Question

Out of 300 people sampled, 237 preferred Candidate A. Based on this, estimate what proportion of the voting population ( π ) prefers Candidate A. Use a 95% confidence level, and give your answers as decimals, to three places. < π

Answer 1

Answer:

And the 95% confidence interval would be given (0.744;0.836).

$0.744 \leq \pi \leq 0.836$

Step-by-step explanation:

1) Data given and notation  

n=300 represent the random sample taken    

X=237 represent the people who prefers the Candidate A

$\hat p=\frac{237}{300}=0.79$ estimated proportion of people who prefers the Candidate A

$\alpha=0.05$ represent the significance level (no given, but is assumed)    

Confidence =95% or 0.95

p= population proportion of people who prefers the Candidate A

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

2) Calculating the interval for the proportion

The confidence interval would be given by this formula

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

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$, with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

And replacing into the confidence interval formula we got:

$0.79 - 1.96 \sqrt{\frac{0.79(1-0.79)}{300}}=0.744$

$0.79 + 1.96 \sqrt{\frac{0.79(1-0.79)}{300}}=0.836$

And the 95% confidence interval would be given (0.744;0.836).

We are confident that about 74.4% to 83.6% of people prefers Candidate A