Two-Sample Inference Procedures with Means

Of the following situations, decide which should be analyzed using one-sample matched pair procedure and which should be analyzed using two-sample procedures? A pharmaceutical company wants to test its new weight-loss drug. Before giving the drug to a random sample, company researchers take a weight measurement on each person. After a month of using the drug, each person’s weight is measured again. Matched pair

Of the following situations, decide which should be analyzed using one-sample matched pair procedure and which should be analyzed using two-sample procedures? A researcher wants to know if a population of brown rats on one city has a greater mean length than a population in another city. She randomly selects rats from each city and measures the lengths of their tails. Two independent samples

Of the following situations, decide which should be analyzed using one-sample matched pair procedure and which should be analyzed using two-sample procedures? A researcher wants to know if a new vitamin supplement will make the tails of brown rats grow longer. She takes 50 rats and divides them into 25 pairs matched by gender and age. Within each pair, she randomly selects one rat to receive the new vitamin. After six months, she measures the length of the rat’s tail. Matched pair

Two independent samples Of the following situations, decide which should be analyzed using one-sample matched pair procedure and which should be analyzed using two-sample procedures? A college wants to see if there’s a difference in time it took last year’s class to find a job after graduation and the time it took the class from five years ago to find work after graduation. Researchers take a random sample from both classes and measure the number of days between graduation and first day of employment Two independent samples

Matched Pairs (Special type of one-sample means)

2) The sample distribution of differences is approximately normal Differences of Paired Means (Matched Pairs) CONDITIONS: 1) The samples are paired. The sample differences can be viewed as a random sample from a population of differences. 2) The sample distribution of differences is approximately normal - the populations of differences is known to be normal, or - the number of sample difference is large (n  30), or - graph data to show approximately normal 3) 10% rule – The sample of differences is not more than 10% of the population of differences.

Differences of Paired Means (Matched Pairs)

Hypothesis Statements: Differences of Paired Means (Matched Pairs) Parameter: md = true mean difference in … Hypothesis Statements: H0: md = hypothesized value Ha: md < hypothesized value Ha: md> hypothesized value Ha: md ≠ hypothesized value

Differences of Paired Means (Matched Pairs) Hypothesis Test:

#18 Summer School. Having done poorly on their Math final exams in June, six students repeat the course in summer school and take another exam in August. If we consider these students to be representative of all students who might attend this summer school in other years, do these results provide evidence that the program is worthwhile? June 54 49 68 66 62 Aug. 50 65 74 64 72

Ho: μd = 0 Page 590: 18 Parameters and Hypotheses μd = the true mean difference in scores between June and August for students who repeated the course in summer school Ho: μd = 0 Ha: μd > 0

Assumptions (Conditions) 1) The samples must be paired and random. The samples are from the same student so they are paired and we will assume the sample differences are a random sample of the population of differences.. 2) The sample distribution should be approximately normal. The normal probability plot is fairly linear and the boxplot shows no outliers, so we will assume that the sample distribution of differences is approximately normal. 3) The sample should be less than 10% of the population. The population should be at least 60 students, which we will assume. 4)  is unknown Since the conditions are met, a t-test for the matched pairs is appropriate.

Calculations 5.333 7.4475 5.333 7.4475  = 0.05

Decision: Since p-value > a, I fail to reject the null hypothesis at the .05 level. Conclusion: There is not sufficient evidence to suggest that the true mean difference in scores is different from June to August. This suggests that program may not be worthwhile.