Q. 33 - The Practice of Statistics for AP

Question

33. Going to the prom Tonya wants to estimate what proportion of her school’s seniors plan to attend the prom. She interviews an $SRS$ of $50$ of the $750$ seniors in her school and finds that $36$ plan to go to the prom.
(a) Identify the population and parameter of interest.
(b) Check conditions for constructing a confidence interval for the parameter.

(c) Construct a 90% confidence interval for p. Show your method.
(d) Interpret the interval in context.

Step-by-Step Solution

Answer

(a) The overall population for the sample is all seniors at Tonya's school, and the parameter of interest is the proportion of seniors who plan to attend the prom.
(b) The confidence interval meets all three conditions.

(c) The amount of confidence interval that contain the true population proportion is from $0.6155$ to $0.8245$.

(d) A $90\%$ confident that the true population proportion of seniors who plan to attend the prom is between $0.6155$ and $0.8245$.

Part (a) Step 1: Given information

To identify the population and parameter of interest.

Part (a) Step 2: Explanation

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.

Part (b) Step 1: Given information

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

Part (b) Step 2: Explanation

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.

Part (c) Step 1: Given information

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

Part (c) Step 2: Explanation

Determine the sample proportion as follows:

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

$=\frac{36}{50}$

$=0.72$
The confidence level is $90\%$.

Then, $90$ percent is converted to decimal.
$\frac{90}{100}=0.90$
From the table,
$z_{\alpha /2}=z 0.05$

$=1.645$

Part (c) Step 3: Explanation

Determine the margin of error as follows: .
Margin of error $={\mathrm{Z}}_{\alpha /2}\times \sqrt{\frac{\text{ samplesize(1-sample size })}{\text{ total population }}}$
$E=z_{q/2}\times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$
$=1.645\times \sqrt{\frac{0.72(1-0.72)}{50}}$
$\approx 0.1045$
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 $0.8245$.

Part (d) Step 1: Given information

To interpret the interval in context.

Part (d) Step 2:Explanation

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 $z_{\alpha /2}=z_{0.05}$ using table A, for confidence level $90\%$as:

$z_{\alpha /2}=z_{0.05}$

$=1.645$.

Determine the margin of errors as follows:

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

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

$\approx 0.1045$

Part (d) Step 3: Explanation

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 $0.8245$.