Answer
Confidence interval: $0.852\lt x ̅\lt0.893$
We are 95% confident that the mean weight of a plain M&M candy is between 0.852 and 0.893 grams.
Work Step by Step
The population from which the data was extracted is normally distributed. Also, there are no outliers.
The mean:
$x ̅=\frac{0.87+0.84+0.88+0.84+0.82+0.91+0.90+0.94+0.86+0.88+0.86+0.87}{12}=0.8725$
The standard deviation:
$s=\sqrt {\frac{(0.87-0.8725)^2+(0.84-0.8725)^2+(0.88-0.8725)^2+(0.84-0.8725)^2+(0.82-0.8725)^2+(0.91-0.8725)^2+(0.90-0.8725)^2+(0.94-0.8725)^2+(0.86-0.8725)^2+(0.88-0.8725)^2+(0.86-0.8725)^2+(0.87-0.8725)^2}{12-1}}=0.033$
$n=12$, so:
$d.f.=n-1=11$
$level~of~confidence=(1-α).100$%
$95$% $=(1-α).100$%
$0.95=1-α$
$α=0.05$
$t_{\frac{α}{2}}=t_{0.025}=2.201$
(According to Table VI, for d.f. = 11 and area in right tail = 0.025)
$Lower~bound=x ̅-t_{\frac{α}{2}}.\frac{s}{\sqrt n}=0.8725-2.201\times\frac{0.033}{\sqrt {12}}=0.852$
$Upper~bound=x ̅+t_{\frac{α}{2}}.\frac{s}{\sqrt n}=0.8725+2.201\times\frac{0.033}{\sqrt {12}}=0.893$
