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Unformatted text preview: Economics 203 Midterm 1 Solutions — Spring 2010 — Form B 1. A. Mean, median, and mode are the three common measures of central tendency, or “where the middle of the data lies”. 2. B. 30 being the median for finance papers means that 50% of finance papers are at least 30 pages long. So, you have data on the proportion of long papers for economics papers and for finance papers. 3. C. The test statistic is (259 . 4 240) / (90 / √ 60) = 1 . 6697. 4. E. We reject the null hypothesis that μ = 240 for large values of the test statistic. Since the population standard deviation is known, the appropriate critical value is z . 05 from the standard normal distribution. We have 1 . 670 > z . 05 and therefore the restaurant should stay open late. 5. B. Notice that w = 10, σ = 90, and z . 025 = 1 . 96. With these results, we obtain n ≈ z 2 . 025 σ 2 /w 2 ≈ 311 . 16. To make sure that we can be at least 95% confident, we round up find that a sample of 312 days is needed. 6. A. Oneway ANOVA is always concerned with population means and being able to identify through the alternative hypothesis if at least one of the population means is different from the others. 7. D. The mean square for error is MSE = SSE/(nk). The SSE corresponds to the ’Within groups’ row of the ANOVA table. Thus 4566867  1687905 = 2878962 is the SSE. The sample size was 67 and you have 3 factor levels. So MSE = 2878962/64 = 44983.8. 8. C. The MSE was found in the previous question. So we only need to find the MST = SST/(k1) = 1687905/2 = 843952.5 and F = MST/MSE = 843952.5/44983.8 = 18.761 9. B. When we perform ANOVA, it is always a onetailed test and we have k1, nk degrees of freedom so the appropriate critical value is 3.140 and since 3.8 is greater than 3.140 we reject the null hypothesis that the means for all three groups are the same. However, this does not imply that big cities have a higher starting salary compared to small and medium cities, it only tells us that at least one of the three means is different. 10. E. The relevant point estimate is found as ˆ p 1 ˆ p 2 = 30 270 20 180 = . 111 . 111 = 0. Since your point estimate is 0, your test statistic will be as well (the numerator is 0). And then go from the test statistic of zero, to the extreme of the nearest rejection region (the right end of the distribution). Since the standard normal distribution is symmetric you know that this area will equal 1/2....
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This note was uploaded on 02/14/2011 for the course ECON 203 taught by Professor Petry during the Spring '09 term at University of Illinois, Urbana Champaign.
 Spring '09
 PETRY
 Economics

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