Mathematics: Asimple random sample of 800 individuals provides 100 yes responses. a. what is the point estimate of the proportion of the population that would provide yes responses (to 2 decimals)? b. what is your estimate of the standard error of the proportion (to 4 decimals)? c. compute the 95% confidence interval for the population proportion (to 4 decimals). ( , )

Asimple random sample of 800 individuals provides

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15.03.2022

a) 0.13

b) $(0.1067,\ 0.1533)$

Step-by-step explanation:

The confidence interval for population proportion is given by :-

$p\pm z_{\alpha/2}\sqrt{\dfrac{p(1-p)}{n}}$

Given : Sample size : n= 800

Number of individuals provides Yes responses = 100

a) The proportion of individuals provides Yes responses = $p=\dfrac{100}{800}=0.125\approx0.13$

hence, the the point estimate of the proportion of the population that would provide Yes responses : p=0.13

Significance level : $\alpha= 1-0.95=0.05$

Critical value : $z_{\alpha/2}=1.96$

Now, the 90% two-sided confidence interval on the proportion of people who regularly have a dental checkup will be :-

$0.13\pm (1.96)\sqrt{\dfrac{0.13(1-0.13)}{800}}\\\\\approx0.13\pm0.0233\\\\=(0.13-0.0233,0.13+0.0233)\\\\=(0.1067,\ 0.1533)$

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