Spring09Midterm1ASols

# Spring09Midterm1ASols - Economics 203 Midterm 1 Solutions...

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Economics 203 Midterm 1 Solutions — Spring 2009 — Form A 1. D. Failing to reject a null hypothesis when the alternative hypothesis is true is known as a type II error. 2. B. 50 and 60 days correspond to 1 and 2 standard deviations above the mean respectively. 95% of the accounts fall between -2 and 2 standard deviations, while 68% fall between -1 and 1 standard deviations. So 95% - 68% = 27% fall in the difference between the two ranges, -2 to -1 standard deviations and 1 to 2 standard deviations. Symmetry tells us that half of this falls in each range, so 27/2 = 13.5% falls between 50 and 60 days. 3. E. The most commonly observed value is known as the mode, not the median. All of the rest of the statements are correct. 4. D. The null hypothesis is that the variances are equal, hence that their ratio is 1; the alternative is that they are unequal, or that their ratio is different from 1. 5. E. The test statistic is found by taking the sample variance of population 1 and dividing by the sample variance of population 2: 8 . 6 12 . 7 = 0 . 6772. 6. C. A 10% level of significance implies 5% in each tail, so we look at the F 0 . 05 , 59 , 49 = 1 . 583 and F 0 . 95 , 59 , 49 = 0 . 639 critical values. Since the observed test statistic is within these bounds, we do not reject the null hypotheis; that is, there is insufficient evidence to show that the variance in the two cities is different. 7. A. The F-distribution is appropriate for testing the equality of two unknown variances, not the χ 2 distribution. 8. A. Your standard normally distributed test statistic is negative; and you have a “less than” alternative, so your p-value is less than 0.5; so the area to be shaded on the graph is smaller than 50%. It should be indicated on the left hand side of the distribution because the test statistic is negative. 9. C. Null and alternative hypotheses always deal in population parameters, so the ones with ¯ x are eliminated. The claim is that the mean for population 2 is greater than that for population 1, hence the alternative is H 1 : μ 1 - μ 2 < 0. The null is that they are equal.

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