false

Textbook Notes
(369,140)

Canada
(162,411)

University of Alberta
(2,708)

BIOL499A
(11)

Blaine Mullins
(11)

Chapter 24

School

University of Alberta
Department

Biology (Biological Sciences)

Course Code

BIOL499A

Professor

Blaine Mullins

Description

Chapter 24: Comparing Two Population Means – Two
Independent Samples
In this chapter, we will talk about:
The difference between independent and dependent samples,
Two-sample hypothesis tests of means when 1 2 is unknown,
Confidence intervals fo 1 2 when 1 2 is unknown
Two-sample hypothesis tests of means when 1 and 2are unknown and unequal,
Confidence intervals fo 1 2when 1and 2 are unknown and unequal,
For comparing two treatments, the two basic types of design are:
Independent Samples,
Matched pairs samples
Goal for Independent samples:
Case 1: 1 2 is unknown: Use t-test with pooling
Case 2: 1and 2are unknown and unequal: Use t-test without
pooling or conservative t-test. Suppose 10 male students and 20 femalestudents were selected from last
year stat 151 classes and we found:
nys0, 75, 12
1 1 1
and
n2 2 2, 70, 9
Do we have enough evidence to conclude that the average of male student
marks is higher than the average of female student marks?
Interested parameter:
1 2
yy
Estimate: 1 2
To test the hypotheH 0 1 2 0, the test statistic is:
Estimate Null Hypothesis y1 2 0
~t distribution
S.E.(Estimate) S.E.(y1 2 ) ) 1 () Var Y1 2 Va1 Y2 Y
= (1)Var Y 1 2 (-1)Var Y )
= Var Y1 2 Var Y )
1 2
=
n1 2
If 1 2 ,then
2 2 2 2
1 2 211
=a1 2
n1 2 1 2 12n
Therefore,
SD Y 11
1 2 nn
1 2
S. ˆ 11 s1
1 2 nn nn
1 2 1 2
Question: Recall 1 1t,n2s12and 20, 9.Should I use
s1 22 or 9as an estimate of Inference about 1 2 for Two Independent Samples: We consider
following two cases:
Case 1: 1 2s unknown
Case 2: 1 and 2are unknown and unequal
Case 1 (Unknown population standard deviations are equal): Suppose that the two
normal population distributions have same standard deviation. Call the common-and still
unknown-standard deviation of both populations . Both sample variances S1 and S 2
2
estimates . The best way to combine these two estimates is:
2 2 )(11 1 2S22 2
Spooled p
n1 2 2
This is called the pooled estimator of 2 because it combines the information in both
samples.
To test the hypothesisH 0: 1 2 0 , the test statistic is:
Y1 2 0
TS ~ t distribution, with df n1 n 2 2 ,ifH 0s true
11
Sp
n1 2
For H : 0 , we reject H ifTS t *
a 1 2 0
*
For H a 1 2 , we reject H 0if TS t
For H : 0 , we reject H if TS t * or TS t *
a 1 2 0
A 100(1)% confidence interval for is given by:
1 2
11
Y1 2t * p
n1 2 Example 24.1: A study was conducted to find out whether the children of
aggressive families have a higher beha vior problem checklist (BPC) score.
The sample mean and standard deviation of BPC scores for 13 children
whose parents were classified as aggressive were 7.92 and 3.45 respectively.
For a sample of 12 children whose parents were classified as non-aggressive,
the mean and standard deviation of BPC score were 5.80 and 2.87,
respectively. Based on the data, a psychiatrist claimed that the children of
aggressive families have higher behavior problem checklist (BPC) scores on
average. Test the psychiatrist's claim at 0.01 .
Solution: Example 24.2: A psychologist was in terested in the effects of two different
kinds of drugs on the mean time to complete a task. She randomly chose 10
subjects and assigned 5 of them to each drug A, and B. The following data
represent the time in minutes required to complete the task.
DruAg 30 24 30 25 26
DruBg 26 25 21 26 27
Assume the time to complete the task for the two drugs are approximately
normally distributed and have equal vari ances. Is there enough evidence to
conclude that the population mean time to complete the task is smaller for
drug B than drug A?
Solution:
2 )(n1As A B B
s p
nA B 2 H0:
1)
Ha:
2) 0.05
YYA B 0
3) TS ~t distribution if H 0is true.
S 11
p nA B
df n n 2
A B
4) We reject H 0if
5) The value of TS is:
6) Since
At 0.05,there enough evid ence to conclude that the
population mean time to complete the task is smaller for drug B than drug A
7) P-value = Example 24.3: The operations manager at a light bulb factory wants to
determine if there is any difference in the average life expectancy of bulbs
manufactured on two different types of machines. A random sample of 35
light bulbs obtained from machine I indicates a sample mean of 375 hours,
and a sample standard deviation 110 hours. A similar sample of 25 from
machine II indicates a sample mean of 362 hours and a sample standard
deviation 125 hours.
(a) Find a 95% confidence interval for the difference of the average life of
bulbs produced by the two types of machines.
Solution:
LargestS 125
Since 2
SmallestS 110
)((1n s 2 2
s p 1 1 22
n1 2 2
1 0.95 0.05 0.025
2
df n1 n 2 2
A 95% confidence interval for 1 2is:
* 11
y1 2 p
n1 2
Conclusion: It is estimated with 95% confidence that the average life of
bulbs produced by first machine is between hours smaller to
hours larger than the average life of bulbs produced by second machine
Since zero is in the confidence interval,
Hence, we conclude that there is no difference in the average life of bulbs
produced by the two machines. (b) Using the 0.05 level of significance, is there any evidence of a
difference in the average life of bulbs produced by the two types of
machines?
Solution:
H :
0
1)
H :
a
2) 0.05
Y1 2 0
3) TS ~ t distribution if H 0 is true.
S 11
p n1 2
df n n 2
1 2
4) We reject H 0if
5) The value of TS is:
6) Since
At 0.05 , there e

More
Less
Unlock Document

Related notes for BIOL499A

Only pages 1,2,3,4 are available for preview. Some parts have been intentionally blurred.

Unlock DocumentJoin OneClass

Access over 10 million pages of study

documents for 1.3 million courses.

Sign up

Join to view

Continue

Continue
OR

By registering, I agree to the
Terms
and
Privacy Policies

Already have an account?
Log in

Just a few more details

So we can recommend you notes for your school.

Reset Password

Please enter below the email address you registered with and we will send you a link to reset your password.