Question

A random sample of elementary school children in New York state is to be selected to estimate the proportion p who have received a medical examination during the past year. A random sample of 200 elementary school children indicated that 18 of them had received a medical examination in the past year.
a)Find point estimate for p and also construct a 95% confidence interval for p.

Answer 1

95 % of confidence interval of p = {0.0503372, 0.129663}

Point estimation

A single price estimate of a parameter is stated as a degree estimate.

A sample mean, for instance, could be a single purpose estimate of the population mean.

confidence interval

This is the vary of values you expect your estimate to fall between if you redo your take a look at at a given level of confidence.

p = 18 /200

95% means using 1.96

18/200 + 1.96*$\sqrt{\frac{18 * 182}{200^{3} } }$

18/200 - 1.96 *$\sqrt{\frac{18 * 182}{200^{3} } }$

18/200 + 1.96 *$\sqrt{\frac{18 * 182}{200^{3} } }$

{0.0503372, 0.129663}

Therefore, the 95% of confidence interval is {0.0503372, 0.129663}

To learn more about point estimation check the given link

https://brainly.com/question/24261796

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