each survey respondent is
asked to rate a specified retailer in terms of six factors: quality of products,
selection, value, checkout efficiency, service, and store layout. An overall
satisfaction score summarizes the rating for each respondent with 100 meaning
the respondent is completely satisfied in terms of all six factors. Sample data
representative of independent samples of Target and Walmart customers are
shown below.

10. Q 7 Pg 613
a. Formulate the null and alternative hypotheses to test whether there is a difference
between the population mean customer satisfaction scores for the two retailers.
b. Assume that experience with the consumer Reports satisfaction rating scale indicates
that a population standard deviation of 12 is a reasonable assumption for both retailers.
Conduct the hypothesis test and report the p-value. At a .05 level of significance what is
your conclusion?
c. Which retailer, if either, appears to have the greater customer satisfaction? Provide a
95% confidence interval for the difference between the population mean customer
satisfaction scores for the two retailers.

11. Solution
𝑍𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑓𝑜𝑟 𝛼
= .05 𝑎𝑛𝑑 𝑡𝑤𝑜 𝑡𝑎𝑖𝑙 𝑖𝑠 1.96; 𝐻𝑒𝑛𝑐𝑒 𝑟𝑒𝑗𝑒𝑐𝑡 𝑡ℎ𝑒 𝑛𝑢𝑙𝑙 𝑎𝑠 2.46 𝑖𝑠 𝑔𝑟𝑒𝑎𝑡𝑒𝑟 𝑡ℎ𝑎𝑛 1.96

13. A simple random sampling survey in respect of monthly earnings
of semi-skilled workers in two cities gives the following statistical
information. Test the hypothesis at 5 per cent level that there is
no difference between monthly earnings of workers in the two
cities.
Pg 227 kothari

14. Difference Between Two
Population Means: 𝜎 1
and 𝜎 2 Unknown

15. 2 2
1 2
1 2 /2
1 2
s s
x x t
n n

  
Where the degrees of freedom for t/2 are:
Interval Estimation of 1 - 2:
 1 and  2 Unknown
 Interval Estimate
2
2 2
1 2
1 2
2 2
2 2
1 2
1 1 2 2
1 1
1 1
s s
n n
df
s s
n n n n
 

 
 

   

   
 
   

16. Hypothesis Tests About  1   2:
 1 and  2 Unknown
 Hypotheses
1 2 0
2 2
1 2
1 2
( )
x x D
t
s s
n n
 


 
 
1 2 0
:
a
H D
 
 
0 1 2 0
:
H D
 
 
0 1 2 0
:
H D
 
 
1 2 0
:
a
H D
 
 
0 1 2 0
:
H D
 
 
1 2 0
:
a
H D
Left-tailed Right-tailed Two-tailed
 Test Statistic
For table value or critical value df = N₁ + N₂ - 2

17. Q14 Pg 620
Are nursing salaries in Tampa, Florida, lower than those in Dallas, Texas? As
reported by the tampa tribune, salary data show staff nurses in Tampa earn less
than staff nurses in Dallas. Suppose that in a follow-up study of 40 staff nurses
in Tampa and 50 staff nurses in Dallas you obtain the following results
a. Formulate hypothesis so that, if the null hypothesis is rejected, we can
conclude that salaries for staff nurses in Tampa are significantly lower than
for those in Dallas. Use a = .05.
b. What is the value of the test statistic? c. What is the p-value? d. What is your
conclusion?

18. Critical value 1.663

19. Q17 Pg 621
Periodically, Merrill Lynch customers are asked to evaluate Merrill Lynch financial
consultants and services. Higher ratings on the client satisfaction survey indicate better
service, with 7 the maximum service rating. Independent samples of service ratings for
two financial consultants are summarized here. Consultant A has 10 years of experience,
whereas consultant B has 1 year of experience. Use 𝛼 = .05 and test to see whether the
consultant with more experience has the higher population mean service rating.
a. State the null and alternative hypotheses.
b. Compute the value of the test statistic.
c. What is the p-value?
d. What is your conclusion?

