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Average mark scored by 32 boys is 72 with standard deviation of 8 while that for 36 girls is 70 with standard deviation of 6. Test at 1% LoS whether the boys perform better than the girls.

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written 7.9 years ago by |

i) $n_1=32$ and $n_2=36$ (>30 it is large sample)

$\overline x_1=72,\overline x_2=70 ;s_1=8 ; s_2=6$

Step 1: Null hypothesis $(H_0): µ_1=µ_0$ (i.e. performance of boys & girls is equal)

Alternate hypothesis $(H_a): µ_1\gtµ_2$ (i.e. boys performs better than girls )(one tailed test)

Step 2: LOS=1% (two tailed test)

$\therefore $ LOS=2% (one tailed test)

$\therefore $ Critical Value $(Z_{\alpha})=2.33$

Step 3: Since Samples are large,

$S.E.=\sqrt{\dfrac {s^2}{n_1}+\dfrac {s_2^2}{n_2}}=\sqrt{\dfrac {8^2}{32}+\dfrac {6^2}{36}}=1.732$

Step 4: Test statistic

$Z_{cal}= \dfrac {\overline x_1-\overline x_2}{S.E.}=\dfrac {72-70}{1.732}=1.1547$

Step 5: Decision

Since $|Z_{cal}|\ltZα, H_0$ is accepted.

$\therefore $ Boys do not perform better than girls.

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