Answer :
Answer:
[tex]n=\frac{0.5(1-0.5)}{(\frac{0.1}{1.96})^2}=96.04[/tex]
Then the minimum sample size in order to satisfy the condition of 0.1 for the margin of error is 97 and since the sample used is n =100 we can conclude that is sufficient and the best answer would be:
D. Yes.
Step-by-step explanation:
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. We know that we require a 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 margin of error for the proportion interval is given by this:
[tex] ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex] (a)
We want a margin of error of [tex]ME =\pm 0.1[/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)
Since we don't have prior info for the population proportion we can use as estimator the value of 0.5. And replacing into equation (b) the values from part a we got:
[tex]n=\frac{0.5(1-0.5)}{(\frac{0.1}{1.96})^2}=96.04[/tex]
Then the minimum sample size in order to satisfy the condition of 0.1 for the margin of error is 97 and since the sample used is n =100 we can conclude that is sufficient and the best answer would be:
D. Yes.