20. 𝑡𝑐𝑟𝑖𝑡𝑓𝑜𝑟 𝑑𝑓 16 𝑎𝑛𝑑 𝛼 = .05 𝑖𝑠 1.746
reject H0. The consultant with more
experience has a higher population
mean rating

21. Difference Between Two
Population Means:
Matched Samples

22. Difference Between Two Population
Means: Matched Samples
With a matched-sample design each sampled item provides a pair of data values
 This design often leads to a smaller sampling error than the independent-sample
design because variation between sampled items is eliminated as a source of
sampling error.

23. 32
30
19
16
15
18
14
10
7
16
25
24
15
15
13
15
15
8
9
11
UPX INTEX Difference
District Office
Seattle
Los Angeles
Boston
Cleveland
New York
Houston
Atlanta
St. Louis
Milwaukee
Denver
Delivery Time (Hours)
7
6
4
1
2
3
-1
2
-2
5
Inferences About the Difference Between Two Population Means:
Matched Samples

24. H0: d = 0
Ha: d 
Let d = the mean of the difference values for the
two delivery services for the population
of district offices
1. Develop the hypotheses.
Inferences About the Difference Between
Two Population Means: Matched Samples
 p –Value and Critical Value Approaches

25. 2. Specify the level of significance.  = .05
Inferences About the Difference Between
Two Population Means: Matched Samples
 p –Value and Critical Value Approaches
3. Compute the value of the test statistic.
d
d
n
i



  

( ... )
.
7 6 5
10
2 7
s
d d
n
d
i




 
( ) .
.
2
1
76 1
9
2 9
2.7 0
2.94
2.9 10
d
d
d
t
s n

 
  

26. 4. Determine the critical value and rejection rule.
Inferences About the Difference Between
Two Population Means: Matched Samples
 Critical Value Approach
For  = .05 and df = 9, t.025 = 2.262.
Reject H0 if t > 2.262
5. Determine whether to reject H0.
Because t = 2.94 > 2.262, we reject H0.
We are at least 95% confident that there is a
difference in mean delivery times for the two
services?

27. Q27 Pg 627
A manufacturer produces both a deluxe and a standard model of an automatic
sander designed for home use. Selling prices obtained from a sample of retail
outlets follow.
a) The manufacturer’s suggested retail prices for the two models show a $10
price differential. Use a .05 level of significance and test that the mean
difference between the prices of the two models is $10.
b) What is the 95% confidence interval for the difference between the mean
prices of the two models

28. 𝑡𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑓𝑜𝑟 𝑑𝑓 6 𝑎𝑛𝑑 𝛼 = .05 two tail t.025 = 2.447
Do not reject H0; we cannot reject the hypothesis that a $10 price differential
exists.

29. *

30. Difference Between Two
Population Proportions

31. If the sample sizes are large, the sampling distribution
of can be approximated by a normal probability
distribution.
The sample sizes are sufficiently large if all of these
conditions are met:
n1p1 > 5 n1(1 - p1) > 5
n2p2 > 5 n2(1 - p2) > 5
Sampling Distribution of p p
1 2


32. p p
p p
n
p p
n
1 2
1 1
1
2 2
2
1 1
 



( ) ( )
where: n1 = size of sample taken from population 1
n2 = size of sample taken from population 2
 Standard Deviation (Standard Error)
Interval Estimation of p1 - p2
 Interval Estimate
1 1 2 2
1 2 / 2
1 2
(1 ) (1 )
p p p p
p p z
n n

 
  

33. Hypothesis Tests about p1 - p2


 
 
 
 
1 2
1 2
( )
1 1
(1 )
p p
z
p p
n n
1 1 2 2
1 2
n p n p
p
n n



 Pooled Estimator of p when p1 = p2 = p
1 2
p p

 Standard Error of when p1 = p2 = p

34. Hypothesis Tests about p1 - p2
 Hypotheses
H0: p1 - p2 < 0
Ha: p1 - p2 > 0  
1 2
: 0
a
H p p
 
0 1 2
: 0
H p p
 
0 1 2
: 0
H p p
 
1 2
: 0
a
H p p
 
0 1 2
: 0
H p p
 
1 2
: 0
a
H p p
Left-tailed Right-tailed Two-tailed
We focus on tests involving no difference between
the two population proportions (i.e. p1 = p2)

35. Q29 pg 632

37. Q32 Pg 632