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I’m sorry about all tests being fail to reject. I did double check these, so I don’t think I made a mistake, but it’s always possible. Let me know. And the drawings would have taken way too much time and not been helpful, so I didn’t do those. 1) H0: µ = 60 H1: µ ≠ 60 Assumptions: sample must be large enough (>30) to use CLT or the data is distributed normally. Distribution: Normal (z test) distribution. Two-tail. Degrees of freedom: NA Critical values: ±1.96 α = .05 which means there’s a 5% probability of rejecting the null hypothesis even though it is true. test statistic = -.981 Since the test statistic (-.981) is not more extreme than the critical value (-1.96), we fail to reject the null hypothesis Since the p-value (.326) is greater than the level of significance (α=.05), we fail to reject the null hypothesis In English: there is insufficient evidence to conclude that the mean reading score for this fifth grade class is different than the national average. 2) H0: σ = 15 H1: σ ≠ 15 Assumptions: data is distributed normally. Distribution: χ2. Two-tail. Degrees of freedom: n-1 = 11 Critical values: 3.82, 21.92 α = .05 which means there’s a 5% probability of rejecting the null hypothesis even though it is true. test statistic = (11 * 16.8372) / 152 = 13.86 Since the test statistic (13.86) is not more extreme than the critical values (3.82 or 21.92), we fail to reject the null hypothesis No p-value In English: there is insufficient evidence to conclude that the standard deviation of reading scores for this fifth grade class is different than the national average. 3) H0: µO = µN H1: µO < µN Assumptions: samples (both) must be large enough (>30) to use CLT or the data is distributed normally. Distribution: Note: we might do this as a Z-test by using the national standard deviation. But I’m going to assume that the standard deviations of the new program may not be known, so it’s going to be a t-test. t-distribution. One-tail. Since I’ve got the old first (on the left above and will be list 1), this is a left tail test. Degrees of freedom: (from calculator) 21 Critical values: -1.72 α = .05 which means there’s a 5% probability of rejecting the null hypothesis even though it is true. test statistic = -1.11 Since the test statistic (-1.11) is not more extreme than the critical value (-1.72), we fail to reject the null hypothesis Since the p-value (.139) is greater than the level of significance (α=.05), we fail to reject the null hypothesis In English: there is insufficient evidence to conclude that the mean reading score for the new teaching program are lower than the older program. 4) H0: σO = σN H1: σO < σN Assumptions: the data is distributed normally. Distribution: F-test. One-tail. Since I’ve got the old variant first (on the left, list #1 in the calculator), this will be the numerator we’re testing if it is bigger than the New which will be the denominator. So the F-stat should be >1 which makes this a right tail test. Degrees of freedom: 11 & 11. Critical values: 2.82 α = .05 which means there’s a 5% probability of rejecting the null hypothesis even though it is true. test statistic = 2.61 Since the test statistic (2.61) is not more extreme than the critical value (2.82), we fail to reject the null hypothesis Since the p-value (.064) is greater than the level of significance (α=.05), we fail to reject the null hypothesis In English: there is insufficient evidence to conclude that the mean reading score for the new variant of the reading program has a lower standard deviation than the old variant of the new program. 5) H0: µ1 = µ2 = µ3 H1: at least one mean (µ) is different than the others. Assumptions: data are distributed normally, independent, with the same standard deviation. Distribution: F-test. One-tail. Right tail. Degrees of freedom: 2 & 33 Critical values: 3.28 α = .05 which means there’s a 5% probability of rejecting the null hypothesis even though it is true. test statistic = 1.01 Since the test statistic (1.01) is not more extreme than the critical value (3.28), we fail to reject the null hypothesis Since the p-value (.373) is greater than the level of significance (α=.05), we fail to reject the null hypothesis In English: there is insufficient evidence to conclude that at least one mean reading score is different than the others.