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Homework #3 1) According to SPSS: Mean s SE = = = 3.54 6.11 0.54 Technically, you should use t for this calculation because σ is unknown. For df, the closest you have in the table is 120, so go ahead and use that… 95% CI = Mean t/2 * (s / sqrt(n)) = 3.54 1.98 * (6.11 / sqrt(130)) = 3.54 1.98 * 0.54 = 3.54 1.06 95% CI = [2.48 – 4.60] But if you used z… 95% CI = Mean z/2 * (s / sqrt(n)) = 3.54 1.96 * (6.11 / sqrt(130)) = 3.54 1.96 * 0.54 = 3.54 1.05 95% CI = [ 2.49– 4.59] Note that the values for the two methods are so close because the critical values are so close because the degrees of freedom are so high. SPSS output below. The CI is a little different due to differences in rounding and because SPSS uses t rather than z and uses the t for 129 degrees of freedom rather than 120 because it can calculate the correct value for t which does not appear in our table. One-Sample Test Test Value = 0 Skip_Hrs t 6.611 df 129 Sig. (2tailed) .000 Mean Difference 3.5423 95% Confidence Interval of the Difference Lower 2.482 Upper 4.602 2) The data from HW 1 Q3): 4.5, 6, 6, 7, 7, 7, 8 Mean = 6.5 s = 1.12 SE = 1.12 / (sqrt(7)) Plug these values into the formula for the CI 90% CI = Mean t/2 * (s / sqrt(n)) = 6.5 1.943 * (.42) = 6.5 .72 90% CI = [5.68 – 7.32] = .42 = Mean t/2 * (s / sqrt(n)) = 6.5 2.447 * (.42) = 6.5 1.03 95% CI = [5.47 – 7.53] Notice that the 95% CI is wider than the 90% CI. 95% CI 3) Here you would have to use z rather than t even though we don’t technically know σ; not ideal. If I ask you to do this on an exam, I’ll be sure to provide you with σ. n = [z/2 * s / E]2 = [1.96 * 6.11 / 1]2 = [11.98]2 = 143.42 Round up to 144 4) Fill in the blanks: According to the central limit theorem, the sampling distrubtion of a population will be approximately normal if n is sufficiently large (n > 30). Also, the population parameter should be equal to the mean of the sampling distribution and the standard error will be equal to s / sqrt (n). 5) A variety of answers would suffice, including the following: To ensure an unbiased estimator (i.e., random sample). To decrease the variability of our estimator (i.e., increase its reliability). To enable us to use the Central Limit Theorem as a way of modeling chance variation in our sample. To increase the likelihood of obtaining a representative sample.