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

Solution:

$\begin{aligned} \bar{X} &=53, \mu=56, N=16 . \\ S &=\sqrt{\frac{\Sigma(X-X)^{2}}{N-1}}=\sqrt{\frac{150}{15}}=\sqrt{10}=3.162 . \\ t &=\frac{\bar{X}-\mu}{S} \sqrt{N} \\ &=\frac{53-56}{3. 162} \sqrt{16}=\frac{3 \times 4}{3 .162}=\frac{12}{3.162}=3 . 8 . \end{aligned}$

The value of t for 16-1=15 degrees of freedom at 5% level of significance is 2.13 . $3. 8>2 . 13$ , hence the result of experiment does not support the hypothesis that the sample is taken from the universe having mean of 56 .

 95% confidence limits for the population mean

$\begin{aligned} \bar{X} \pm \frac{S}{\sqrt{N}} t .05 &=53 \pm \frac{3.162}{\sqrt{16}} \times 2.131 \\ &=53 \pm \frac{3.162 \times 2.131}{4} \\ &=53 \pm 1.685 \\ &=54.685 \text { to } 51.315 \end{aligned}$

 99% confidence limits of the population mean :

$\begin{aligned} \bar{X} \pm \frac{S}{\sqrt{N}} t .01 &=53 \pm \frac{3.162}{\sqrt{16}} \times 2.947 \\ &=53 \pm \frac{3.162 \times 2.947}{4}=53 \pm 2.329 \\ &=50.671 \text { to } 55.329 \end{aligned}$