1. One Sample β Example 1. Over the past five hundred years or so, the amount of time it has taken Santa Claus to climb down chimneys, deposit presents and get back to his sleigh has been normally distributed with a mean of 10 seconds per chimney and a standard deviation of 3 seconds. This year, since he's beginning to feel his age, Santa has engaged in a rigorous physical fitness program which he believes will allow him to significantly reduce his time. To test his belief, he plans to have one of the elves time him on a random sample of 36 chimneys and then to conduct the hypothesis test with α ≤ .025. If, in fact, his true mean time is now 8 seconds, what is the probability that Santa will incorrectly conclude that his exercise had no effect?

2. One Sample β Example µ = 10 σ = 3 n = 36 µA = 8 α = .025 Z = 1.96 • First step is to obtain the value of X that must be exceeded in order to reject HO. Z = -1.96 = X – 10 3/√36 X = -1.96 (1/2) + 10 = 9.02

3. One Sample β Example 2. Second step is to work out the probability that sample mean does NOT exceed that value, given that the new population mean is 8. Z = 9.02 – 8.0 = 2.04 3/√36 P(Z ≤ 2.04) = .4793. Desired P = .4 - .4793) = .0207

4. Two Sample β Example 1 2. The Christmas exam scores of two sections of a course are compared to see if a significant difference exists between their means. It is assumed that students were originally randomly assigned to their sections. Section 1Section 2 X1= 66.0 X2= 62.0 s1= 10.75 s2 = 12.46 n1 = 50 n2 = 38

5. Two Sample β Example 1 a) Perform the appropriate test on the data using α ≤ .05: b) Suppose that, over a period of years, section 1 has actually outperformed section 2 by 2.8 points, on average. What is the probability that this year's sample data would allow you to correctly reject the null hypothesis of no difference between the sections (still using α ≤ .05 and a 2-tailed alternative hypothesis)?

6. Two Sample β Example 1 HO: µ1 – µ2 = 0 HA: µ1 – µ2 ≠ 0 Zcrit = Z.025 = 1.96 Zobt = (66 – 62) – 0 = 1.58 10.75 + 12.46 50 38 Do not reject HO √ 2 2

7. Two Sample β Example 1 b. The alternative hypothesis would specify that (µ1 – µ2) = 2.8. The question asks, with the actual difference in the populations being 2.8, and with sample sizes n1 and n2, and given s1 and s2, what is the probability that we exceed the critical value for X1 – X2 (and thus reject HO)?

8. Two Sample β Example 1 • First we work out the critical value of X1 – X2. 1.96 = (X1 – X2) – 0 2.53 0 + 1.96 (2.53) = (X1 – X2) = 4.96

9. These are sampling distributions of the difference between the sample means .025 0 4.96 .3023 .1977 2.8 4.96

10. Two Sample β Example 2 3. One of the reasons more people buy Florida orange juice is that Florida oranges have historically had higher levels of vitamin C (on average 120 mg. per orange with a standard deviation of 10) than California oranges (on average 110 mg. per orange with a standard deviation of 15). This year, California orange growers are using a new fertilizer in an effort to increase the level of vitamin C in their oranges. The alternative hypothesis that you plan to test is that because of the new fertilizer the difference in vitamin C concentrations between California and Florida oranges is now less than 10 mg.

11. Two Sample β Example 2 To carry out this test you plan to sample 40 oranges from each State. If, because of the fertilizer, the true difference is now only 2 mg., what is the probability that you would be able to reject the null hypothesis that the true difference has not changed (α ≤ .01)?

12. Two Sample β Example 2 HO: µF – µC = 10 HA: µF – µC < 10 Zcrit = Z.01 = 2.33 Zobt = (X1 – X2) – 10 = 2.33 10 + 15 40 40 (X1 – X2) = 10 – 2.33 (2.85) = 3.36 √ 2 2

13. Two Sample β Example 2 Z = 3.36 – 2.0 = .48 2.85 P (Z < .48) = .1844 Desired P = .5 + .1844 = .6844

14. These are sampling distributions of the difference between the sample means 3.36 10 .5 .1844 2 3.36