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Business Statistics for Managerial Decision. Comparing two Population Means. Comparing Two means. How do small businesses that fail differ from those that succeed? - PowerPoint PPT Presentation
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Business Statistics for Managerial Decision
Comparing two Population Means
Comparing Two means How do small businesses that fail differ
from those that succeed? Business school researchers compare two
samples of firms started in 2000, one sample of failed businesses and one of firms that are still going after two years.
This study compares two random samples, one from each of two different populations.
Two-Sample problems The goal of inference is to compare the
responses in two groups Each group is considered to be a sample
from a distinct population. The responses in each group are
independent of those in other group
Two-Sample problems Notation
We have two independent samples, from two distinct populations (such as failed businesses and successful businesses).
We measure the same variable (such as initial capital) in both samples
We call the variable x1 in the first population and x2 in the second population.
Population Variable Mean Standard deviation
1 x1 1 1
2 x2 2 2
Two-Sample problems We want to compare the two population means,
either by giving a confidence interval for 1-2 or by testing the hypothesis of no difference, H0:1=2.
We base inference on two independent SRSs, one from each population.
Sample Sample
Population Sample size mean standard deviation
1 n1 s1
2 n2 s2
1x
2x
The Two-Sample z Statistic The natural estimator of the difference 1-2 is the
difference between the sample means . To base inference on this statistic we need to know
its sampling distribution. The mean of the difference is the difference of the
means 1-2.
Because the samples are independent, their sample means and are independent.
The variance of the is the sum of their variances which is
21 xx
21 xx
21 xx
2
22
1
21
nn
1x 2x
The Two-Sample z Statistic Suppose that is the mean of a SRS of size n1
drawn from a N(1, 1) population and that is the mean of an independent SRS of size n2 drawn from a N(2, 2) population. Then the two-sample z statistic
has the standard Normal (0, 1) sampling distribution.2
22
1
21
2121 )()(
nn
xxz
1x2x
The Two-Sample t Procedures In practice, the two population standard deviations 1 and 2 are
not known We estimate them by sample standard deviations s1 and s2 from
our two samples. The two-sample t statistic:
This statistic does not have a t distribution. We can approximate the distribution of the two-sample t statistic
by using the t(k) distribution with an approximation for the degrees of freedom k.
2
22
1
21
2121 )()(
ns
ns
xxt
The Two-Sample t Procedures We use the approximation to find approximate
value of t* for confidence intervals and to find approximate P-values for significance tests.
This can be done in two ways: Scatterwait approximation to calculate a value of k
from data. In general, the resulting k will not be a whole number.
Use degrees of freedom k equal to the smaller of n1-1 and n2-1.
The Two-Sample t Significance Test Draw a SRS of size n1 from a Normal population with
unknown mean 1 and an independent SRS of size n2 from another Normal population with unknown 2. To test the hypothesis H0: 1-2= 0, compute the two-sample t statistic
And use P-values or critical values for the t(k) distribution, where the degree of freedom k are the smaller of n1-1 and n2-1.
2
22
1
21
2121 )()(
ns
ns
xxt
Example: Is our product effective?
A company that sells educational materials reports statistical studies to convince customers that its materials improve learning. One new product supplies ”directed reading activities” for class room use. These activities should improve the reading ability of elementary school pupils.
Example: Is our product effective?
A consultant arranges for a third-grad class of 21 students to take part in these activities for an eight-week period. A control classroom of 23 third-graders follows the same curriculum without the activities. At the end of the eight weeks, all students are given a Degree of Reading Power (DRP) test, which measures the aspects of reading ability that the treatment is designed to improve. The data appear in table 7.3.
Example: Is our product effective?
Example: Is our product effective?
A back to back stemplot suggests that there is a mild outlier in the control group but no deviation from Normality serious enough to forbid use of t procedure.
Example: Is our product effective? The summary statistics are
Group n s
Treatment 21 51.48 11.01Control 23 41.52 17.15
We hope to show that the treatment (group 1) is better than the control (group 2), therefore the hypotheses are
H0: 1= 2
Ha: 1 > 2
x
Example: Is our product effective?
The two-sample t statistic is
31.2
23)15.17(
21)01.11(
52.4148.5122
2
22
1
21
21
ns
ns
xxt
Example: Is our product effective? The P-value for the one-sided test is
The degrees of freedom k are equal to the smaller of n1-1= 21-1=20 and n2-1=23-1=22 comparing t= 2.31 with entries in t-table for 20 degrees of freedom, we see that P lies between .02 and .01.
Conclusion The data strongly suggest that directed reading activity
improves the DRP score.
)31.2( tP
The Two Sample t Confidence Interval The same ideas that we used for the two-sample t significance test
can apply to give us two-sample t confidence interval. Draw a SRS of size n1 from a Normal population with unknown
mean 1 and an independent SRS of size n2 from another Normal population with unknown mean 2. The confidence interval for 1- 2 given by
t* is the value for t(k) density curve with area C between –t* and t*. The value of the degrees of freedom k is approximated by software or we use the smaller of n1-1 and n2-1.
