Confidence Intervals for a Population Proportion

What can we conclude about the proportion of all college students who have taken a social media break for mental health reasons?

This is our point estimate: a single value used to estimate the population proportion $p$.

A confidence interval is an interval that is used to estimate the value of a parameter.

How do we determine the margin of error for a given level of confidence?

We know: $p-1.96·{\sigma }_{\widehat{p}}<\widehat{p}<p+1.96·{\sigma }_{\widehat{p}}$

Rearranging to isolate $p$: $\widehat{p}-1.96·{\sigma }_{\widehat{p}}<p<\widehat{p}+1.96·{\sigma }_{\widehat{p}}$

This is precisely in the form $\widehat{p}\pm$ Margin of Error

Constructing a confidence interval for a population proportion requires the sampling distribution to be approximately normal.

Point Estimate: ${\displaystyle \widehat{p}=\frac{x}{n}=\frac{126}{300}=0.42}$

Confidence Level: $95\%$

Critical Value: $1.96$

Standard Error: ${\displaystyle \sqrt{\frac{\widehat{p}(1-\widehat{p})}{n}}=\sqrt{\frac{0.42\left(0.58\right)}{300}}=0.0285}$

Margin of Error: (Critical Value)·(Standard Error) $=1.96\left(0.0285\right)=0.056$

Confidence Interval: Point Estimate ± Margin of Error

$0.42\pm 0.056$$0.364<p<0.476$

Population

$p=\mathrm{?}$

What we want

Critical Value: depends on the confidence level

$\alpha$ is the area outside the middle. At 95%, $\alpha =0.05$, and $\alpha /2=0.025$ is the area in each tail.

The larger the confidence level, the larger the critical value.

$x=126$

$n=300$

confidence level 95%

Incorrect interpretation: "There is a 95% probability that $p$ is in this interval."

Once the interval is computed, $p$ is either in it or it's not. There's no probability involved for any single interval.

The confidence level describes how often the method succeeds, not the probability for any particular interval.

(a)We are 95% confident that the proportion of all college students who took a social media break is between 0.364 and 0.476.

(b)There is a 95% probability that $p$ is between 0.364 and 0.476.

(c)Between 36.4% and 47.6% of the 300 students surveyed took a social media break.

(d)If we took another sample of 300, there's a 95% chance its $\widehat{p}$ would be between 0.364 and 0.476.

Among 242 households surveyed, 107 reported using a meal kit delivery service in the past year. Construct a 90% confidence interval. Industry analysts claim 43% of households have tried meal kit services. Does the confidence interval contradict this claim?

Construct a 90% confidence interval. Industry analysts claim 43% of households have tried meal kit services. Does the confidence interval contradict this claim?

In a survey of 432 office workers, 108 said they prefer fully remote work over hybrid or in-person options. Construct a 99% confidence interval. Workplace consultants claim more than 32% of workers prefer fully remote. Does the confidence interval contradict this claim?

Construct a 99% confidence interval. Workplace consultants claim more than 32% of workers prefer fully remote. Does the confidence interval contradict this claim?

An online retailer reviewed 1,000 returns and found that 6 were flagged as potentially fraudulent. Construct a 95% confidence interval. Industry analysts estimate that 2% of online returns are fraudulent. Does the confidence interval contradict this claim?

A data center received a shipment of AI accelerator chips. A quality control engineer tested a random sample of 200 chips and found that 24 failed to meet performance specifications. Construct a 95% confidence interval. The supplier claims that fewer than 10% of chips fail specifications. Does the confidence interval contradict this claim?

Construct a 95% confidence interval. The supplier claims that fewer than 10% of chips fail specifications. Does the confidence interval contradict this claim?