Question

Is honey a cough remedy? Refer to the Archives of Pediatrics and Adolescent Medicine (Dec. 2007) study of honey as a children’s cough remedy, Exercise 8.23 (p. 470). The data (cough improvement scores) for the 33 children in the DM dosage group and the 35 children in the honey dosage group are reproduced in the table below. In Exercise 8.23, you used a comparison of two means to determine whether “honey may be a preferable treatment for the cough and sleep difficulty associated with childhood upper respiratory tract infection.” The researchers also want to know if the variability in coughing improvement scores differs for the two groups. Conduct the appropriate analysis, using $\alpha =0.10$

Short answer

we reject the null hypothesis.

Step-by-step solution

Specifying the hypothesis

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

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

The null hypothesis are given by

$H_0:{\mu }_1-{\mu }_2=0$

The alternative hypothesis are given by

$H_a:{\mu }_1-{\mu }_2>0$

Compute mean and standard deviation

The mean for first group is given by

$$\begin{gathered} {\overline{x}}_1=\frac{{\sum }_{i=1}^n{X}_i}{n} \\ =\frac{375}{35} \\ =10.71 \end{gathered}$$

The mean for second group is given by

$$\begin{gathered} {\overline{x}}_2=\frac{{\sum }_{i=1}^n{X}_i}{n} \\ =\frac{275}{33} \\ =8.33 \end{gathered}$$

The sd for first group is given by

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

The sd for second group is given by

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

Test statistic

The test statistic is computed as

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

Conclusion

For $\alpha =0.10$

Rejection region for right tailed test is given by

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

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

Therefore, we reject the null hypothesis.