Chapter 18 - Hypothesis testing and confidence intervals for two population means –
Independent samples
Inferences about Two Means with Unknown Population Standard Deviations – Independent
Samples – Population Standard Deviations not Assumed Equal (Non-Pooled t-Test)
Assumptions
 The samples are obtained using simple random sampling
 The samples are independent
 The populations from which the samples are drawn are normally distributed or the
sample sizes are large ( n1
 30, n2  30 )
The procedure is robust, so minor departures from normality will not adversely affect the
results. If the data have outliers, the procedure should not be used.
3) In the Spacelab Life Sciences 2 payload, 14 male rats were sent to space.
Upon their return, the red blood cell mass (in milliliters) of the rats was
determined. A control group of 14 male rats was held under the same
conditions (except for space flight) as the space rats and their red blood
cell mass was also determined when the space rats returned. The project,
led by Dr. Paul X. Callahan, resulted in the data listed below.
Part 1 - Construct a 95% confidence interval about 1   2
Part 2 - Test the claim that the flight animals have a different red blood cell
mass from the control animals at the 5% level of significance.
Flight
8.5 8.6
9
4
Control
8.6 6.9
5
9
7.4
3
7.2
1
6.8
7
7.8
9
9.7
9
6.8
5
7.0
0
8.8
0
9.3
0
8.0
3
6.3
9
7.5
4
8.4
0
9.6
6
7.6
2
7.4
4
8.5
5
8.
7
7.3
3
8.5
8
9.8
8
9.9
4
7.1
4
9.1
4
First: Verify assumptions. Because the sample sizes are small, we must
verify normality and that the samples does not contain any outliers.
Construct a normal probability plot and a boxplot in order to observe if the
conditions for testing the hypothesis are satisfied.
For each one of the samples, do this with your calculator! You are expecting a “close to linear”
normal probability plot.
Part 1 - Construct a 95% confidence interval about
1   2 . (Are you using z or t?
Why?)
With two populations we’ll be using the calculator only
Population 1: flight rats
(ml)
Notice that
Population 2: control rats
Variable: red blood cell mass
x1  7.88 ml. and x2  8.43 ml.
Are the x-bars different by chance, or are they significantly different?
The point estimate is
x1  x2  7.88  8.43 = -.55
To construct the interval use 2-SampTInterval, Data option and get
-1.335 <
1   2
<
.23655
(Why are using T instead of z?)
c) What does the interval suggest about the difference between the mean red blood cell mass of the
two groups? Circle one of the following statements and explain your choice.
1  2
1  2
1  2
Since the interval contains zero, with 98% confidence we conclude that the mean red blood
cell mass of the two groups may be equal
Part 2 - Test the claim that the flight animals have a different red blood cell
mass from the control animals at the 5% level of significance. (Are you
using z or t? Why?)
a) Set both hypothesis
1  2 
1  2 
1  2  0
1  2  0
This a two tailed test
b) Sketch graph, shade rejection region, label, and indicate possible locations of the point estimate in
the graph.
You do this
The point estimate is
x1  x2  7.88  8.43 = -.55
***You should be wondering: Are the x-bars different by chance, or significantly
different? The p-value found below will help you in answering this.
c) Use a feature of the calculator to test the hypothesis. Indicate the feature used and the results:
Use 2-Samp-TTest and get
Test statistic t = -1.437
2*P( x1  x2
 -.55) =2* P(t < -1.437) = p-value = .1627
***How likely is it observing such a difference between the x-bars (or a more extreme
one) when the mean of the two populations is equal?
very likely, likely, unlikely, very unlikely
*** Is the difference between the x-bars different to zero by chance, or is it significantly
different?
d) What is the initial conclusion with respect to Ho and H1?
Reject Ho and support H1
Fail to reject Ho, we don’t have enough evidence to support H1
e) Write the conclusion using words from the problem
We don’t have enough evidence to support the claim that the two groups have different red
blood cell mass. Flight is not affecting the red blood cell mass of the rats.
4) Neurosurgery Operative Times
Several neurosurgeons wanted to determine whether a dynamic system (Zplate) reduced the operative time relative to a static system (ALPS plate). R.
Jacobowitz, Ph.D.. an ASU professor, along with G. Visheth, M.D., and
other neurosurgeons, obtained the data displayed below on operative
times, in minutes for the two systems.
Dynamic:
370
345
Static:
360
450
510
505
445
335
295
280
315
325
490
500
430
445
455
455
490
535
Part 1 - At the 1% significance level, do the data provide sufficient evidence
to conclude that the mean operative time is less with the dynamic system
than with the static system?
Part 2 - Obtain a 98% confidence interval for the difference between the
mean operative times of the dynamic and static systems.
First: Verify assumptions
Do this with your calculator
Part 1 - At the 1% significance level, do the data provide sufficient evidence
to conclude that the mean operative time is less with the dynamic system
than with the static system?
Let’s think about it:
Populations and variable:
Operative times in minutes for the Dynamic and Static systems
Notice that
x1  394.6 minutes,
x2  468.3 minutes
Is x1-bar lower than x2-bar by chance or significantly lower?
a) Set both hypothesis
Populations and variable:
Operative times in minutes for the Dynamic and Static systems
1  2 
1  2 
1  2  0
1  2  0
This is a left tailed test
b) Sketch graph, shade rejection region, label, and indicate possible locations of the point estimate in
the graph.
You do this
The point estimate is
x1  x2  394.6  468.3  73.7
***You should be wondering: Is x1-bar lower than x2-bar by chance, or is it significantly
lower? The p-value found below will help you in answering this.
c) Use a feature of the calculator to test the hypothesis. (Are you using z or t? Why?)
We are not given the population standard deviations. We are using t.
Indicate the feature used and the results:
Use 2-Samp-TTest and get
Test statistic t = -2.68
P( x1  x2
 73.7 ) = P(t < -2.68) = p-value = .008 < .01 (significance level)
***How likely is it observing such a difference between the x-bars (or a more extreme
one) when the mean of the two populations are equal?
very likely, likely, unlikely, very unlikely
*** Is the difference between the x-bars lower than zero by chance, or is it significantly
lower?
Such a point estimate would be a more likely event in the case in which
1
is lower than
2 . This is why we conclude ***********(see conclusion, part (e))
d) What is the initial conclusion with respect to Ho and H1?
****Reject Ho and support H1
Fail to reject Ho, we don’t have enough evidence to support H1
e) Write the conclusion using words from the problem
********The data provide sufficient evidence to conclude that the mean operative time is less
with the dynamic system than with the static system. (t = -2.68, p = .008)
Part 2 - Obtain a 98% confidence interval for the difference between the mean
operative times of the dynamic and static systems. Are the results consistent with the
results of the hypothesis test? Explain. (Are you using z or t? Why?)
To construct the interval use 2-SampTInterval, Data option and get
- 143.9 <
3.456
Notice that the interval is completely below zero, this supports that
1  2
1   2
< -