Caelan

2021-10-22

To find: The $95\mathrm{%}$ confidence interval for the comparison.

odgovoreh

Calculation:
The study is conducted to see that whether a sample of children of different age groups consumed an adequate amount of calcium or not. In the study, samples are considered from two independent populations (different age groups).
First, the proportion of children aged 5 to 10 year who met the calcium requirement is defined by the formula:
${\stackrel{^}{p}}_{1}=\frac{\text{Count}}{\text{Sample size}}$
Substitute the values in the above formula:
${\stackrel{^}{p}}_{1}=\frac{\text{Count}}{\text{Sample size}}$
$=\frac{861}{1055}$
$=0.8161$
Now, the proportion of children aged 11 to 13 years who met the calcium requirement is defined by the formula:
${\stackrel{^}{p}}_{2}=\frac{\text{Count}}{\text{Sample size}}$
Substitute the values in the above formula:
${\stackrel{^}{p}}_{2}=\frac{\text{Count}}{\text{Sample size}}$
$=\frac{417}{974}$
$=0.4281$
The $95\mathrm{%}$ confidence interval is defined by the formula:
$CI=\left({p}_{1}-{p}_{2}\right)±{z}_{\frac{\alpha }{2}}×PSK\sqrt{\frac{{p}_{1}\left(1-{p}_{1}\right)}{{n}_{1}}+\frac{{p}_{2}\left(1-{p}_{2}\right)}{{n}_{2}}}$
Substitute the values in the above formula:
$CI=\left({p}_{1}-{p}_{2}\right)±{z}_{\frac{\alpha }{2}}×PSK\sqrt{\frac{{p}_{1}\left(1-{p}_{1}\right)}{{n}_{1}}+\frac{{p}_{2}\left(1-{p}_{2}\right)}{{n}_{2}}}$
$=\left(0.8161-0.4281\right)±1.96×\sqrt{0.8161\left(1-0.8161\right)}\left\{1055\right\}+\frac{0.4281\left(1-0.4281\right)}{974}$
$=0.388±1.96×0.01983$
$=0.388±0.03886$
$=\left(0.3492,0.4268\right)$
Hence, the required confidence interval is (0.3492, 0.4268).

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