Question

Thanks in advance.

In Excel or StatCrunch, in a survey of 800 adults, 352 reported that they took a vacation in the last 12 months. Calculate a 95% confidence interval for the population proportion. Assume a random sample.

Answer 1

Sol:

sample proportion=p^=x/n=352/800= 0.44

z crit for 95%=1.96

95% confidence interval for the population proportion

p^-z*sqrt(p^(1-p^)/n,p^+z*sqrt(p^(1-p^)/n

0.44-1.96*sqrt(0.44*(1-0.44)/800),0.44+1.96*sqrt(0.44*(1-0.44)/800)

0.4056,0.4744

lower limit=0.4056

upper limit=0.4744

we are 95% confident that the population proportion lies in between 0.4056 and 0.4744