Spring 2010 Midterm 1 Form A Solutions

# Spring 2010 Midterm 1 Form A Solutions - Economics 203...

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Economics 203 Midterm 1 Solutions — Spring 2010 — Form A 1. A. Mean, median, and mode are the three common measures of central tendency, or “where the middle of the data lies”. 2. B. The empirical rule says that approximately 68% of observations fall within one standard deviation of the mean, and approximately 95% within two standard deviations, and virtually all of the observations will be within three standard deviations of the mean. Using symmetry we infer that 34% of the stores will be between -1 and 0 standard deviations, or between 19 and 23 employees. Similarly, approximately 47.5% of the stores will be between 0 and +2 standard deviations, or between 23 and 31 employees. Adding these gives 81.5%. 3. E. H 0 : σ 2 7 . 4 vs. H 1 : σ 2 < 7 . 4 would also be correct. 4. C. The test statistic is χ 2 = ( n - 1) s 2 σ 2 = (50 - 1)(5 . 1) 7 . 4 = 33 . 770 5. B. The test statistic is z = (71 - 64 . 5) - 5 4 2 / 14+3 2 / 17 = 1 . 1599 6. B. α is the probability of rejecting the null hypothesis when it is true. A higher α means we reject the null hypothesis more often, but we make the mistake of rejecting it when it is true more often also. 7. A. The one-tailed test referred to in the output shows a signiﬁcant diﬀerence; this refers to 1. The two-tailed test is not signiﬁcant, as the p-value given is greater than 5%. So we do not reject the null hypothesis in 2. ¯ x 1 > ¯ x 2 , so we would not reject the null in 3 at any sensible level of α . 8. D. First compute the grand mean: ¯ ¯ x = n 1 ¯ x 1 + n 2 ¯ x 2 + n 3 ¯ x 3 + n 4 ¯ x 4 n 1 + n 2 + n 3 + n 4 = 30(3 . 52)+16(3 . 4)+20(2 . 8)+24(3) 30+16+20+24 = 288 90 = 3 . 2. Then the sum of squares for treatments is SST = n 1 x 1 - ¯ ¯ x ) 2 + n 2 x 2 - ¯ ¯ x ) 2 + n 3 x 3 - ¯ ¯ x ) 2 + n 4 x 4 - ¯ ¯ x ) 2 = 30(3 . 52 - 3 . 2) 2 + 16(3 . 4 - 3 . 2) 2 + 20(2 . 8 - 3 . 2) 2 + 24(3 - 3 . 2) 2 = 7.872 9. C. The ANOVA test is always a one-sided greater than test, so we use α , not α/ 2. The degrees of freedom are k - 1 and n - k . Hence the correct choice is F 0 . 05 , 3 , 86 10. B. Because this is an unknown but equal variance t-test, we ﬁrst calculate the pooled variance: s 2 p = (20 - 1) s 2 1 +(10 - 1) s 2 2 20+10 - 2 = 1660. Then the test statistic is t = 123 - 141 1660(1 / 20+1 / 10) = - 1 . 1407 11. C. For the pooled variance t-test, the degrees of freedom are simply n 1 + n 2 - 2, or 28. 12. D. This is a one-sided test, with a rejection region to the right. The test statistic is negative, however, so the

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Spring 2010 Midterm 1 Form A Solutions - Economics 203...

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