Determine a 99 confidence interval for the mean

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$116,500 and the sample standard deviation of $53,870. Determine a 99% confidence interval for the mean amount. e) If the interval in d) should have width less than $1000 and assuming the sample standard deviation will be $53,870, how many subscribers should have responded?
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2 3. Suppose we wish to numerically integrate A = 2 1 sin( ) x dx π . Consider a very large frame { } 1 2 , ,... U u u = that is the finite set of possible values returned by a uniform (1, π ) random number generator and let ( ) i i y f u = . Then the population average i i U y N multiplied by some factor (that you need to determine) is very close to the integral. Any difference is frame (study) error. If we select a random sample of size 1000, then the sample average of 1 2 1000 , ,..., y y y is an estimate of the population average, and thus proportional to the integral. You can adapt the R commands: u<-runif(1000) and y<-sin(u^2) to generate the sample values. Find an estimate for A and construct a 95% confidence interval. 4. Suppose in a survey we sample using simple random sampling n items from each of two (roughly) equal sized subpopulations (e.g. males and females). By introduce suitable notation for the sample summaries, derive a symbolic estimate and 95% confidence interval for the difference in population means for the two subpopulations. You can use results given in the course notes.
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