Question

Marketing companies are interested in knowing the population percent of women who make the majority of household purchasing decisions. When designing a study to determine this population proportion, what is the minimum number you would need to survey to be 90% confident that the population proportion is estimated to within 0.05? (Round your answer up to the nearest whole number.)

Answer 1

Answer: 271

Step-by-step explanation:

As per given description in the question, we have

Critical value for 90% confidence interval = $z_{\alpha/2}=1.645$

[using z-value calculator ]

Margin of error : E= 0.05

Since prior estimate of population proportion is unknown so we take p= 0.5

Formula we use to find the sample size :

$n=p(1-p)(\dfrac{z_{\alpha/2}}{E})^2$  [we assume p= 0.50]

i.e. $n=0.5(1-0.5)(\dfrac{1.645}{0.05})^2$

Simplify :

$\Rightarrow\ n=270.6025\approx271$  

Therefore , the minimum sample size required = 271