T-test for two sample assuming equal variances Calculator
Note: This calculator is for testing using sample values. If you only have sample mean and sd TWO SAMPLE TEST FROM SAMPLE MEAN AND SD calculator
You can see a sample solution below. Enter your data to get the solution for your question
$$ \displaylines{---} $$
$$ \displaylines{First\;we\;have\;to\;find\;mean\;and\;sample\;standard\;deviation\;of\;sample\;1
\\ \\
} $$
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$$ \displaylines{Mean = \frac{\sum_{i=1}^{n}X1_{i}}{n}
\\ \\ \,=\frac{2+4+6+7+8}{5}
\\ \\ \,=\frac{27}{5}
\\ \\Mean = 5.4
} $$
$$ x $$ | $$ \bar{x} $$ | $$ x-\bar{x} $$ | $$ [x-\bar{x}]^{2} $$ |
$$ 2 $$ | $$ 5.4 $$ | $$ -3.4 $$ | $$ 11.56 $$ |
$$ 4 $$ | $$ 5.4 $$ | $$ -1.4 $$ | $$ 1.96 $$ |
$$ 6 $$ | $$ 5.4 $$ | $$ 0.6 $$ | $$ 0.36 $$ |
$$ 7 $$ | $$ 5.4 $$ | $$ 1.6 $$ | $$ 2.56 $$ |
$$ 8 $$ | $$ 5.4 $$ | $$ 2.6 $$ | $$ 6.76 $$ |
$$ Total $$ | $$ $$ | $$ $$ | $$ 23.2 $$ |
$$ \displaylines{} $$
$$ \displaylines{Sample \;variance = (\sigma)^{2} = \frac{\sum_{i=0}^{n}(x_{i}-\bar{x})^{2}}{n-1}
\\ \\ \Rightarrow
\frac{23.2}{4}
\\ \\ \Rightarrow
5.8
\\ \\ \sigma = \sqrt{variance}
\\ \\ \Rightarrow
\mathbf{\color{Red}{2.408319}}
} $$
$$ \displaylines{} $$
$$ \displaylines{Now,\;we\;have\;to\;find\;mean\;and\;sample\;standard\;deviation\;of\;sample\;2
\\ \\
} $$
$$ \displaylines{} $$
$$ \displaylines{Mean = \frac{\sum_{i=1}^{n}X2_{i}}{n}
\\ \\ \,=\frac{3+5+6+7+5+9}{6}
\\ \\ \,=\frac{35}{6}
\\ \\Mean = 5.833333
} $$
$$ x $$ | $$ \bar{x} $$ | $$ x-\bar{x} $$ | $$ [x-\bar{x}]^{2} $$ |
$$ 3 $$ | $$ 5.833333 $$ | $$ -2.833333 $$ | $$ 8.027776 $$ |
$$ 5 $$ | $$ 5.833333 $$ | $$ -0.833333 $$ | $$ 0.694444 $$ |
$$ 6 $$ | $$ 5.833333 $$ | $$ 0.166667 $$ | $$ 0.027778 $$ |
$$ 7 $$ | $$ 5.833333 $$ | $$ 1.166667 $$ | $$ 1.361112 $$ |
$$ 5 $$ | $$ 5.833333 $$ | $$ -0.833333 $$ | $$ 0.694444 $$ |
$$ 9 $$ | $$ 5.833333 $$ | $$ 3.166667 $$ | $$ 10.02778 $$ |
$$ Total $$ | $$ $$ | $$ $$ | $$ 20.833334 $$ |
$$ \displaylines{} $$
$$ \displaylines{Sample \;variance = (\sigma)^{2} = \frac{\sum_{i=0}^{n}(x_{i}-\bar{x})^{2}}{n-1}
\\ \\ \Rightarrow
\frac{20.833334}{5}
\\ \\ \Rightarrow
4.166667
\\ \\ \sigma = \sqrt{variance}
\\ \\ \Rightarrow
\mathbf{\color{Red}{2.041241}}
} $$
$$ \displaylines{} $$
$$ \displaylines{Now\;we\;can\;start\;significant\;test
\\ \\
} $$
$$ \displaylines{} $$
$$ \displaylines{ \mathbf{\color{Red}{left\;tail\;test}}
\\ \\
x1 = 5.4
\\ \\
x2 = 5.833333
\\ \\
n1 = 5
\\ \\
n2 = 6
\\ \\
s1 = 2.408319
\\ \\
s2 = 2.041241
\\ \\
\alpha = 0.05
\\ \\
H_{0}: \mu_{1} < \mu_{2}
\\ \\
H_{a}: H_{0} is false
\\ \\
Assuming\;equal\;variances,\;So\;pooled\;standard\;deviation\;=\;s_{p}
\\ \\ \Rightarrow
\sqrt{\frac{(n_{1}-1)s_{1}^{2} + (n_{2}-1)s_{2}^{2}}{n_{1}+n_{2}-2}}
\\ \\ \Rightarrow
\sqrt{\frac{(5-1)2.408319^{2} + (6-1)2.041241^{2}}{5+6-2}}
\\ \\ \Rightarrow
\sqrt{\frac{44.033326}{9}}
\\ \\ \Rightarrow
\sqrt{4.892592}
\\ \\ \Rightarrow
2.21192
\\ \\
t=
\frac{x1-x2}{s_{p} \sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}}
\\ \\
\\ \\ \Rightarrow
\frac{5.4-5.833333}{2.21192 \sqrt{\frac{1}{5}+\frac{1}{6}}}
\\ \\
\\ \\ \Rightarrow
\frac{-0.433333}{1.339384}
\\ \\
\\ \\ \Rightarrow
-0.323532
\\ \\
df=n1+n2-2=
5 + 6 -2
\\ \\ \Rightarrow
9
\\ \\
Critical\;values\;are\; 2.262157 -2.262157
\\ \\
p= 0.376842
\\ \\
The\;result\;is\;not\;significant\;at\;p\;<\;0.05.\;So\;Failed\;to\;Reject\;H_{0}
} $$
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$$ \displaylines{} $$