Question

A particular brand of diet margarine has analysed to determine the level of polyunsaturated fatry acids (in percentages).A sample of six packages resulted in the following data 16.8,17.2,17.2,16.9,16.5,17.1 .Find the 95%confidence interval for the mean level of polyunsaturated fatty accid for this brand of diet margarine

Answer 1

mean $(\bar{x})=\dfrac{\sum x}{n}=\dfrac{1019}{6}=16.9833$

Standard deviation $(S)=\sqrt{\dfrac{1}{(n-1)}(\sum X^2-\dfrac{(\sum x)^2}{n})}$

           $= \sqrt{\dfrac{1}{5}(11731.11-\dfrac{(101.91)^2}{6})}=0.3189$

at $\alpha=0.05, z_{\frac{\alpha}{2}}=z_{0.05}=1.96$

So 95% confidence interval is $=\bar{x}\pm z_{0.025}\times \dfrac{S}{\sqrt{n}}$

               $=16.9833\pm(1.96\times \dfrac{0.3189}{\sqrt{6}} \\[9pt] =(16.9833\pm 0.255) \\ =(16.728, 17.2383)$