91影视

It is given in the question that,

The margin of error$E=0.04$

Guessed value$\stackrel{鈭�}{p}=75\%$

Confidence interval $=90\%$

The con铿乨ence level was not stated in the report. Use what you have learned to 铿乶d the con铿乨ence level, assuming that the NCAA took an SRS.

The confidence interval is $90\%$.

Convert $90\%$into a decimal.

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

For confidence interval $0.90$, use table A.

$z_{伪/2}=1.645$

Guessed value $\stackrel{鈭�}{p}=75\%$

convert $75\%$into decimal.

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

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

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

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

$=318$

Therefore the sample size is$318$

it is given in the question that,

The margin of error$E=0.04$

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

Confidence interval $=90\%$

It was not possible to calculate the number of students who were asked to respond but did not. How does this fact affect the way that you interpret the results?

The confidence interval is $90\%$.

Convert $90$% into a 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 role="math" $n=\frac{{\left[z_{伪/2}\right]}^2\hat{p}(1鈭�\hat{p})}{E^2}$

role="math" localid="1649869305731" $n\text{'}=\frac{{\left[z_{伪/2}\right]}^2\hat{p}(1鈭�\hat{p})}{E^2}$

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

$=423$

Therefore, the new sample size is $423$.

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

$n^{\mathrm{鈥�}}鈭�n=423鈭�318$

$=105$

Hence, the sample size is increased by $105$.