2
22
1
21
21 *)(n
s
n
stxx
Example:How much improvement? We will find a 95% confidence interval for the mean improvement in the
entire population of third-graders. The interval is
Using t(20) distribution, t-table gives t* = 2.086
We estimate the mean improvement in DRP scores to be about 10 point, but with a margin of error of almost 9 points.
23
)15.17(
21
)01.11(*)52.4148.51(
*)(
22
2
22
1
21
21
t
n
s
n
stxx
)9.18,0.1(99.896.9
31.4086.296.923
)15.17(
21
)01.11(086.2)52.4148.51(
22
The Pooled Two-sample t Procedures
There is one situation in which a t statistic for comparing two means has exactly a t distribution.
Suppose that the two Normal population distribution have the same standard deviation.
Call the common standard deviations . Both sample variances s1
2 and S22 estimate 2.
The best way to combine these two estimates is to average them with weights equal to their degrees of freedom.
The resulting estimate of 2 is
2
)1()1(
21
222
2112
nn
snsnsp
The Pooled Two-sample t Procedures Sp
2 is called the pooled estimator of 2. When both populations have variance 2, the addition rule for
variance says that has variance equal to the sum of the individual variances
Now we can substitute sp2 in the test statistic, and the resulting t
statistic has a t distribution.
21)(
11ˆ21 nn
sES pxx
21 xx
)11
(21
2
2
2
1
2
nnnn
The Pooled Two-sample t Procedures
Draw a SRS of size n1 from a Normal population with unknown mean 1 and an independent SRS of size n2 from another Normal population with unknown mean 2. Suppose that the two populations have the same unknown standard deviation. A level C confidence interval for 1- 2 is
Here t* is the value for the t(n1+n2 -2) density curve with area C between -t* and t*.
2121
11*)(
nnstxx p
The Pooled Two-sample t Procedures
To test the hypothesis H0: 1=2, compute the pooled two-sample t statistic
and use P-values from the t(n1+ n2 - 2) distribution.
21
21
11nn
s
xxt
p
Healthy Companies versus Failed Companies
In what ways are companies that fail different from those that continue to do business?
To answer this question, one study compared various characteristics of 68 healthy and 33 failed firms.
One of the variables was the ratio of current assets to current liabilities.
The data appear in table 7.4.
Healthy Companies versus Failed Companies
Healthy Companies versus Failed Companies
First let’s Look at the data. Histograms for the two groups
of firms superimposed with a Normal curve with mean and standard deviation equal to the sample values is given.
The distribution for the healthy firms looks more Normal than the distribution for the failed firms.
Healthy Companies versus Failed Companies
The back to back stemplot confirms our findings from the previous plots that there are no outliers or strong departure from Normality that prevent us from using the t procedure for these data.
Example: Do mean asset/liability ratio differ?
Take group 1 to be the firms that were healthy and group 2 to be those that failed. The question of interest is whether or not the mean ratio of current assets to current liabilities is different for the two groups. We therefore test
H0:1 =2
Ha:1 2
Example: Do mean asset/liability ratio differ?
Here are the summary statistics:
Group Firm n s
1 Health 68 1.7256 .6393 2 Failed 33 0.8236 .4811
The sample standard deviations are fairly close.We are willing to assume equal population standard deviations. The pooled sample variance is
and
x
35141.023368
)4811.0)(32()6393.0)(67(
2
)1()1(
22
21
222
2112
nn
snsnsp
5928.35141.0 ps
Example: Do mean asset/liability ratio differ?
The pooled two-sample t statistic is
The P-value is Where t has t(99) distribution. In t-table we have entries for100 degrees of
freedom. We will use the entries for 100. Our calculated value of t is larger than the t-value corresponding to p = .0005 entry in the table. Doubling 0.0005 , we conclude that the two sided P-value is less than .001.
17.7
331
681
5928.0
8236.07256.1
11
21
21
nns
xxt
p
)17.7(2 tPP
Example: How different are mean Asset/liability ratios?
P-value is rarely a complete summary of a statistical analysis. To make a judgment regarding the size of the difference between the two groups of firms, we need a confidence interval.
The difference in mean current assets to current liabilities ratios for healthy versus failed firms is
151.18236.07256.121 xx
Example: How different are mean Asset/liability ratios?
For a 95% margin of error we will use the critical value t* = 1.984 from the t(100) distribution. The margin of error is
This will gives the following 95% confidence interval
249.33
1
68
1)5928.0)(984.1(
11*
21
nn
st p
)40.1,90.0(
249.0151.1
11*)(
2121
nn
stxx p
Example: How different are mean Asset/liability ratios?
We report that the successful firms have current assets to current liabilities ratio that average 1.15 higher than failed firms, with margin of error 0.25 for 95% confidence.
Alternatively, we are 95% confident that the difference is between 0.90 and 1.40.