Question

A new post-surgical treatment is being compared with a standard treatment. Seven subjects receive the new treatment, while seven others (the controls) receive the standard treatment. The recovery times, in days, are given below.
Treatment: 12 13 15 19 20 21 24
Control: 18 23 24 30 32 35 39

Required:
Find a 98% confidence interval for the difference in the mean recovery times between treatment and control.

Answer 1

Answer:

$(17.714-28.714) -2.681 \sqrt{\frac{4.461^2}{7} +\frac{7.387^2}{7}}= -19.745$

$(17.714-28.714) +2.681 \sqrt{\frac{4.461^2}{7} +\frac{7.387^2}{7}}= -2.255$

Step-by-step explanation:

For this case we have the following info given:

Treatment: 12 13 15 19 20 21 24

Control: 18 23 24 30 32 35 39

We can find the sample mean and deviations with the the following formulas:

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

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

And repaplacing we got:

$\bar X_T = 17.714$ the sample mean for treatment

$\bar X_C = 28.714$ the sample mean for treatment

$s_T= 4.461$ the sample deviation for treatment

$s_C= 7.387$ the sample deviation for control

$n_T= n_C= 7$ the sample size for each sample

The degrees of freedom are given by:

$df= 7+7-2= 12$

The confidence interval for the difference of means is given by:

$(\bar X_T -\bar X_C) \pm t_{\alpha/2} \sqrt{\frac{s^2_T}{n_T} +\frac{s^2_C}{n_C}}$

The confidence is 98% so then the significance is $\alpha=0.02$ and $\alpha/2 =0.01$. Then the critical value would be:

$t_{\alpha/2}=2.681$

And replacing we got:

$(17.714-28.714) -2.681 \sqrt{\frac{4.461^2}{7} +\frac{7.387^2}{7}}= -19.745$

$(17.714-28.714) +2.681 \sqrt{\frac{4.461^2}{7} +\frac{7.387^2}{7}}= -2.255$