Refer to the Archives of Paediatrics and Adolescent Medicine (Dec. 2007) study of honey as a children’s cough remedy, Exercise 2.31 (p. 86). Children who were ill with an upper respiratory tract infection and their parents participated in the study. Parents were instructed to give their sick child dosage of liquid “medicine” before bedtime. Unknown to the parents, some were given a dosage of dextromethorphan (DM)—an over-the-counter cough medicine—while others were given a similar dose of honey. (Note: A third group gave their children no medicine.) Parents then rated their children’s cough symptoms, and the improvement in total cough symptoms score was determined for each child. The data (improvement scores) for the 35 children in the DM dosage group and the 35 in the honey dosage group are reproduced in the next table. Do you agree with the statement (extracted from the article), “Honey may be a preferable treatment for the cough and sleep difficulty associated with childhood upper respiratory tract infection”? Use the comparison of the two means methodology presented in this section to answer the question.

The data is given below:

Honey Dosage: $\begin{array}{l}\mathbf{\text{121115111013104151691410610811}}\\ \mathbf{\text{12128129111510159138121089512}}\end{array}$

DM Dosage: $\begin{array}{l}\mathbf{\text{46947779121011634978}}\\ \mathbf{\text{1212412137101394410159126}}\end{array}$

Step by step solution

01

Step-by-Step Solution Step 1: Calculate the mean and standard deviation of both the groups

The mean and standard deviation of both the groups are

02

Conduct a z-test

Null Hypothesis: There is no difference between the Honey and Control groups.

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

Alternate Hypothesis: Honey is a preferable treatment. $H_a:{\mu }_1-{\mu }_2>0$

Level of significance $\left(\alpha \right)=0.01$

$$\begin{gathered} z=\frac{(\overline{x_1}-{\overline{x}}_2)-({\mu }_1-{\mu }_2)}{\sqrt{\left(\frac{{{\sigma }_1}^2}{n_1}+\frac{{{\sigma }_2}^2}{n_2}\right)}} \\ =\frac{(10.71-8.33)-0}{\sqrt{\left(\frac{{\left(2.86\right)}^2}{35}+\frac{{\left(3.26\right)}^2}{33}\right)}} \\ =\frac{2.38}{0.7455} \\ =3.19 \end{gathered}$$

The critical value is $2.33$.

As the value $z$ is more than the critical value, the null hypothesis should be rejected.

Therefore, the data provide sufficient evidence to indicate that ${\mu }_1-{\mu }_2>0$.

So, honey may be a preferable treatment for the cough and sleep difficulty associated with childhood upper respiratory tract infection.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!