Homework #1
due in two weeks
Two-sample-test
In this homework, you need to develop your own problem, collect your data, analyze
your data by Minitab software and answer the following questions:
1. What is your problem? Why is it related to two-sample-test?
2. What are the sample sizes large enough for each sample to detect the mean
3.
difference? That is, you need to find the  (You are going to assume same sample
sizes and two-sample-t-test to be used here)
According to 2., how would you collect your data? What is your sampling plan?
4.
5.
6.
After collecting your data, are these data in normal distribution?
Are these two samples have same variances?
What are your data analysis methods? What are the Minitab functions used?
7.
What is the p-value in your test? What is the confidence interval for the mean
difference?
What are the results?
Example:
1. Two machines are used for filling plastic bottles with a net volume of 16.0
ounces. The quality engineering department suspects that both machines fill to
2.
the same net volume, whether or not this volume is 16.0 ounces. In the past, the
customers would complain their quality if the weight difference is larger than
0.25 ounces. And the variances for these two machines are assumed to be the
same σ1=σ2=0.12. This problem could become a two-sample-t-test or two
sample-z-test.
If one want to have a power of test closely to 99%, the sample size should be
Power and Sample Size
2-Sample t Test
Testing mean 1 = mean 2 (versus not =)
Calculating power for mean 1 = mean 2 + difference
Alpha = 0.05
Difference
0.25
Assumed standard deviation = 0.12
Sample
Target
Size
Power
10
0.99
Actual Power
0.992617
The sample size is for each group.
3.
An experiment is performed by taking a random sample from the output of each
machine.
4.
Machine 1
16.03
16.04
16.05
16.05
16.02
Machine 2
16.01
15.96
15.98
16.02
15.99
16.02
15.97
15.96
16.01
15.99
16.03
16.04
16.02
16.01
16.00
Probability Plot of Machine 1, Machine 2
Normal -95% CI
99
Variable
Machine 1
Machine 2
95
90
Mean StDev N AD
P
16.02 0.03028 10 0.274 0.582
15.99 0.02404 10 0.474 0.186
Percent
80
70
60
50
40
30
20
10
5
1
15.90
15.95
16.00
16.05
16.10
16.15
Data
These two samples are normal
5.
Test and CI for Two Variances: Machine 1, Machine 2
Method
Null hypothesis
Sigma(Machine 1) / Sigma(Machine 2) = 1
Alternative hypothesis
Sigma(Machine 1) / Sigma(Machine 2) not = 1
Significance level
Alpha = 0.05
Statistics
Variable
N StDev
Variance
Machine 1 10 0.030
0.001
Machine 2 10 0.024
0.001
Ratio of standard deviations = 1.260
Ratio of variances = 1.587
95% Confidence Intervals
CI for
Distribution
CI for StDev
Variance
of Data
Ratio
Ratio
Normal
(0.628, 2.527)
(0.394, 6.387)
Continuous
(0.495, 2.202)
(0.245, 4.850)
Tests
Test
Method
DF1
DF2 Statistic
P-Value
F Test (normal)
9
9
1.59
0.503
Levene's Test (any continuous)
1
18
0.18
0.673
These two samples has equal variance.
6.
Since these two samples have equal variance and are from normal distributions,
we will use two-sample-t-test to see if any difference between two machines.
Two-Sample T-Test and CI: Machine 1, Machine 2
Two-sample T for Machine 1 vs Machine 2
N
Mean
StDev SE Mean
Machine 1 10 16.0150
0.0303
0.0096
Machine 2 10 15.9900
0.0240
0.0076
Difference = mu (Machine 1) - mu (Machine 2)
Estimate for difference:
0.0250
95% CI for difference: (-0.0007, 0.0507)
T-Test of difference = 0 (vs not =): T-Value = 2.05
P-Value = 0.056 DF = 18
Both use Pooled StDev = 0.0273
7.
These two machines' fillings have no difference.
revised: 2010/12/29