A random sample of 50 people at McDonalds revealed a mean am

A random sample of 50 people at McDonalds revealed a mean amount spent of $6.80 and a standard deviation of amount spent $0.90. Find the 92% confidence interval for p. How large a sample size is needed if you wanted to reduce the above bound by 20% and increase your confidence coefficient to 005?

Solution

mean = 6.8 and s.d = 0.9.............n = 50..

92% C.I = [ 6.8 - ( 1.787758 * 0.9 ) / sqrt ( 50) , 6.8 + ( 1.787758 * 0.9 / sqrt ( 50 ) ]
= [ 6.572456 , 7.027544 ]
margin of error = 0.2275444..bound = 2*0.2275444

we want to reduce the bound by 20%..now bound = 2* 0.2275444 * (.8) = 0.364071

Now, margin of error = 0.364071 / 2 = 0.1820355..

t-score for 95% confidence and 49 d.f = 2.009575...

margin of error = t-score * s.d / sqrt(n).....
i.e, 0.1820355 = (2.009575 * 0.9) / sqrt(n).....

i.e, n = 98.71457..i.e, n = 99...
a sample size of 99 persons is requirred now!