Question

From an industrial area 70 companies were selected at random and 45 of them were panning for expansion next year. Find 95% confidence limit for the proportion of companies planning for expansion?

Answer 1

Answer:

Confidence limit = [52.8%, 75.2%]

Step-by-step explanation:

$P=\frac{45}{70}= 0.64$

$(1-P)=1-0.64=0.36$

$n= 70$

$P$ ± $z$ $\sqrt{\frac{P(1-P)}{n} }$

where the value $z$ will be taken from the z-table for 95% confidence interval

1-0.95= 0.05/2= 0.025

0.95+0.025= 0.0975

From the z-table the value of $z$ corresponding to 0.0975 is 1.96

$0.64$ ± $1.96$ $\sqrt{\frac{0.64*0.36}{70} }$

$0.64$ ± $1.96$ $(0.057)$

$0.64$ ± $0.112$

$64%$% ± $11.2$%

so the confidence interval is

$64+11.2=75.2$%

$64-11.2=52.8$%

$[52.8, 75.2]$