Answer-
A 95% confidence interval for the true percent of movie goers is 36.41% to 44.25%
Solution-
Given,
n = 600 (sample size)
x = 252 (number of people who bought)
Confidence interval = 95%, so z = 1.96
We know that,
[tex]\mu = M\pm Z(SE)[/tex]
where,
M = sample mean
Z = Z statistic determined by confidence level
SE = standard error of mean
Calculating the values,
[tex]M=\frac{x}{n} =\frac{242}{600} =0.4033[/tex]
[tex]Z=1.96[/tex] from the tables
[tex]SE=\sqrt{\frac{M\times (1-M)}{n}}[/tex]
[tex]\Rightarrow SE=\sqrt{\frac{0.4033\times (1-0.4033)}{600}}=0.02[/tex]
Putting all the values in the formula,
[tex]\mu = M\pm Z(SE)[/tex]
[tex]\mu = 0.4033\pm 1.96(0.02)[/tex]
[tex]=0.4033\pm 0.0392[/tex]
[tex]=0.4425, 0.3641[/tex]
[tex]= 44.25\%,36.41\%[/tex]
