Question

I want to estimate the population mean number of times that adults go to the movies a year. In this prompt, I spoke to 12 different people to find out how many times a year they go to the movies. In this survey, I found that these selected people visited the movies 3,5,2,4,4,2,1,3,4,7,0,1 times a year.

Calculate a 90% confidence interval for the population mean.

Answer 1

x
3
5
2
4
4
2
1
3
4
7
0
1
total=36
mean=x bar=sum of observation/number           of observation (n)
=36/12=3
mean=3
s=sqrt{(1/n)(x-x bar)^2}
(x-x bar)^2
0
4
1
1
1
1
4
0
1
16
9
4
total=42

s=sqrt{(1/n)(x-x bar)^2}
=sqrt{(1/12)(42)}
=sqrt(3.5)=1.8708286
90% confidance interval for mean is,

P{X^bar-t_{n-1($\alpha /2$)}(S/$sqrt$(n))<mu<X^bar +t_{n-1($\alpha /2$)}(S/sqrt(n))}=0.90

Upper limit=X^bar+t_11(0.05)(S/sqrt(n))

Lower limit=X^bar-t_11(0.05)(S/sqrt(n))

From t-table critical value is 2.20

3-2.20(1.8708286/sqrt(12))<mu<3+2.20(1.8708286/sqrt(12))

1.811863<mu<4.188136

Upper limit=4.188136

Lower limit=1.811863