Question

(Hypothetical.) Suppose five measurements of the volume of a liquid yield the following data (in milliliters).

Answer 1

Answer:

The 95% confidence interval for the measurements is [48.106, 53.494].

Step-by-step explanation:

The average M of this sample is

$M=\dfrac{1}{5}\sum_{i=1}^5x_i=\dfrac{1}{5}(52+48+49+52+53)\\\\\\M=\dfrac{254}{5}=50.800$

The standard deviation s of this sample is

$s=\sqrt{\dfrac{1}{(n-1)}\sum_{i=1}^5(x_i-M)^2}$

$s=\sqrt{\dfrac{1}{4}\cdot [(52-50.8)^2+(48-50.8)^2+(49-50.8)^2+(52-50.8)^2+(53-50.8)^2]}$

$s=\sqrt{\dfrac{1}{4}\cdot [(1.44)+(7.84)+(3.24)+(1.44)+(4.84)]}=\sqrt{\dfrac{18.8}{4}}=\sqrt{4.7}\\\\\\s=2.168$

The degrees of freedom are

$df=n-1=5-1=4$

Then, the critical value of t for a 95% CI and 4 degrees of freedom is t=2.776.

The margin of error of the CI is:

$E=t\cdot s/\sqrt{n}=2.776\cdot 2.17/\sqrt{5}=6.024/2.236=2.694$

Then, the lower and upper bounds of the CI are:

$LL=50.800-2.694=48.106\\\\UL=50.800+2.694=53.494$

The 95% confidence interval for the measurements is [48.106, 53.494].