Answer :
Option-B is correct that interval is the lower boundary of 31.99 and upper boundary of 40.59 more grams of sugar content than adult's cereals.
Given that,
We have to find the confidential interval.
We know that,
It is reasonable to presume that the populations are distributed normally.
[tex]\bar x_{C}[/tex] = 46.67, [tex]s_C[/tex] = 7.514, [tex]n_C[/tex] = 20
[tex]\bar x_{A}[/tex] = 10.38, [tex]s_A[/tex] = 7.112, [tex]n_A[/tex] = 30
DF = ([tex]\frac{s_C^2}{n_C}[/tex] + [tex]\frac{s_A^2}{n_A}[/tex])² ÷ ([tex](\frac{(\frac{s_C^2}{n_C})^2}{n_C-1}+ \frac{(\frac{s_A^2}{n_A})^2}{n_A-1})[/tex])
By substituting the values we get
DF = 39
At 95% confidence interval the critical value is [tex]t_{0.025,39}[/tex] = 2.023
The 95% confidence interval is
= [tex]\bar x_{c} - \bar x_{A} }{\pm t_{0.025, 42} \times \sqr{\frac{s_C^2}{n_C} + \frac{s_A^2}{n_A}}}[/tex]
= [tex]{(46.67 - 10.38)}\pm{- 2.023 \times \sqrt{(\frac{(7.514)^2}{20} + \frac{(7.112)^2}{30}}}[/tex]
= 36.29 ± 4.30
= 31.99, 40.59
Therefore, Option-B is correct that interval is the lower boundary of 31.99 and upper boundary of 40.59more grams of sugar content than adult's cereals.
To know more about interval visit:
https://brainly.com/question/31851317
#SPJ4
