121k views
2 votes
Hi, can you help me answer this question, please, thank you!

1 Answer

3 votes

It is given that standard deviation = 2

Margin error =0.5 mm

At 99% confidence interval the Z is;


\begin{gathered} \text{ }\alpha\text{ = 1-99\%} \\ \alpha\text{ = 1 - 0.99} \\ \alpha=0.01 \\ (\alpha)/(2)=0.005 \\ Z_{(\alpha)/(2)}=2.58 \end{gathered}

Sample size express as;


\begin{gathered} n=(Z_{(\alpha)/(2)}*(\sigma)/(E))^2 \\ n=(2.58*(2)/(0.5))^2 \\ n=106.50 \\ n=107 \end{gathered}

Answer : A) 107

...

User Junichiro
by
8.1k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories