Download Life Sciences Stats: Mid-semester Exam 3 Solutions - Hypothesis Testing - Prof. David Galv and more Exams Mathematics in PDF only on Docsity! Math 20340: Statistics for Life Sciences Fall 2008 Mid-semester Exam 3 — Solutions 1. Achievement test scores of all high school seniors in a certain state have mean 60. A random sample of 100 students from one large high school had a mean score of 58, with sample standard deviation 8. A researcher want to see if this data provides sufficient evidence to suggest that this high school is inferior. a): State the null and alternative hypotheses for this test. Solution: Let µ be the average score for the entire high school. H0 : µ = 60 Ha : µ < 60 b): Calculate the value of the test statistic. Solution: Test statistic is x̄−60 8/ √ 100 = −2.5 c): Calculate the p-value of the test statistic. Solution: p-value is P (z ≤ −2.5) = .0061. d): Calculate the critical value for rejecting the null at 5% significance. Solution: It’s a one-sided test, so critical value is −1.645. e): Is there sufficient evidence to suggest that this high school is inferior, at 5% significance? Solution: Yes; p-value is smaller that .05, test statistic is more negative than −1.645. f): Is there sufficient evidence to suggest that this high school is inferior, at 1% significance? Solution: Yes; p-value is smaller that .01. 1 2. A coin is called fair if, when tossed repeatedly, the proportion of times it comes up Heads is in the long run .5. Charlie has a theory that the coin used by the umpire at the start of Saturday’s game against Syracuse was not fair. To test his theory, he tosses the coin 100 times. It comes up Heads 61 times. At 5% significance, is there enough evidence to support Charlies’s feeling that the coin is not fair? State clearly your null and alternative hypothesis. Solution: Let p be the proportion of times that this particular coin comes up Heads. H0 : p = .5 Ha : p 6= .5 Test statistic: p̂−.5√ .5∗.5/100 = 2.2 (p̂ = .61) p-value: P (z > 2.2 or z < −2.2) = .0278. There’s evidence to accept Charlie’s claim at 5% significance, but not at 1% significance. 3. Salmon grown at a commercial hatchery have weights that are normally distributed. When the “Norwegian method” is used, the adult salmon have mean weight 7.6 pounds. A hatchery claims that a modification to the method that they are using increases the average weight. Suppose a random sample of 16 fish grown using the new method yielded an average weight of 8.1 pounds with a sample standard deviation of 1.2 pounds. a): State the null and alternative hypotheses for this test. Solution: Let µ be average weight of salmon using new process. H0 : µ = 7.6 Ha : µ > 7.6 b): Write down the test statistic. Solution: Test statistic: 8.1−7.6 1.2/ √ 16 = 1.666... c): What is the distribution of the test statistic? Solution: It’s a t distribution with 15 degrees of freedom d): What is the critical value for rejecting the null at 5% significance? Solution: From the t table, t.05 = 1.753 (it’s a one-sided test) e): Is there strong enough evidence to accept the hatchery’s claim at 5% level of significance? Solution: No, because the test statistic is not as large as the p-value. 2
