To find: The 95% confidence interval for the comparison.

Question

odgovoreh · Accepted Answer

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:

{\hat{p}}_{1} = \frac{Count}{Sample size}

Substitute the values in the above formula:

{\hat{p}}_{1} = \frac{Count}{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:

{\hat{p}}_{2} = \frac{Count}{Sample size}

Substitute the values in the above formula:

{\hat{p}}_{2} = \frac{Count}{Sample size}

= \frac{417}{974}

= 0.4281

The

95 %

confidence interval is defined by the formula:

C I = (p_{1} - p_{2}) \pm z_{\frac{α}{2}} \times P S K \sqrt{\frac{p_{1} (1 - p_{1})}{n_{1}} + \frac{p_{2} (1 - p_{2})}{n_{2}}}

Substitute the values in the above formula:

C I = (p_{1} - p_{2}) \pm z_{\frac{α}{2}} \times P S K \sqrt{\frac{p_{1} (1 - p_{1})}{n_{1}} + \frac{p_{2} (1 - p_{2})}{n_{2}}}

= (0.8161 - 0.4281) \pm 1.96 \times \sqrt{0.8161 (1 - 0.8161)} {1055} + \frac{0.4281 (1 - 0.4281)}{974}

= 0.388 \pm 1.96 \times 0.01983

= 0.388 \pm 0.03886

= (0.3492, 0.4268)

Hence, the required confidence interval is (0.3492, 0.4268).

To find: The 95\% confidence interval for the comparison.

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