Question

Taco Palace wanted to determine what proportion of its customers prefer soft-shell tacos. Out of 350 customers, 133 of them (that is, 38% of them) chose soft-shell tacos. What is the 95% confidence interval of the true proportion of customers who prefer soft-shell tacos? Round your answer to the nearest hundredth (or nearest whole percent).

Answer 1

We have to find the 95% confidence interval for the proportion.

We first need the sample proportion, the standard error and the critical value of z.

The sample proportion is p=0.38.

`p=X/n=133/350=0.38`

The standard error of the proportion is:

`\begin{gathered} \sigma_p=\sqrt{(p(1-p))/(n)}=\sqrt{(0.38*0.62)/(350)} \\ \sigma_p=√(0.000673)=0.0259 \end{gathered}`

The critical z-value for a 95% confidence interval is z=1.96.

Now we can can calculate the margin of error (MOE) as:

`MOE=z\cdot\sigma_p=1.96\cdot0.0259=0.0509`

Then, the lower and upper bounds of the confidence interval are:

`\begin{gathered} LL=p-z\cdot\sigma_p=0.38-0.0509=0.3291 \\ UL=p+z\cdot\sigma_p=0.38+0.0509=0.4309 \end{gathered}`

The 95% confidence interval for the population proportion is (0.3291, 0.4309).

When expressed as percentage, we can conclude that the 95% confidence interval is (33%, 43%).

Answer: the 95% confidence interval is (33%, 43%).