Question

A survey was taken to determine the proportion of recent BCIT Marketing graduates who achieved a starting salary of at least $50,000 per year.

A random sample of 100 graduates resulted in a confidence interval of 12% ≤Π≤ 36%.

What is the confidence level used for this estimate of the population proportion? Use z, not t

Give the confidence level as a percent (without the % symbol), rounded to the nearest whole percent:

Answer 1

Solution:

n = 100

Given a confidence interval for p is

12% ≤ Π ≤ 36%

0.12  ≤ Π ≤ 0.36

Upper limit = 0.36

Lower limit = 0.12

$\hat p$ = (Upper + Lower)/2 = (0.36+0.12)/2 = 0.24

Margin of error = E = (Upper - Lower)/2 = (0.36-0.12)/2 = 0.12

But ,

E =
2012
*  
Vp(1-P)/n

$\therefore$ 0.12 =   *   $url=https://latex.codecogs.com/gif.latex?%5Csqrt%7B%7D&t=1596156601&ymreqid=f4764063-0d40-e2b0-1cde-f00024012f00&sig=pxcgzcyCx.UZUhNvsIAj0g--~D$[0.24 *(1 - 0.24)/100]

2012

$\therefore$ 0.12 =    * 0.042708313

2012

$\therefore$ =  2.81

2012

Use z table .Go to column of z . See 2.8 in this column and then see corresponding probability in the row at .01

It is 0.9975

So ,  1- $\alpha$ /2 = 0.9975

$\alpha$/2 = 0.0025

$\alpha$ = 0.005

Now ,

confidence level = 1 - $\alpha$ = 1 - 0.005 = 0.995 =99.5% = 99%

Answer : 99