Example Bowker and Lieberman (1972) gave the results for ...
Hypothesis Test for Comparing Two Population Means
Using Independent Random Samples (Pooled-t Test)
Example Bowker and Lieberman (1972) gave the results for tests of two thermostats used in irons that were made by an old supplier and a new supplier. Twenty three thermostats from each supplier were tested at a setting of 550 F. The actual temperature, measured with a thermocouple, are given below. Do the data provide evidence of differences in the mean temperatures for the thermostats from the two suppliers? Conduct test at α=.05.
new old
530.3 559.7
549.4 554.8
551.7 544.6
549.9 550.7
536.7 551.1
538.8 538.8
559.1 554.5
538.6 538.4
565.4 552.9
550.0 555.0
554.7 558.4
569.1 560.3
559.3 534.7
544.0 545.0
566.3 538.0
556.9 563.1
558.8 553.8
543.3 564.6
555.0 553.0
551.1 548.3
554.9 535.1
554.9 544.8
536.1 548.7
Entering the data in Minitab:
|New |Old |
|530.3 |559.7 |
|549.4 |554.8 |
|. |. |
|. |. |
|536.1 |548.7 |
Is Normality Assumption Reasonable?
In Minitab: STAT> BASIC STATISTICS> NORMALITY TEST> VARIABLE New
[pic]
Since the plotted points fall approximately along a straight line, we can conclude that the normal distribution assumption is reasonable.
In Minitab: STAT> BASIC STATISTICS> NORMALITY TEST> VARIABLE Old
[pic]
Since the plotted points fall approximately along a straight line, we can conclude that the normal distribution assumption is reasonable.
Is Equal Variances Assumption Reasonable?
IN MINITAB:
STAT> BASIC STATISTICS > 2 VARIANCES > EACH SAMPLE IS IN ITS OWN COLUMN >
Ho: [pic]= [pic]
Ha: [pic][pic][pic]
α=.05
Test
Method DF1 DF2 Statistic P-Value
Bonett 1 — 0.75 0.385
Levene 1 44 0.36 0.549
p-value > α thus we fail to reject H0 at α=.05. We can conclude that the assumption of equal variances is reasonable. So we can go ahead and conduct pooled-t test.
Ho: μnew=μold
Ha: μnew[pic]μold
MINITAB COMMANDS: STAT > BASIC STATISTICS > 2-SAMPLE t >
EACH SAMPLE IS IN ITS OWN COLUMN Sample 1 New Sample 2 Old > OPTIONS > Click ASSUME EQUAL VARIANCES > ALTERNATIVE > NOT EQUAL
Two-sample T for New vs Old
N Mean StDev SE Mean
New 23 551.1 10.3 2.1
Old 23 549.93 8.80 1.8
Difference = μ (New) - μ (Old)
Estimate for difference: 1.13
95% CI for difference: (-4.56, 6.82)
T-Test of difference = 0 (vs ≠): T-Value = 0.40 P-Value = 0.691 DF = 44
Both use Pooled StDev = 9.5745
p-value > α thus we fail to reject Ho at α=.05. We do not have sufficient evidence to conclude that there exists a difference in the mean temperatures for the thermostats from the two suppliers.
................
................
In order to avoid copyright disputes, this page is only a partial summary.
To fulfill the demand for quickly locating and searching documents.
It is intelligent file search solution for home and business.
Related download
- one sample t test
- copying independent t test output from spss to ms word
- 2 sample t test nc state university
- chapter 8 introduction to hypothesis testing
- t test two sample assuming equal variance tutorial
- 2 sample t test independent samples
- hypothesis testing an example
- pooled t test
- example bowker and lieberman 1972 gave the results for
Related searches
- take five lottery results for tonight
- ny lottery results for friday
- ny lottery results for sunday
- ny lottery results for saturday
- lottery numbers results for wednesday
- quest lab results for physicians
- blood test results for lupus
- enter lab results for interpretation
- lottery results for saturday night
- euromillions lottery results for today
- lottery results for last night
- interpreting dna results for genealogy