Question

A laboratory tested twelve chicken eggs and found that the mean amount of cholesterol was 185 milligrams with s = 17.6 milligrams. A confidence interval of 173.8 mg < μ < 196.2 mg is constructed for the true mean cholesterol content of all such eggs. It was assumed that the population has a normal distribution. What confidence level does this interval represent?

Answer 1

The appropriate solution is "95%".

Explanation:

The given values are:

Lower bound,

= 173.8

Upper bound,

= 196.2

Mean of cholesterol,

= 138 milligrams

S = 17.6 mg

Now,

The margin of error (E) will be:

=
`(Upper \ bound-Lower \ bound)/(2)`

=
`(196.2-173.8)/(2)`

=
`(22.4)/(2)`

=
`11.2`

As we know,

⇒
`E=t_(\alpha 12)* (S)/(√(n) )`

On putting the values, we get

⇒
`11.2=t_(\alpha 12)* (17.6)/(√(12) )`

On applying cross-multiplication, we get

⇒
`t_(\alpha 12)=2.2044`

∴
`(\alpha)/(2) =0.025`

`\alpha = 0.025* 2`

`=0.05`

hence,

The confidence level will be:

⇒
`CI=1-\alpha`

⇒
`=1-0.05`

⇒
`=0.95`

⇒
`=95`%