Question

1. A survey of 196 randomly selected college students at a local university found that only 78% of them checked their college email on a regular basis. The standard deviation is 7.4. Construct the confidence interval for the population proportion with a 90% confidence level.

Answer 1

Solution :

Given that,

n = 196

Point estimate = sample proportion = = 78%=0.78

1 -   = 1-0.78=0.22

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 *((0.78*0.22) /196 )

E = 0.05

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.78 -0.05 < p < 0.78+0.05

0.73< p <0.83