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:

Sample size is [tex]n=423[/tex]

Step-by-step explanation:

Given that,

Margin of error [tex]=4[/tex]%

Confidence level [tex]=90[/tex]%

Suppose, sample proportion[tex]=0.5[/tex]

        i.e. [tex]\hat{P}=0.5[/tex]

We know that,

     Margin of error [tex]=2^*\sqrt{\frac{\hat{p}(1-\hat{p})}{n} }[/tex]

 ∴ [tex]1.64\sqrt{\frac{0.5(0.5)}{n} } \leq 4\%[/tex]

[tex]\Rightarrow 1.64\sqrt{\frac{0.25}{n} } \leq 0.04[/tex]

[tex]\Rightarrow \frac{0.5}{\sqrt{n} } \leq \frac{0.04}{1.64}[/tex]

[tex]\Rightarrow \frac{0.5}{\sqrt{n} } \leq 0.0243[/tex]

[tex]\Rightarrow \sqrt{n}\geq \frac{0.5}{0.0243}[/tex]

[tex]\Rightarrow \sqrt{n}\geq 20.57[/tex]

squaring on both side,

∴ [tex]n=423.1249[/tex]

Hence, the sample size is,

   [tex]n=423[/tex]

Hence, the correct option is [tex](b).[/tex]

Answer:

423

Step-by-step explanation: