Handout_15b(2) - Apr 2510 HYPOTHESIS TESTING: ILLUSTRATIVE...

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Apr 25’10 HYPOTHESIS TESTING: ILLUSTRATIVE #15b EXAMPLES INVOLVING t DISTRIBUTION As we have already noted, the population variance is usually un known in actual situations involving statistical inference about a population mean. When sampling is from a normally (or approximately normally) distributed population with an unknown variance , the relevant test statistic for testing a null hypothesis H 0 is _ _ t = (x – µ 0 )/(s/√n) (15b.1) _ where s is a sample (not population!) deviation, n is a sample size, x is a sample mean, and µ 0 is the hypothesized value of the mean. When H 0 is true, the statistic (15b.1) is distributed as Student’s t with n – 1 degrees of freedom. The hypothesis testing procedure resembles the ten-step procedure that have been described in the handouts #15, 15a for the situation when a p opulation va riance is known, but now we deal with t statistic (15b.1) instead of z statistic z = (x – µ 0 )/(σ/√n ) (15a.1). The following example illustrates the hypothesis testing procedure involving the t statistic. Example 1c (problem 7.2.13 on pg. 234 of the textbook) Can we conclude that the mean maximum voluntary ventilation for apparently healthy college seniors in not 110 liters per minute? Let α = 0.01 . A sample of 20 individuals yields the following values: 132, 33, 91, 108, 67, 169, 54, 203, 190, 133, 96, 30, 187, 21, 63, 166, 84, 110, 157, 138 Solution . Researchers can conclude that the mean maximum voluntary ventilation is different from 110 liters if they can reject the null hypothesis that the mean maximum voluntary ventilation is equal to 110 liters . Let us follow the ten-step decision-making procedure of two - sided hypothesis test which is similar to the procedure described earlier in handouts #15, 15a. 1. Data . The available data on the values of maximum voluntary ventilation of 20 individuals randomly selected from the population of interest allows us to compute a sample mean of 111.6 liters and a sample deviation of 56.30313 liters [use the MS Excel functions = AVERAGE() and = STDEV() ] 2. Assumptions . It is assumed that the population in question is approximately normally distributed with an unknown variance. 3. Hypotheses . The null hypothesis H 0 to be tested is that the mean age of the population is equal to 110 liters. The alternative hypothesis H A is that the mean age of the population is not equal to 110 liters. Symbolically these hypotheses may be written as H 0 : µ = µ 0 and H A : µ ≠ µ 0 or
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This note was uploaded on 11/24/2011 for the course MATH 3000 taught by Professor Kzaer during the Spring '05 term at St. Johns College MD.

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Handout_15b(2) - Apr 2510 HYPOTHESIS TESTING: ILLUSTRATIVE...

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