Question

The Student Recreation Center wanted to determine what sort of physical activity was preferred by students. In a survey of 89 random students, 68 indicated that they preferred outdoor exercise over exercising in a gym. The 95% confidence interval estimating the proportion of all students at the university who prefer outdoor exercise is given by which of the following?

Question 9 options:

	1) ( 0.69002 , 0.83807 )
	2) ( 0.14774 , 0.32417 )
	3) ( -0.67583 , 0.85226 )
	4) ( 0.67583 , 0.85226 )
	5) ( 0.71904 , 0.80905 )

Answer 1

Solution :  

Given that,

n = 89

x = 68

Point estimate = sample proportion = = x / n = 68 / 89 = 0.76404

1 - = 1 - 0.76404 = 0.23596

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

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

= 1.96 * (((0.76404 * 0.23596) / 89)

= 0.08821

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.76404 - 0.08821 < p < 0.76404 + 0.08821

0.67583 < p < 0.85226

The 95% confidence interval for the population proportion p is : 0.67583 , 0.85226

option 4) is correct