91影视

Let ${渭}_1$ be the mean improvement for children receiving the honey dosage.

Let ${渭}_2$ be the mean improvement for children receiving the DM dosage.

The null hypothesis are given by

$H_0:{渭}_1-{渭}_2=0$

The alternative hypothesis are given by

$H_a:{渭}_1-{渭}_2>0$

The mean for first group is given by

$$\begin{gathered} {\stackrel{炉}{x}}_1=\frac{{鈭�}_{i=1}^n{X}_i}{n} \\ =\frac{375}{35} \\ =10.71 \end{gathered}$$

The mean for second group is given by

$$\begin{gathered} {\stackrel{炉}{x}}_2=\frac{{鈭�}_{i=1}^n{X}_i}{n} \\ =\frac{275}{33} \\ =8.33 \end{gathered}$$

The sd for first group is given by

$$\begin{gathered} sd=\sqrt{\frac{{鈭�}_{i=1}^n{(X_i-\stackrel{炉}{X})}^2}{n-1}} \\ =\sqrt{\frac{277.1435}{34}} \\ =\sqrt{8.1512} \\ =2.85 \end{gathered}$$

The sd for second group is given by

$$\begin{gathered} sd=\sqrt{\frac{{鈭�}_{i=1}^n{(X_i-\stackrel{炉}{X})}^2}{n-1}} \\ =\sqrt{\frac{339.3337}{32}} \\ =\sqrt{10.60} \\ =3.255 \end{gathered}$$

The test statistic is computed as

$$\begin{gathered} z=\frac{\left({\stackrel{炉}{x}}_1-{\stackrel{炉}{x}}_2\right)-0}{\sqrt{\frac{{蟽}_1^2}{n_1}+\frac{{蟽}_2^2}{n_2}}} \\ =\frac{10.71-8.33}{\sqrt{\frac{{2.85}^2}{35}+\frac{{3.255}^2}{33}}} \\ =\frac{2.38}{\sqrt{0.232+0.321}} \\ =\frac{2.38}{.7436} \\ =3.200 \end{gathered}$$

For $伪=0.10$

Rejection region for right tailed test is given by

$$\begin{gathered} z_{伪}=z_{0.10} \\ =1.282 \end{gathered}$$

The test statistic is greater than tabulated value. i.e. the calculated value falls in rejection region.

Therefore, we reject the null hypothesis.