Question

2. In a survey of 2000 students, 1638 of them thought that the prices in the school cafeteria were too high.

2a. Find a point estimate for the proportion of the school that thinks the prices in the school cafeteria are too high.

2b. Which requirement might not be met here?

2c. Find a 90% confidence interval for the proportion of the school that thinks the prices in the school cafeteria are too high.

Answer 1

Solution :

Given that,

n = 2000

x = 1638

a

Point estimate = sample proportion = = x / n = 1638/2000=0.819

1 -   = 0.181

b

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05 = 1.645 ( Using z table )

Margin of error = E Z/2 *(( * (1 - )) / n)

= 1.645 *((0.819*0.181) / 2000)

E = 0.014

c

A 90% confidence interval proportion p is ,

- E < p < + E

0.819 - 0.014 < p < 0.819 +0.014

0.805< p < 0.833

(0.805 , 0.833)