ii) A random sample of 15 items gives the mean 6.2 and variance 10.24. Can it be regarded as drawn from a normal population with mean 5.4 at 5% LOS

i. Given $\bar{X_1} = 72$, $\bar{X_2} = 70$, $n_1$=32, $n_2$=36, $s_1$=8, $s_1$=8

I. Null hypothesis ($H_0$)= $\mu_1$= $\mu_2$

Alternative hypothesis ($H_{\alpha}$)= $\mu_1$ $\neq$ $\mu_2$

II. Calculation of statistics:

$ \bar{X_1}- \bar{X_2} = 72-70 = 2$

$ S.E = \sqrt{\frac {s_1^2}{n_1}+ \frac{s_2^2}{n_2}}$

$ = \sqrt{\frac {8^2}{32}+ \frac{8^2}{36}}$

$ = \sqrt{\frac {64}{32}+ \frac{64}{36}}$

$ = \sqrt{3}$

$ Z = \frac {\bar{X_1}- \bar{X_2}}{S_t}$

$ Z = \frac {2}{\sqrt{3}} = 1.15$

$ \bar{Z} = 1.15$

III. Level of significance: $\alpha$ = 5%

IV. Critical value: The value of $Z_{\alpha}$ at 5% level of significance from the take is 1.96

V. Decision: Since the computed value of $ \mid Z \mid$ = 1.15 is less than the critical value $Z_{\alpha} = 1.96$, the hypothesis is accepted.

$ \therefore$ Boys don not perform better than the girls.

ii. Given, n=15 variance = $S^2 = 10.24 s= 3.2 \bar{X}=6.2\ \mu = 5.2$

I The null hypothesis $(H_0) = \mu = 5.4$

Alternative hypothesis $(H_{\alpha}) = \mu \neq 5.4$

II Calculation of test statistic

$ t = \frac{\bar x - \mu}{\frac{S}{\sqrt{n-1}}}$

$ t = \frac{\bar x - \mu}{S.E}$

let $ S.E = \frac{S}{\sqrt{n-1}}$

$ S.E = \frac{32}{\sqrt{15-1}}$

S.E = 0.8552

$ \therefore t = \frac{\bar x - \mu}{S.E}$

$ = \frac{\6.2- 5.4}{0.8552}$

t= 0.9354

III. Level of significance: at 5%

$ \therefore \alpha = 0.05$

Degree of freedom = n-1

=15-1

Degree of freedom = 14

IV. Critical value: The value of $t_{\alpha}$ at 5% level of significance and 14 degree of freedom from the take is 2.145

V. Decision: Since the computed value of $ \mid t \mid$ = 0.9354 is less than the take value $t_{\alpha} = 2.145$, the null hypothesis is accepted.

$ \therefore$ The sample is drawn from a normal population with mean 5.4