91影视

It is given in the question that,

margin of error $E=0.04$

guessed value $\stackrel{鈭�}{p}=75\%$

confidence interval =$90\%$

The confidence interval is $90\%$

convert $90\%$ into decimal.

$\frac{90}{100}=0.90$

For confidence interval $0.90$, use table A

convert $75\%$into decimal.

$\frac{75}{100}=0.75$

Now, find the sample size. Use the formula $n=\frac{{\left[z_{a/2}\right]}^2\stackrel{鈫�}{p}(1鈭�\stackrel{鈫�}{p})}{E^2}$

$n=\frac{{\left[z_{a/2}\right]}^2\stackrel{鈫�}{p}(1鈭�\stackrel{鈫�}{p})}{E^2}$

$n=\frac{{1.645}^2脳0.75脳(1鈭�0.75)}{{0.04}^2}$

$=318$

Therefore, the sample size is$318$

It is given in the question that,

margin of error $E=0.04$

guessed value $\stackrel{鈭�}{P}=75\%$

confidence interval =$90\%$

By how much do the two sample sizes differ?

Convert $90\%$into decimal.

$\frac{90}{100}=0.90$

For confidence interval $0.90$, use table A.

$z_{伪/2}=1.645$

From part (a) sample size n = $318$

Find the new sample size. Use the formula $n=\frac{{\left[z_{o/2}\right]}^2\stackrel{鈫�}{p}(1鈭�\stackrel{鈫�}{p})}{E^2}$

$n^{\mathrm{鈥�}}=\frac{{\left[z_{a/2}\right]}^2\stackrel{鈫�}{p}(1鈭�\stackrel{鈫�}{p})}{E^2}$

$n^{\mathrm{鈥�}}=\frac{{1.645}^2脳0.5脳(1鈭�0.5)}{{0.04}^2}$

$=423$

Thus, the new sample size is $423$

For the difference between the sample sizes, subtract $n$from $n\text{'}$

$n^{\mathrm{鈥�}}鈭�n=423鈭�318$

$=105$