Statistics - Student T Test
T-test is small sample test. It was developed by William Gosset in 1908. He published this test under the pen name of "Student". Therefore, it is known as Student's t-test. For applying t-test, the value of t-statistic is computed. For this, the following formula is used:
Formula
${t} = \frac{Deviation\ from\ the\ population\ parameter}{Standard\ Error\ of\ the\ sample\ statistic}$
Where −
Test of Hypothesis about population
Formula
${t} ={\bar X - \frac{\mu}{S}.\sqrt{n}} , \\[7pt]
\, where\ {S} = \sqrt{\frac{\sum{(X-\bar X)}^2}{n-1}}$
Example
Problem Statement:
An irregular sample of 9 qualities from an ordinary populace demonstrated a mean of 41.5 inches and the entirety of square of deviation from this mean equivalent to 72 inches. Show whether the supposition of mean of 44.5 inches in the populace is reasonable.(For ${v}={8},\ {t_.05}={2.776}$)
Solution:
${\bar x = 45.5}, {\mu = 44.5}, {n=9}, {\sum{(X-\bar X)}^2 = 72} $
Let us take the null hypothesis that the population mean is 44.5.
$ i.e. {H_0: \mu = 44.5}\ and\ {H_1: \mu \ne 44.5} , \\[7pt]
\ {S} = \sqrt{\frac{\sum{(X-\bar X)}^2}{n-1}}, \\[7pt]
\ = \sqrt{\frac{72}{9-1}} = \sqrt{\frac{72}{8}} = \sqrt{9} = {3}$
Applying t-test:
$ {|t|} = {\bar X - \frac{\mu}{S}.\sqrt{n}} , \\[7pt]
\ {|t|} = \frac{|41.5 - 44.5|}{3} \times \sqrt {9}, \\[7pt]
\ = {3}$
Degrees of freedom = $ {v = n-1 = 9-1 = 8 }$. For ${v = 8, t_{0.05}}$ for two tailed test = ${2.306}$. Since, the calculated value of $ {|t|}$ > the table value of $ {t}$, we reject the null hypothesis. We conclude that the population mean is not equal to 44.5.
