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### Assi1(Soln)

Course: STAT 2225, Fall 2009
School: Sveriges...
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2225 Stat Both Sections Spring 2006 Stat 2225 Both Sections Assignment #1 (Answer Key) [Total 40 marks] 1. [8 marks] (a) Quite obvious that the sample size is 245; therefore it is a large sample case and by Central Limit Theorem; the z-score can be used to develop the required confidence intervals. [2] (b) For 90% confidence level, we have 12,000 1.645 2200 , i.e. 11,768.79 to 12,231.21. Therefore, 245 we...

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2225 Stat Both Sections Spring 2006 Stat 2225 Both Sections Assignment #1 (Answer Key) [Total 40 marks] 1. [8 marks] (a) Quite obvious that the sample size is 245; therefore it is a large sample case and by Central Limit Theorem; the z-score can be used to develop the required confidence intervals. [2] (b) For 90% confidence level, we have 12,000 1.645 2200 , i.e. 11,768.79 to 12,231.21. Therefore, 245 we have 90% confidence that the population average debt (or the actual average debt of all colleges and universities students in USA) upon graduation is in between \$11,768.79 and \$12,231.21. [2] (c) For 95% confidence level, we have 12,000 1.96 2200 , i.e. 11,724.51 to 12,275.48. 245 [1] (d) The interval width will increase as can been seen from the part (b) and (c). (Note: The higher the confidence level, the bigger the z-score!). [2] (e) We should draw more samples. (Because we cant control the either one of the standard deviations.) [1] 2. [6 marks] (a) Because the question tells you to treat it as large-sample case; therefore the formula to be used will be the similar to the one used in #1 with the only difference is instead of the population standard deviation (), use sample standard deviation (s). The result is: 22.4 1.96 5 = 21.15 to 23.65 61 Therefore, we have 95% confidence that the population average number of weekly contacts for all the sale personnel is in between 21.15 and 23.65. (Do you need rounding?) [2] (b) According to the given information, the formula to be used will be, x t / 2 s n . For 95% confidence level and with 60 degrees of freedom (df), the t-score = 2.000. Hence the required confidence interval is 22.4 2.000 5 = 21.12 to 23.68 61 [2] (c) Because the sample size is more than 30, therefore by Central Limit Theorem, the normal distribution can be used instead of t-distribution, and will provide a good approximation. However, keep in mind that if the conditions of using t-distribution are fulfilled, t can be used because it will provide a better estimation. [2] 3. [5 marks] (a) With the given, the required sample size is n = (1.96) 2 (0.5)(1 0.5) = 600.25. Use n = 601. [1] (0.04) 2 P. 1 of 3 Stat 2225 Both Sections Spring 2006 (b) The point estimate of the proportion of Broadway audiences that do not live in New York City is, p= 445 = 0.7404 601 [1] (c) With a sample size n = 601, np = 444.98 >5 and n(1 p ) = 156.02 >5, we can consider it is a large sample case; therefore we can use the z to find the required confidence interval, as 0.7404 1.96 (0.7404)(1 0.7404) , i.e. 0.7054 to 0.7755 601 Therefore, we have 95% confidence that the population proportion of Broadway audiences that do not live in NY City is in between 0.7054 and 0.7755. 4. [3] [6 marks] (a) Because the sample size is 100, therefore it is considered as a large sample case. By the Central Limit Theorem, we can use the normal distribution, i.e. the z-score as the test statistic. [2] (b) With a 0.05 level of significance and it is a one-tailed test, the decision point is z0.05 = 1.645, and the rejection rule is Reject H0 if z < -1.645. [2] (c) As mentioned in part (a), this is a large sample case. In particular, the population standard deviation is unknown, we then use the sample standard deviation. We have z= x 0 9300 10,192 = = 1.98 4500 10 s n [1] (d) We reject the H0 (Oh, forget to ask you to state the hypotheses!) because from part (b), we know z < -1.645. That means, we have enough evidence to conclude the population average sales price of used cars is less than \$10,192. [1] 5. [8 marks] (a) The sample mean is x = 58 + 52 + ..... + 55 = 59.3750 8 [1] (b) The sample SD is s = 123.87 = 4.2066 8 1 [1] (c) The question isnt clear about whether the variable Working hours is normally distributed or not. We now first assume it is. Because the sample size is only eight, therefore it is considered as a small sample case and we dont know the population SD, we should use the t-distribution, i.e. the t-score as the test statistic. [2] (d) As mentioned in part (c), the test statistic used should be the t, and the value is t= x 0 59.3750 55 = = 2.9417 s n 4.2066 8 [1] P. 2 of 3 Stat 2225 Both Sections Spring 2006 (e) From the table in the text, we can find, with 7 degrees of freedom, t0.025 = 2.365 (look at the right tail) and t0.01 = 2.998 . Because 2.9417 is in betwee...

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