91Ó°ÊÓ

To identify the population and parameter of interest.

The complete set of people being investigated is referred to as the population. As a result, the sample population includes all seniors at Tonya's school. The examined subject is the parameter of interest.
The population proportion of seniors planning to attend the prom is the metric of interest here.
As a result, the total population for the sample is all seniors at Tonya's school, and the parameter of interest is the population fraction of seniors who plan to attend the prom.

To check conditions for constructing a confidence interval for the parameter.

Three conditions must be met in order to find a confidence interval.
Random, Independent, and Normal are the conditions.
The criteria of randomness is met because the sample was drawn at random from the senior class.
The requirement for independence is satisfied because the sample size is less than 10% of the population size.
The study found 36 success stories and 14 failure stories.
Because the chances of success and failure are both greater than ten.
So, the typical condition is met.
As a result, all three confidence interval conditions are met.

To construct a $90\%$confidence interval for $p$.

Determine the sample proportion as follows:

Sample proportion $=\frac{\text{samplesize}}{\text{total populatiom}}$
$\hat{p}=\frac{x}{n}$

$=\frac{36}{50}$

Then, $90$ percent is converted to decimal.
$\frac{90}{100}=0.90$
From the table,
$z_{\mathrm{Î\pm }/2}=z0.05$

$=1.645$

Determine the margin of error as follows: .
Margin of error $={\mathrm{Z}}_{\mathrm{Î\pm }/2}×\sqrt{\frac{\text{samplesize(1-sample size})}{\text{total population}}}$
$E=z_{q/2}×\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$
$=1.645×\sqrt{\frac{0.72(1-0.72)}{50}}$
$≈0.1045$
The amount of margin of error might change the sample proportion.
As a result, from the smallest to the largest sample percentage, the confidence that contains genuine proportion varies.
$\hat{p}-E<p<\hat{p}+E$
$0.72-0.1045<p<0.72+0.1045$
$0.6155<p<0.8245$
As a result, the confidence interval containing the true population proportion ranges from $0.6155$ to $0.8245$.

To interpret the interval in context.

The sample proportion is calculated by dividing the number of successes by the sample size:
$\hat{p}=\frac{x}{n}$

$=\frac{36}{50}$

$=0.72$
Determine $z_{\mathrm{Î\pm }/2}=z_{0.05}$using table A, for confidence level $90\%$as:

$z_{\mathrm{Î\pm }/2}=z_{0.05}$

$=1.645$.

Determine the margin of errors as follows:

$E=z_{\mathrm{Î\pm }/2}×\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

$=1.645×\sqrt{\frac{0.72(1-0.72)}{50}}$

$≈0.1045$

Determine the confidence interval as follows:
$0.6155=0.72-0.1045$

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

$=0.72+0.1045$

$=0.8245$

Asa result, $90\%$confident that the true population proportion of seniors who plan to attend the prom is between $0.6155$and $0.8245$.