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Chapter 25 Paired Samples and Blocks. Independent vs. Dependent Samples We would like to determine if students taking an ACT prep course will score better.

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Presentation on theme: "Chapter 25 Paired Samples and Blocks. Independent vs. Dependent Samples We would like to determine if students taking an ACT prep course will score better."— Presentation transcript:

1 Chapter 25 Paired Samples and Blocks

2 Independent vs. Dependent Samples We would like to determine if students taking an ACT prep course will score better than students not taking the course. A random sample of 25 students was chosen who took the course and a random sample of another 25 students was chosen who did not take the course. At the end of the prep course, both groups were given the ACT.

3 Independent vs. Dependent Samples Group #1 – score on ACT from students taking prep course Group #2 – score on ACT from students NOT taking prep course Observations taken from two independent samples Use the difference of means test 2 sample t-test Lets change the study just a little bit...

4 Independent vs. Dependent Samples We would like to determine if students can improve their ACT score by taking a prep course. A random sample of 25 students was chosen. They first took the ACT test. Then they spent 6 weeks taking the prep course. At the end of the 6 weeks, they took the ACT test again.

5 Independent vs. Dependent Samples Group #1 – score on ACT before prep course Group #2 – score on ACT after prep course Two observations taken for each subject The samples are tied together or are dependent Use the matched pairs test or The t-test on the difference of means

6 Important Difference If the values come from:If the values come from: –two independent samples use inference for (μ 1 – μ 2 ) = difference in means for two groups –dependent samples Matched Pairs (e.g. values collected twice from same subject) use inference for μ d = mean difference between two values

7 Lets test your ability to distinguish between the two

8 1. To test the effect of background music on productivity, the workers are observed. For one month they had no music. For another month they had background music. A worker's productivity measurement with music is paired with a productivity measurement for the same worker without musicrandomize order.

9 2. A random sample of 10 workers in Plant A are to be compared to a sample of 10 workers in Plant B. If we pick at random a worker in Plant A, we have no information that would allow us to match that worker to another worker in Plant B. Hence we would treat these as independent samples.

10 3. A new weight reducing diet was tried on ten women. The weight of each woman was measured before the diet, and again after being on the diet for ten weeks. A woman's weight before using the diet is paired with a weight for the same woman after the diet.

11 4. To compare the average weight gain of pigs fed two different rations, nine pairs of pigs were used. The pigs in each pair were litter-mates. As it says, litter-mates are paired. Some hog farmers question whether this is a worthwhile pairing. Randomly assign members of pairs to two foods.

12 5. To test the effects of a new fertilizer, 100 plots are treated with one fertilizer, and 100 plots are treated with the other. If we pick at random a plot treated with the new fertilizer, we have no information that would allow us to match that plot with one of the plots where the old fertilizer was used. Hence we would treat these as independent samples.

13 6. A sample of college teachers is taken. We wish to compare the average salaries of male and female teachers. We would like to pair teachers with the same degrees, years of experience, publications, etc., but this is usually impossible and we have no choice but to take independent samples.

14 7. A new fertilizer is tested on 100 plots. Each plot is divided in half. Fertilizer A is applied to one half and B to the other. This is called a split plot design. The results obtained on one half of the plot are paired with the results obtained on the other half of the same plot. Randomly assign fertilizers to the two halves.

15 8. Consumers Union wants to compare two types of calculators. They get 100 volunteers and ask them to carry out a series of 50 routine calculations (such as figuring discounts, sales tax, totaling a bill, etc.). Each calculation is done on each type of calculator, and the time required for each calculation is recorded. The time it takes one individual to do a particular calculation on one calculator is paired with the time it took the same person to do the same calculation on the other calculator. Randomly assign order.

16 Inference for μ d {the formulas} d = pairwise difference {find the difference for each and avg them} Confidence Interval Hypothesis Test (degrees of freedom = n-1)

17 Matched Pairs Hypothesis Test Example Measure effectiveness of exercise program in lowering blood cholesterol levels –SRS of men from a population –Measure cholesterol before program starts –Go through exercise program for 12 weeks –Measure cholesterol after program ends Is the exercise program effective?

18 Matched Pairs Hypothesis Test Example - Data Subject Pre Post

19 When to use Paired T-test Two observations from each subject Post value depends on pre value Observations are dependent Do not have independent samples Cannot use inference for μ 1 - μ 2 –Violates independent samples assumption

20 Matched Pairs Hypothesis Test Example - Data Look at pre-program levels minus post-program levels Subject Pre Post Diff

21 Matched Pairs Because it is the differences we are interested in, we will treat them as the data and ignore the original two groups. A matched pairs t-test is just a one-sample t-test for the mean of the pairwise differences.

22 Find the 1-var stats for the differences Let μ d = mean difference in cholesterol levels in population Sample mean difference is Sample standard deviation of differences is s d = n = number of differences = 10

23 Hypothesis Statements If exercise program has no effect, mean difference will be 0 If exercise program is effective, mean difference will be positive H O : μ d = 0 H A : μ d > 0

24 Significance Level Significance level = 95% Alpha Level = 0.05 Test statistic:

25 Conditions Random sample Nearly Normal Population Samples < 10% Population Not independent - two measures on same subjects

26 Matched Pairs T-test t = 1.42 p-value = Run the test using the calculator

27 Compare p-value to alpha Since p-value > α, we will fail to reject the null hypothesis.

28 Conclusion: There is no evidence that the exercise program reduced the mean cholesterol level of men in this population.

29 Matched Pairs Confidence Interval Example For a random sample of 12 European cities, the average high temperatures in January and July are given on the following slide. Find a 90% confidence interval for the mean temperature difference between summer and winter in Europe. Assume the Nearly Normal assumption is satisfied.

30 Matched Pairs Confidence Interval Example Jan July Diff

31 Matched Pairs Confidence Interval Example Check assumptions Random - ok n 10% N - ok Nearly Normal - ok NOT independent

32 Matched Pairs Confidence Interval Example Mean and Stdev

33 Matched Pairs Confidence Interval Example

34 Matched Pairs Confidence Interval Example interpretation of interval We are 90% confident that the true temperature difference between average summer and winter highs in European cities is contained in the interval (32.33, 41.31)


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