The mean percentage of a population of people eating out at least once a week is 57% with a standard deviation of 3.5%. Assume that a sample size of 40 people was surveyed from the population an infinite number of times. 95% of the sample mean occurs between % and %.
i am also doing this. you would have to remember to move the decimal enable to get normal numbers for the standard of deviation and mean. so [tex]= .035/ \sqrt{40} =.035/6.324= .005534472 *2= .011[/tex] then subtract and add that from your mean the standard of error formula is = deviation/ the square root of the sample size then the confidence interval of 95% would have you multiply it by 2 or 2.02 for a more accurate result then you add and subtract that from the mean which would give you your answer 55.89 to 58.11
