Question

Ming works as a quality assurance analyst at a bottling factory. She wants to use a one-sample z interval to estimate what proportion of 500 ml bottles are underfilled. She wants the margin of error to be no more than 4% at 90% confidence. What is the smallest sample size required to obtain the desired margin of error?
a) 271
b) 423
c) 651
d) 888

Answer 1

Answer:

Sample size is $n=423$

Step-by-step explanation:

Given that,

Margin of error $=4$%

Confidence level $=90$%

Suppose, sample proportion$=0.5$

        i.e. $\hat{P}=0.5$

We know that,

     Margin of error $=2^*\sqrt{\frac{\hat{p}(1-\hat{p})}{n} }$

 ∴ $1.64\sqrt{\frac{0.5(0.5)}{n} } \leq 4\%$

$\Rightarrow 1.64\sqrt{\frac{0.25}{n} } \leq 0.04$

$\Rightarrow \frac{0.5}{\sqrt{n} } \leq \frac{0.04}{1.64}$

$\Rightarrow \frac{0.5}{\sqrt{n} } \leq 0.0243$

$\Rightarrow \sqrt{n}\geq \frac{0.5}{0.0243}$

$\Rightarrow \sqrt{n}\geq 20.57$

squaring on both side,

∴ $n=423.1249$

Hence, the sample size is,

   $n=423$

Hence, the correct option is $(b).$