Is honey a cough remedy? Refer to the Archives of Pediatrics and Adolescent Medicine (December 2007) study of honey as a remedy for coughing, Exercise 2.31 (p. 86). Recall that the 105 ill children in the sample were randomly divided into groups. One group received a dosage of an over-the-counter cough medicine (DM); another group received a dosage of honey (H). The coughing improvement scores (as determined by the children’s parents) for the patients in the two groups are reproduced in the accompanying table. The pediatric researchers desire information on the variation in coughing improvement scores for each of the two groups.

a. Find a 90% confidence interval for the standard deviation in improvement scores for the honey dosage group.

b. Repeat part a for the DM dosage group.

c. Based on the results, parts a and b, what conclusions can the pediatric researchers draw about which group has the smaller variation in improvement scores? (We demonstrate a more statistically valid method for comparing variances in Chapter 8.)

Honey Dosage	11 12 15 11 10 13 10 13 10 4 15 16 9 14 10 6 10 11 12 12 8 12 9 11 15 10 15 9 13 8 12 10 9 5 12
DM Dosage	4 6 9 4 7 7 7 9 12 10 11 6 3 4 9 12 7 6 8 12 12 4 12 13 7 10 13 9 4 4 10 15 9

The given problem is a study of remedy of cough. Here for the study two types of remedies are considered, one is honey and cough medicine. The coughing improvement scores for the two treatments are recorded. Given data is as follows

a.

From the honey dosage improvement scores,

Let,

$n=35,\overline{x}=10.714,s=2.86$

n=35

The sample mean for honey dosage group is calculated using the following formula:

$\begin{array}{rcl}\overline{x}& =& \frac{1}{n}\sum _{i=1}^n{x}_i\\ & =& \frac{1}{35}\left(11+12+...+12\right)\\ & =& 10.714\\ & & \end{array}$

Also, the sample standard deviation for honey dosage group is calculated using the following formula:

$\begin{array}{rcl}s& =& \sqrt{\frac{1}{n}\sum _{i=1}^n{\left(x_i-\overline{x}\right)}^2}\\ & =& \sqrt{\frac{1}{35}\left[{\left(11-10.71\right)}^2+...+{\left(12-10.71\right)}^2\right]}\\ & =& 2.86\\ & & \end{array}$

Therefore, the 90% confidence interval can be calculated using the formula,

$\sqrt{\frac{\left(n-1\right)s^2}{{\chi }_{\alpha /2}^2}}⩽\sigma ⩽\sqrt{\frac{\left(n-1\right)s^2}{{\chi }_{\left(1-\alpha /2\right)}^2}}$

The degrees of freedom is 34, at 0.10 level of significance, from the table value,

${\chi }_{0.05}^2=48.60$and${\chi }_{0.995}^2=21.66$

Substitute the values to get the required confidence interval,

$\begin{array}{rcl}\sqrt{\frac{\left(35-1\right){\left(2.86\right)}^2}{48.60}}& ⩽& \sigma ⩽\sqrt{\frac{\left(35-1\right){\left(2.86\right)}^2}{21.66}}\\ \sqrt{\frac{34\left(8.1796\right)}{48.60}}& ⩽& \sigma ⩽\sqrt{\frac{34\left(8.1796\right)}{21.66}}\\ 2.39& ⩽& \sigma ⩽3.58\\ & & \end{array}$

Therefore, the 90% confidence interval for$\sigma$ is$\left(2.39,3.58\right)$

b.

From the DM dosage improvement scores,

Let,

n=33

The sample mean for honey dosage group is calculated using the following formula:

$\begin{array}{rcl}\overline{x}& =& \frac{1}{n}\sum _{i=1}^n{x}_i\\ & =& \frac{1}{33}\left(4+6+...+9\right)\\ & =& 8.33\\ & & \end{array}$

Also, the sample standard deviation for honey dosage group is calculated using the following formula:

$\begin{array}{rcl}s& =& \sqrt{\frac{1}{n}\sum _{i=1}^n{\left(x_i-\overline{x}\right)}^2}\\ & =& \sqrt{\frac{1}{33}\left[{\left(4-8.33\right)}^2+...+{\left(9-8.33\right)}^2\right]}\\ & =& 3.26\\ & & \end{array}$

Therefore, the 90% confidence interval can be calculated using the formula,

$\sqrt{\frac{\left(n-1\right)s^2}{{\chi }_{\alpha /2}^2}}⩽\sigma ⩽\sqrt{\frac{\left(n-1\right)s^2}{{\chi }_{\left(1-\alpha /2\right)}^2}}$

The degrees of freedom is 32, at 0.10 level of significance, from the table value,

${\chi }_{0.05}^2=46.19$and${\chi }_{0.995}^2=20.07$

Substitute the values to get the required confidence interval,

$\begin{array}{rcl}\sqrt{\frac{\left(33-1\right){\left(3.26\right)}^2}{46.19}}& ⩽& \sigma ⩽\sqrt{\frac{\left(33-1\right){\left(3.26\right)}^2}{20.07}}\\ \sqrt{\frac{32\left(10.6276\right)}{46.19}}& ⩽& \sigma ⩽\sqrt{\frac{32\left(10.6276\right)}{20.07}}\\ 2.71& ⩽& \sigma ⩽4.12\\ & & \end{array}$

Therefore, the 90% confidence interval for$\sigma$ is$\left(2.71,4.12\right)$

c.

From the above two confidence intervalsfor coughing improvement scores, It is observed that the width of the confidence interval for honey dosage is less when compare the same with the cough dosage and two interval coincides with each other.

So, it can say that the honey dosage is having smaller variation in the improvement scores.