91影视

It is given in the question that, the margin of error $E=0.03$

sample proportion $\stackrel{鈭�}{p}=44\%$

confidence interval =$90\%$

How large an SRS is required to obtain a margin of error of $0.03$(that is, localid="1649855663210" $卤3$%) in a localid="1649855669803" $99$% con铿乨ence interval?

The confidence interval is $99\%$

convert $99\%$into decimal.

$\frac{99}{100}=0.99$

For confidence interval $0.99,$use tableA.

$z_{伪/2}=2.575$

Sample proportion $\stackrel{鈭�}{p}is44\%$

Convert $44\%$into decimal.

$\frac{44}{100}=0.44$

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

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

$n=\frac{{2.575}^2脳0.44脳(1鈭�0.44)}{{0.03}^2}$

$=1816$

Hence, the required sample size is$1816$

It is given in the question that, the margin of error

sample proportion $\stackrel{鈭�}{p}=0.5$

confidence interval =

By how much do the two sample sizes differ?

The confidence interval is

convert into decimal.

For confidence interval use tableA.

$z_{伪/2}=2.575$

From part (a) sample size localid="1649858053972" $n=1816$

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

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

$n^{\mathrm{鈥�}}=\frac{{2.575}^2脳0.5脳(1鈭�0.5)}{{0.03}^2}$

$=1842$

Thus, the new sample size is localid="1649858390720" $1842$.

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

localid="1650094699527" $n^{\mathrm{鈥�}}鈭�n=1842鈭�1816$

=$26$