the police department in your city was asked by the mayor s office to estimate the cost of

QUESTION

the police department in your city was asked by the mayors office to estimate the cost of crime. the police began their study with burglary records, taking a random sample of 500files since there were too many crime records to calculate statistics for all of the crimes committed. if the average dollar loss in a buglary, for this sample size 500 is $678 with a standard deviation of $560, construct the 95% confidence interval for the true average dollar loss in burglaries rounded to the nearest dollar amount. we can be 95% confident that the lower limit of our interval of dollar loss in crimes committed is?
z = (score mean) / s.d. Lets look at a distribution with mean = 100, s.d. = 20. A score of 120 would have a “z score” of: z = (120 100) / 20 = 20 / 20 = 1.00 If another distribution had a mean of 50 and s.d. of 10, a score of 60 would have a z score of: (60 50) /10 = = 10/10 = 1 In some sense, a score of “120” on one distribution is “equivalent” to a score of 60 on another. You can think of this as having one line 72 inches and another 2 yards. We can standardize the length in feet and find that each one is 6 feet long. Now, there is one number you must memorize. 1 z score is the 84th percentile, which is 34% above the mean which is 50%. This memorization will allow you to use any z table (normal curve table). You see, different tables report the percentages different ways. When using any table, go to the z score of “1” and look at the number. If it is “84”, the table reports the percentages from the left tail; if it is “34”, the table reports the percentage from the mean. Alternatively, you can look at the value 0 and see if it is 0 or 50. Now, we know the curve is symmetrical; and we know the mean is the 50th percentile; so lets play with it. If 1 s.d. is 34%, then, it must be the 84th %ile, because we have to add the 50% in for the mean. If 1 s.d. is 84, then there must be 16% ABOVE 1 s.d. Going the “other way”, -1 s.d. must be 34% from the mean, or 84% from 1

s.d to the “right edge”, so it must be the 16th %ile. You should “play” with some more values, flipping back and forth above the value, below the value, using the left side of the mean and the right side of the mean. Drawing pics of the situation helps. As you go through this course (and the next one) you will find that there are MANY “standard deviations”. Although they have different names and different formulas, they ARE JUST “standard deviations”. Hence, they will be called “standard error”, standard deviation of the mean, standard error of a score, standard error of sampling, standard error of prediction, etc. All are used exactly the same way in relation to the normal curve. NOW, lets look at YOUR problem Sample mean= 678 Standard Deviation= 560 n = 500 95% ci = z = 1.96 we go to our basic formula (score mean) / sd = z (score 678) / sd = 1.96 sd(pop) / sqrt(n) = 560 / sqrt(500) = 560 / 22.36 =25 back to the formula (score 678) / 25 = 1.96 (score 678) = 49 score = 678 + 49 Thats the upper limit. Now, you can repeat the computations using a z of negative 1.96 (-1.96) or simply note that the confidence interval is symmetrical, and SUBTRACT 49 from 678

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