Question

Suppose $250$randomly selected people are surveyed to determine if they own a tablet. Of the $250$surveyed, $98$reported owning a tablet. Using a$95$% confidence level, compute a confidence interval estimate for the true proportion of people who own tablets

Short answer

A $95\%$two-sided cl estimate for the true proportion of people who own tablets is $0.331\le p\le 0.452$.

Step-by-step solution

Given Information

Given in the question that

n = 250 surveyed

x = 98 reported

Explanation

An approximate $100(1-a)$percent confidence interval on the proportion p of the population that belongs to this class if$\stackrel{\wedge }{p}$is the proportion of observations in a random sample of size n that belong to a class of interest is:

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

Where ${\mathit{z}}_{\frac{\mathbf{\alpha }}{\mathbf{2}}}$is the upper $\frac{a}{2}$percentage point of the standard normal distribution

This approach is dependent on the typical approximation to the binomial being adequate. To be suitably cautious, $np$ and $n(1-p)$ must both be bigger than or equal to $5.$ Other methods must be employed in instances where this approximation is incorrect, especially when n is small. $x=98$ of the $n=250$ people polled said they owned a tablet. The point estimate of the population percentage $p$ is then:

localid="1650531647747" $$\begin{gathered} \hat{p}=\frac{x}{n} \\ =\frac{98}{250} \\ =0.392 \end{gathered}$$ (2)

and

localid="1650531657569" $$\begin{gathered} np=250\times 0.392 \\ =98>5 \end{gathered}$$

localid="1650531668206" $$\begin{gathered} n(1-p)=250\times (1-0.392) \\ =152>5 \end{gathered}$$

Step 3: Explanation

For 95 % two-sided confidence interval,

$\frac{\alpha }{2}=\frac{1-0.95}{2}=0.025$

and

$z_{\frac{\alpha }{2}}=1.96$

Therefore, from Equations ($1$) and ($2$) a $95$% two-sided CI estimate for the true proportion of people who own tablets is

$0.392-1.96\sqrt{\frac{0.392(1-0.392)}{250}}\le p\le 0.392+1.96\sqrt{\frac{0.392(1-0.392)}{250}}$

which simplifies to

$0.331\le p\le 0.452$