Question

A vendor claims that the variance for measurement of tiles that his factory produce were 13 square feet. A sample of nine tiles was measured in square feet and the results of the tiles produced were recorded as below:

204.5 206.3 202.4 207.8 203.1 206.2 203.8 206.6 205.8

(Assuming the sample comes from a normal population)

i) Calculate the point estimate of the population mean. (2 marks)

ii) Compute the variance and standard deviation. (3 marks)

iii) Determine the estimate of 95% confidence interval for the population mean of the tiles. (5 marks)

Answer 1

$i:\\\bar{x}=\frac{204.5+206.3+202.4+207.8+203.1+206.2+203.8+206.6+205.8}{9}=205.167\\ii:\\s^2=\frac{204.5^2+206.3^2+202.4^2+207.8^2+203.1^2+206.2^2+203.8^2+206.6^2+205.8^2-9\cdot 205.167^2}{9-1}=3.09362\\s=\sqrt{s^2}=\sqrt{3.09362}=1.75887\\iii:\\\left( \bar{x}-\frac{1}{\sqrt{n}}t_{\frac{1+\gamma}{2},n-1}s,\bar{x}+\frac{1}{\sqrt{n}}t_{\frac{1+\gamma}{2},n-1}s \right) =\\=\left( 205.167-\frac{1}{\sqrt{9}}\cdot 2.306\cdot 1.75887,205.167+\frac{1}{\sqrt{9}}\cdot 2.306\cdot 1.75887 \right) =\\=\left( 203.815,206.519 \right) \\$