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# Business Statistics for Managerial Decision - PowerPoint PPT Presentation

Business Statistics for Managerial Decision. Comparing two Population Means. Comparing Two means. How do small businesses that fail differ from those that succeed?

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## Business Statistics for Managerial Decision

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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

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
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.

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.
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.
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?
• 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 11.01

Control 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

Example: Is our product effective?
• The two-sample t statistic is
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.
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.
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.
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 s12 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
The Pooled Two-sample t Procedures
• Sp2 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.
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*.

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.

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
• 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

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.
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
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
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.