Question

Help BAN122_HW2_Butler_Darren Tell me what you want to do Developer Excel File Edit View Insert Format Tools Data Window AutoSave OFF HEVODO Home Insert Draw Page Layout Formulas Data Review View Calibri (Body) 11 AA = = = ste BIV A E G24 x fx A B C F G General a Wrap Text Merge & Center $ % , 8- Conditional Formatting as Table S LM a nown Question 1: Many medical professionals believe that eating too much red meat increases the risk of heart disease and cancer. Suppose you would like to conduct a survey to determine the annual consumption of beef by a typical American and want to use four pounds as the desired margin of error for a confidence interval of the population mean. Assume the population standard deviation is 20 pounds. Determine the sample size when: a) A 90% confidence interval is desired for the mean amount of beef consumed. b) A 99% confidence interval is desired for the mean amount of beef consumed. 8 Suppose you want to develop a confidence interval for the proportion of Americans who over-consume red meat with a desire margin of error of 0.1. Determine the sample size needed when: c) A 90% confidence interval is desired for the proportion of Americans who over-consume red meat. d) A 95% confidence interval is desired for the proportion of Americans who over-consume red meat.

I just need to confirm that I must first solve for n(sample size); then whether or not I use normal distribution (that is I'm assuming it's greater than 30); and the accompanying excel function as norm.s.inv??? Also any other additional assistance you deem pertinent...

Answer 1

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22 Margin of error is given as 4 and ois 20 Since o is known, Cl is based on Z-test (Za/ 2 0 /sqrt(n) = 4 For 90% CI, a=0.10 and a/2=0.05 2005 = 1.645 For 99% CI, a=0.01 and a/2=0.005 Zoos =2.58 (1.645*20)/sqrt(n)=4 (1.645*20)/4=sqrt(n) 8.225=sqrt(n) n=67.65-68 (2.58*20)/sqrt(n)=4 (2.58*20)/4=sqrt(n) 12.9=sqrt(n) n=166.41 167