91Ó°ÊÓ

We have been given that,

For men:

Number of men =$840$

Mean score of men =$272.40$

Standard Deviation of men =$59.2$

For women:

Number of women =$1077$

Mean score of women =$274.73$

Standard Deviation of women =$57.5$

$\mathrm{σ}=0.10$

X₁=Observed mean in first sample

X₂=Observed mean in second sample

${\mathrm{μ}}_1$=mean of first population

${\mathrm{μ}}_2$=proportion in second population

Observed value of difference in proportions=X₁-X₂

Expected value of difference in proportions=${\mathrm{μ}}_{1-}{\mathrm{μ}}_2=0$

Standard Deviation= localid="1654275753578" $\sqrt{\frac{{{\mathrm{σ}}_1}^2}{n_1}+\frac{{{\mathrm{σ}}_2}^2}{n_2}}$

$\left|z\right|=\frac{Observedvalue-Expectedvalue}{S\tan dardDeviation}$

localid="1654275760409" $\left|z\right|=\frac{X_1-X_2-0}{\sqrt{{\displaystyle \frac{{{\mathrm{σ}}_1}^2}{n_1}+\frac{{{\mathrm{σ}}_2}^2}{n_2}}}}$

$\left|z\right|=\frac{272.40-274.73-0}{2.68}$

$\left|z\right|=-0.86$

So, the calculated value =$-0.86$

The tabulated value at $90$% confidence level =$Â\pm 1.645$

We have to find ,

whether there is a difference in mean score for male and female young adults.

Null hypotheses:There is no difference in mean score for male and female young adults.

Alternative hypotheses:There is difference in mean score for male and female young adults.

Calculated value of z =$-0.86$

Tabulated value of z =$Â\pm 1.645$

Since,$Z_{cal}<Z_{tab}$

Null hypotheses is not rejected.

Conclusion:There is no difference in mean score for male and female young adults.