Of 1000 randomly selected cases of lung cancer, 823 resulted in death within 10 years.
a. Calculate a 95% two-sided confidence interval on the death rate from lung cancer.
b. Using the point estimate of p obtained from the preliminary sample, what sample size is needed to be 95% confident that the error in estimating the true value of p is less than 0.03?
c. How large must the sample be if you wish to be at least 95% confident that the error in estimating p is less than 0.03, regardless of the true value of p?

Respuesta :

Answer:

a) [tex]0.823 - 1.96\sqrt{\frac{0.823(1-0.823)}{1000}}=0.799[/tex]

[tex]0.823 + 1.96\sqrt{\frac{0.823(1-0.823)}{1000}}=0.847[/tex]

The 95% confidence interval would be given by (0.799;0.847)

b) [tex]n=\frac{0.823(1-0.823)}{(\frac{0.03}{1.96})^2}=621.79[/tex]  

And rounded up we have that n=622

c) [tex]n=\frac{0.5(1-0.5)}{(\frac{0.03}{1.96})^2}=1067.11[/tex]  

And rounded up we have that n=1068

Step-by-step explanation:

Part a

[tex]\hat p=\frac{823}{1000}=0.823[/tex]

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2 =0.025[/tex]. And the critical value would be given by:

[tex]z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96[/tex]

The confidence interval for the mean is given by the following formula:  

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

If we replace the values obtained we got:

[tex]0.823 - 1.96\sqrt{\frac{0.823(1-0.823)}{1000}}=0.799[/tex]

[tex]0.823 + 1.96\sqrt{\frac{0.823(1-0.823)}{1000}}=0.847[/tex]

The 95% confidence interval would be given by (0.799;0.847)

Part b

The margin of error for the proportion interval is given by this formula:  

[tex] ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]    (a)  

And on this case we have that [tex]ME =\pm 0.03[/tex] and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

[tex]n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}[/tex]   (b)  

And replacing into equation (b) the values from part a we got:

[tex]n=\frac{0.823(1-0.823)}{(\frac{0.03}{1.96})^2}=621.79[/tex]  

And rounded up we have that n=622

Part c

[tex]n=\frac{0.5(1-0.5)}{(\frac{0.03}{1.96})^2}=1067.11[/tex]  

And rounded up we have that n=1068