91影视

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,

$n=35,\stackrel{炉}{x}=10.714,s=2.86$

n=35

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

$\begin{array}{rcl}\stackrel{炉}{x}& =& \frac{1}{n}\underset{i=1}{\overset{n}{鈭�}}{x}_i\\ & =& \frac{1}{35}\left(11+12+...+12\right)\\ & =& 10.714\\ & & \end{array}$

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

$\begin{array}{rcl}s& =& \sqrt{\frac{1}{n}\underset{i=1}{\overset{n}{鈭�}}{\left(x_i-\stackrel{炉}{x}\right)}^2}\\ & =& \sqrt{\frac{1}{35}\left[{\left(11-10.71\right)}^2+...+{\left(12-10.71\right)}^2\right]}\\ & =& 2.86\\ & & \end{array}$

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

$\sqrt{\frac{\left(n-1\right)s^2}{{蠂}_{伪/2}^2}}猢�蟽猢�\sqrt{\frac{\left(n-1\right)s^2}{{蠂}_{\left(1-伪/2\right)}^2}}$

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

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

Substitute the values to get the required confidence interval,

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

Therefore, the 90% confidence interval for$蟽$ 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:

$\begin{array}{rcl}\stackrel{炉}{x}& =& \frac{1}{n}\underset{i=1}{\overset{n}{鈭�}}{x}_i\\ & =& \frac{1}{33}\left(4+6+...+9\right)\\ & =& 8.33\\ & & \end{array}$

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

$\begin{array}{rcl}s& =& \sqrt{\frac{1}{n}\underset{i=1}{\overset{n}{鈭�}}{\left(x_i-\stackrel{炉}{x}\right)}^2}\\ & =& \sqrt{\frac{1}{33}\left[{\left(4-8.33\right)}^2+...+{\left(9-8.33\right)}^2\right]}\\ & =& 3.26\\ & & \end{array}$

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

$\sqrt{\frac{\left(n-1\right)s^2}{{蠂}_{伪/2}^2}}猢�蟽猢�\sqrt{\frac{\left(n-1\right)s^2}{{蠂}_{\left(1-伪/2\right)}^2}}$

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

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

Substitute the values to get the required confidence interval,

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

Therefore, the 90% confidence interval for$蟽$ 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.