Question

The latest poll from ESPN asked fans what their favorite sport was, considering pro, college and highschool levels. Out of 1000 fans, 32% of fans said their favorite sport was college basketball. Construct a 95% confidence interval.

Answer 1

Answer:

Confidence interval is $0.2911<p<0.3489$

Step-by-step explanation:

Given that 32% of fans sport is basketball as favorite sport. So total number of fans is,

$32\%\times 1000$

$= \dfrac{32}{100}\times 1000$

$=320$

So 320 fans favorite sport is basket ball.

Now find sample proportion ,

$\widehat{p}=\dfrac{number\:of\:people\:in\:the\:sample}{total\:number\:of\:people}$

Substituting the value,

$\widehat{p}=\dfrac{320}{1000}$

$\widehat{p}=0.320$

Formula for confidence interval is,  

$CI=\widehat{p}\pm z\sqrt{\dfrac{\widehat{p}\left (1-\widehat{p}\right )}{n}}$

To calculate the z value.

$Confidence Level=1-\alpha$

$0.95=1-\alpha$

$0.05=\alpha$

Now divide above value by 2,  

$\dfrac{0.05}{2}=\dfrac{\alpha}{2}$

$0.025=\dfrac{\alpha}{2}$

To find the value of z score calculate the area under the curve as follows

$A=\dfrac{1+CL}{2}$

$A=\dfrac{1+0.95}{2}$

$A=\dfrac{1.95}{2}$

$A=0.975$

On z table, find the value of 0.975. So the row corresponding to 0.975 is 1.9 and column corresponding to 0.975 is 0.06. Adding these two values the z score is 1.96.

or find the value from excel by using command as follows,  

=NORM.S.INV(0.025)= -1.959963985

Now substituting the value,  

$CI=0.320\pm 1.96\sqrt{\dfrac{0.320\left (1-0.320\right )}{1000}}$

$CI=0.320\pm 0.02891$

Lower end of the interval is,

$CI=0.320- 0.02891$

$CI=0.2911$

Upper end of the interval is,

$CI=0.320+ 0.02891$

$CI=0.3489$

Therefore 95% confidence interval is $0.2911<p<0.3489$