Question #320140

Solving Problems Involving Sample Size Determination


Apply the knowledge you obtained from the discussion by solving the given problems below.


1. A 95% confidence interval has to be created to estimate the mean volume of contents in the

bottles of Honey-Z. Assume that the population standard deviation is 30 ml. Calculate the sample

size needed if the margin of error is no more than:


a. 6 ml


b. 5 ml


c. 4ml


d. 3 ml


2. How large sample size is required to determine the mean IQ of seventh graders of Luis Palma

National High School within +5 points, population standard deviation of 25, and with the following

confidence level?

a. 98%


b. 95%.


c. 90%.


d. 80%.


1
Expert's answer
2022-04-01T02:06:31-0400

1:σnz1+γ2En(σz1+γ2)2E2=(301.960)2E2=3457.44E2a:n3457.4462=96.04n=97b:n3457.4452=138.298n=139c:n3457.4442=216.09n=217d:n3457.4432=384.16n=3852:σnz1+γ2Enσ2E2z1+γ22=25252z1+γ22=25z1+γ22a:n252.32632=135.3n=136b:n251.9602=96.04n=97c:n251.6452=67.64n=68d:n251.2822=41.06n=421:\\\frac{\sigma}{\sqrt{n}}z_{\frac{1+\gamma}{2}}\leqslant E\Rightarrow n\geqslant \frac{\left( \sigma z_{\frac{1+\gamma}{2}} \right) ^2}{E^2}=\frac{\left( 30\cdot 1.960 \right) ^2}{E^2}=\frac{3457.44}{E^2}\\a:n\geqslant \frac{3457.44}{6^2}=96.04\Rightarrow n=97\\b:n\geqslant \frac{3457.44}{5^2}=138.298\Rightarrow n=139\\c:n\geqslant \frac{3457.44}{4^2}=216.09\Rightarrow n=217\\d:n\geqslant \frac{3457.44}{3^2}=384.16\Rightarrow n=385\\2:\\\frac{\sigma}{\sqrt{n}}z_{\frac{1+\gamma}{2}}\leqslant E\Rightarrow n\geqslant \frac{\sigma ^2}{E^2}{z_{\frac{1+\gamma}{2}}}^2=\frac{25^2}{5^2}{z_{\frac{1+\gamma}{2}}}^2=25{z_{\frac{1+\gamma}{2}}}^2\\a:n\geqslant 25\cdot 2.3263^2=135.3\Rightarrow n=136\\b:n\geqslant 25\cdot 1.960^2=96.04\Rightarrow n=97\\c:n\geqslant 25\cdot 1.645^2=67.64\Rightarrow n=68\\d:n\geqslant 25\cdot 1.282^2=41.06\Rightarrow n=42


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