Question

A random sample of size 16 has 53 as mean. The sum of the squares of the deviation

taken from mean is 150. Can this sample be regarded as taken from the population having

56 as mean? Obtained 95% and 99% level of confidence limit of the mean of population

Answer 1

Solution;

$N=16$

$\bar{X}=53$

$\mu=56$

$S=\sqrt{\frac{\sum(X-\bar{X})}{N-1}}$

$S=\sqrt{\frac{150}{15}}=\sqrt{10}=3.162$

t computed is;

$t=\frac{\bar{X}-\mu}{S}\sqrt{N}$

$t=\frac{|53-56|}{3.162}\sqrt{16}=3.795$

t critical is;

df=16−1=15

$\alpha=0.05$

$t_{critical}=2.131$

Hence;

Since $t_{computed}>t_{critical}$ ,the result of the experiment does not support the hypothesis that the sample is taken from the universe having a mean 56.

95% confidence limit ;

$\bar{X}\displaystyle _-^+\frac{S}{\sqrt{N}}t_{0.05}=53\displaystyle_-^+\frac{3.162}{\sqrt{16}}×2.131$

$53\displaystyle_-^+1.6846$

$51.316<\bar{X}<54.684$

99% confidence limit;

$\bar{X}\displaystyle _-^+\frac{S}{\sqrt{N}}t_{0.01}=53\displaystyle_-^+\frac{3.162}{\sqrt{16}}×2.947$

$53\displaystyle_-^+2.330$

$50.67<\bar{X}<55.33$