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Tests of Hypotheses Involving Two Populations. 12.1 Tests for the Differences of Means Comparison of two means: and The method of comparison depends on.

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Presentation on theme: "Tests of Hypotheses Involving Two Populations. 12.1 Tests for the Differences of Means Comparison of two means: and The method of comparison depends on."— Presentation transcript:

1 Tests of Hypotheses Involving Two Populations

2 12.1 Tests for the Differences of Means Comparison of two means: and The method of comparison depends on the design of the experiment The samples will either be Independent or Dependent

3 Independent Samples – implies data values obtained from one sample are unrelated to values from the other sample Dependent Samples – implies subjects are paired so that they are as much alike as possible The purpose of pairing is to explain subject to subject variability. (In some studies we apply both treatments to the same subject) If subject to subject variability is large relative to the expected treatment differences then a dependent sample design should be considered

4 12.3 Tests for Differences of Means of Paired Samples Observe n pairs of observations observation for treatment i in pair j difference between treatments 1 and 2 in pair j

5 PairTreatment 1Treatment 2Difference 1 2 3............ n Sample Mean Sample Variance Sample Standard Deviation

6 For dependent sample design all analysis is based on differences: The differences are a sample from a distribution with mean and an unknown variance We can calculate Confidence Intervals and perform Hypothesis Tests in the same way as with one sample

7 Point Estimate for is The Test Statistic is which has a t-distribution with n-1 degrees of freedom Confidence Interval for

8 Example A marketing expert wishes to prove that a new product display will increase sales over a traditional display Treatment 1:New Method Treatment 2:Traditional Method 12 stores are available for the study There is considerable variability from store to store We will divide the 12 stores into six pairs such that within each pair the stores are as alike as possible Measurement will be cases sold in a one month period

9 Pair New Method Treatment 1 Traditional Method Treatment 2Difference 113112 231292 32021 419172 542393 626224 Sample Mean Sample Variance Sample Standard Deviation

10 Perform a Hypothesis test using Conclude with 95% confidence that the new method produces larger sales. Perform and interpret a 95% confidence interval We are 95% confident that the mean increase in sales is between 0.244 cases and 3.756 cases using the new product display.

11 12.2 Tests of Differences of Means Using Small Samples from Normal Populations When the Population Variances Are Equal but Unknown Notation for Two Independent Samples Sample 1 (Treatment 1)Sample 2 (Treatment 2) Sample Size Data Sample Mean Sample Variance Sample Standard Deviation

12 The point Estimate for (Independent Samples) is This is the same as what we did with dependent samples The sampling distribution for

13 If we sample from populations with means of and, and standard deviations of and respectively, then is approximately standard normal for large n. The form of a confidence interval and a hypothesis test for two independent samples depends on what we assume

14 If we sample from populations 1 and 2, the samples are independent, and then has a t-distribution with Where is pooled estimate of This means

15 Thus for a Hypothesis Test the Test Statistic is and a Confidence Interval is calculated using

16 Example A study is designed to compare gas mileage with a fuel additive to gas mileage without the additive. A group of 10 Ford Mustangs are randomly divided into two groups and the gas mileage is recorded for one tank of gas. Treatment 1 (with additive) Treatment 2 (without additive) Sample Size Data26.3, 27.4, 25.1, 26.8, 27.124.5, 25.4, 23.7, 25.9, 25.7 Sample Mean Sample Variance

17 Do we have enough evidence at to prove that the Additive increases gas mileage. Assume Conclude with 95% confidence that the additive improves gas mileage Compute and interpret a 95% Confidence Interval for We are 95% confident that the additive will increase gas mileage by an amount between 0.177 and 2.823 miles.

18 What if we do not assume ? It then makes no sense to compute A reasonable variable is Unfortunately the distribution of T is unknown.

19 For large and, T is approximately standard normal. For small and, the distribution of T can be approximated by a t-distribution using a complicated formula for the degrees of freedom.


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