Question

The water diet requires you to drink two cups of water every half hour from the time you get up until you go to bed, but otherwise allows you to eat whatever you like. Four adult volunteers agree to test the diet. They are weighed prior to beginning the diet and after six weeks on the diet. Their weights (in pounds) are:
Subject A B C D

Weight before diet 180 125 240 150

Weight after diet 170 130 215 152


We want to estimate the mean difference (before – after) in weight. What is the appropriate 95% confidence interval?

Answer 1

Answer:

$7-3.182 \frac{13.638}{\sqrt{4}}= -14.698$

$7+3.182 \frac{13.638}{\sqrt{4}}= 28.698$

Step-by-step explanation:

For this case we have the following info given:

Weight before diet 180 125 240 150  

Weight after diet 170 130 215 152

We define the random variable $D = before-after$ and we can calculate the inidividual values:

D: 10, -5, 25, -2

And we can calculate the mean with this formula:

$\bar X= \frac{\sum_{i=1}^n X_i}{n}$

And the deviation with:

$s= \sqrt{\frac{\sum_{i=1}^n (X_i- \bar X)^2}{n-1}}$

And after replace we got:

$\bar D= 7, s_d = 13.638$

And the confidence interval for this case would be given by:

$\bar D \pm t_{\alpha/2} \frac{s_d}{\sqrt{n}}$

The degrees of freedom are given by:

$df = n-1= 4-1=3$

For the 95% of confidence the value for the significance is $\alpha=0.05$ and the critical value would be $t_{\alpha/2}= 3.182$. And replacing we got:

$7-3.182 \frac{13.638}{\sqrt{4}}= -14.698$

$7+3.182 \frac{13.638}{\sqrt{4}}= 28.698$