A simple 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). ( , )

Respuesta :

Answer: a) 0.13

b) [tex](0.1067,\ 0.1533)[/tex]

Step-by-step explanation:

The confidence interval for population proportion is given by :-

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

Given : Sample size : n= 800

Number of individuals provides Yes responses =  100

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

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

Significance level :[tex]\alpha= 1-0.95=0.05[/tex]

Critical value : [tex]z_{\alpha/2}=1.96[/tex]

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

[tex]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)[/tex]

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