Question

In a survey of 329 people who attended a local professional soccer team’s most recent home game, 141 said that they were fans of the visiting team. Create a 95% confidence interval for the population proportion of attendees who were fans of the visiting team. Use Excel to create the confidence interval, rounding to four decimal places.

Answer 1

Answer: (0.3751,0.4821)

Step-by-step explanation:

Given : Level of significance : $1-\alpha:0.95$

Then , significance level : $\alpha: 1-0.95=0.05$

Since , sample size : $n=329$ , i.e. a large sample (n<30).

Then we use z-test.

Using excel (by going in formulas for more statistics and then statistics), Critical value : $z_{\alpha/2}=1.96$

Also, the proportion of people said that they were fans of the visiting team :-

$\hat{p}=\dfrac{ 141}{329}\approx 0.4286$

The confidence interval for population proportion is given by :-

$\hat{p}\pm z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

$0.4286\pm(1.96)\sqrt{\dfrac{0.4286(1-0.4286)}{329}}\\\\\approx0.4286\pm0.053\\\\=(0.4286-0.0534, 0.4286+0.053=(0.3751,\ 0.4821)$

Hence,  a 95% confidence interval for the population proportion of attendees who were fans of the visiting team= (0.3751,0.4821)