A random sample of size 16 has 53 as mean. The sum of the squares of the deviations taken from mean is 150. If the population mean is 56 then the value of t-test statistic is:

-3.79

The correct answer is Option (1) → -3.79 **

Given: $n=16,\;\bar{x}=53,\;\sum (x_i-\bar{x})^2=150,\;\mu_0=56$.

Sample variance: $s^2=\frac{\sum (x_i-\bar{x})^2}{n-1}=\frac{150}{15}=10$.

Sample standard deviation: $s=\sqrt{10}$.

Standard error: $\displaystyle \frac{s}{\sqrt{n}}=\frac{\sqrt{10}}{4}$.

Test statistic:

$t=\displaystyle\frac{\bar{x}-\mu_0}{s/\sqrt{n}}=\frac{53-56}{\sqrt{10}/4}=\frac{-3}{\sqrt{10}/4}=-\frac{12}{\sqrt{10}}=-\frac{6\sqrt{10}}{5}\approx -3.7948$.

Answer: $t=-\frac{6\sqrt{10}}{5}\approx -3.7948$