Question

Suppose a student club on campus is surveying students regarding the latest changes to on-campus food menus. The survey yields that out of the 196 students surveyed, 111 rated the changes as "very positive" or "mostly positive". Using this information, determine the upper bound of a 95% confidence interval for the proportion of positive opinions on menu changes for the entire student body.

Answer 1

Solution :

Given that,

Point estimate = sample proportion = = x / n = 111 / 196 = 0.566

1 - = 1 - 0.566 = 0.434

Z = Z0.05 = 1.645

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

= 1.645 (((0.566 * 0.434) / 196 )

=

A 95% upper confidence interval for population proportion p is ,

+ E   

= 0.566 + 0.058 = 0.624

= upper bound = 0.624