R8.4. We love football! A recent Gallup Poll conducted telephone interviews with a random sample of adults aged 18 and older. Data were obtained for $1000$ people. Of these, $37\%$ said that football is their favorite sport to watch
on television.
(a) Define the parameter $p$in this setting. Explain to someone who knows no statistics why we can’t just say that $37\%$of all adults would say that football is their favorite sport to watch on television.
(b) Check conditions for constructing a confidence interval for $p$.
(c) Construct a 95% confidence interval for p.
(d) Interpret the interval in context.

To define the parameter $p$ in this setting. Then to explain who knows no statistics why we can’t just say that $37\%$ of all adults would say that football is their favorite sport to watch on television.

Let, the number of trials is $\left(n\right)=1000$.
The sample proportion is $\left(\hat{p}\right)=0.37$.
The confidence interval is calculated as follows:
$\hat{p}-z_{a/2}\times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}<p<\hat{p}+z_{a/2}\times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$
The population proportion $p$ indicates the percentage of adults aged $18$ and up who consider football to be their favorite sport.

To check conditions for constructing a confidence interval for $p$.

The following requirements are met:
(1) The sample was chosen at random.
(2) The total number of successes and failures exceeds $10$.

(3) The sample size is $10\%$ less than the population size.
As a result, all of the parameters have been met.

To construct a $95\%$confidence interval for $p$.

Using a Ti-83 calculator, the following confidence interval was calculated:

As a result, the confidence interval is $(0.341,0.3999)$.

To interpret the interval in context.

Let, $\hat{p}=37\%$

$=0.37$

Then,

$n=1000$

Using table $II$for confidence level $95\%$,

$z_{\alpha /2}=1.96$

Determine the margin of error as:

$E=z_{\alpha /2}\times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

$=1.96\times \sqrt{\frac{0.37(1-0.37)}{1000}}$

$\approx 0.0299$

Determine the confidence interval as:
$0.3401=0.37-0.0299$

$=\hat{p}-E<p<\hat{p}+E$

$=0.37+0.0299$

$=0.3999$

Asa result, $95\%$confident that the true population proportion of adults who say that football is their favorite sport to watch on television is between $0.3401$and $0.3999.$