3 years ago

Can someone help me with this question please

Mathematics

1 answer:

rusak2 [61]3 years ago

6 0

The correct answer is c

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$n=\frac{0.3(1-0.3)}{(\frac{0.01}{1.96})^2}=8067.36$

And rounded up we have that n=8068

Step-by-step explanation:

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

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

We want for this case a 95% of confidence desired, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by:

$z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96$

The margin of error for this case is $ME =\pm 0.01$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

The estimated proportion for this case is $\hat p=0.3$ . And replacing into equation (b) the values from part a we got:

$n=\frac{0.3(1-0.3)}{(\frac{0.01}{1.96})^2}=8067.36$

And rounded up we have that n=8068

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