Question

Of the 201 employees selected from an ice cream factory, 30 of them admit to grabbing a spoon and sampling when nobody is looking. Construct a 95% for the proportion of employees who grab a spoon and eat when nobody is looking.

Answer is formatted like this,

”We are 95% confident that the proportion of all employees who grab a spoon and eat are between ___ and _____

Answer 1

Solution:

$\hat p=\frac{30}{201}=\frac{10}{67}$

$n=201$

$\hat q=1-\frac{10}{67}=\frac{57}{67}$

For 95% confidence interval, $z=1.96$

Now, 95% confidence interval$=(\hat p\pm z\sqrt{\frac{\hat p \hat q}{n}})$

$=(\frac{10}{67}\pm 1.96\sqrt{\frac{\frac{10}{67}. \frac{57}{67}}{201}}) \\=(0.0999,0.1985)$

Thus, We are 95% confident that the proportion of all employees who grab a spoon and eat are between 0.0999 and 0.1985.