A random sample of size 16 from a normal population showed a mean of 103.75 cm and sum of squares of the deviations from the mean 843.75 cm^{2}. Can we say that the population has a mean of 108.75?

Solution:

$\bar{X}=103.75$

$\mu=108.75$

$(X_{i}-\bar{X})=843.75$ and $n=16$

Calculate standard deviation,

$s^{2}=\cfrac{\sum(X_{i}-\bar{X})^{2}}{n}=\cfrac{843.75}{16}=52.73$

Step 1: The null hypothesis $H_{0}=\mu=108.75$

Alternative hypothesis $H_{a}=\mu \ne 108.75$

Step 2: Calculation of the test statistic.

$t=\cfrac{\bar{X}-\mu}{s/\sqrt{n-1}}= \cfrac{103.75-108.75}{\sqrt{52.74}/\sqrt{16-1}} = \cfrac{-5}{1.875} = -2.67$

$\therefore |t| = |-2.67|=2.67$

Step 3: Level of significance = 5%

i.e., $\alpha=0.05$

Step 4: Critical Value $\Longrightarrow$ The value of $t_{\alpha}$ for 5% level of significance and degrees of freedom.

$V=16-1=15$ from the table is 2.131 (percentage point distribution table)

Step 5: Since the computed value of $|t|=2.67$ is greater than the table value $t_{\alpha}=2.131,$ the null hypothesis is rejected.

We cannot say that the population mean is 108.75 .