Question

Fifteen out of the 145 randomly selected study-abroad students who were surveyed in a state can speak the French language. Construct and interpret a 98% confidence interval for the proportion of all study-abroad students in the state who can speak French.

Answer 1

Solution :

Given that,

n = 145

x = 15

Point estimate = sample proportion = = x / n = 0.103

1 - = 0.897

At 98% confidence level the z is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02 / 2 = 0.01

Z/2 = Z0.01 = 2.326

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

= 2.326 * (((0.103 * 0.897) / 145)

= 0.059

A 98% confidence interval for population proportion p is ,

- E < p < + E

0.103 - 0.059 < p < 0.103 + 0.059

0.044 < p < 0.162

A 98% confident that the proportion of all study-abroad students in the state who can speak French is between 0.044 and 0.162.