Question

In a survey of 603 adults, 98 said that they regularly lie to people conducting surveys. Create a 99% confidence interval for the proportion of adults who regularly lie to people conducting surveys. Use Excel to create the confidence interval, rounding to four decimal places.

Answer 1

Answer:

The population proportion is estimated to be with 99% confidence within the interval (0.1238, 0.2012).

Step-by-step explanation:

The formula for estimating the population proportion by a confidence interval is given by:

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

Where:

$\hat{p}$ is the sample's proportion of success, which in this case is the people that regularly lie during surveys,

$z_{\alpha /2}$ is the critical value needed to find the tails of distribution related to the confidence level,

$n$ is the sample's size.

First we compute the $\hat{p}$ value:

$\hat{p}=\frac{successes}{n}=\frac{98}{603}=0.1625$

Next we find the z-score at any z-distribution table or app (in this case i've used StatKey):

$z_{\alpha /2}=2.576$

Now we can replace in the formula with the obtained values to compute the confidence interval:

$\hat{p}\pm z_{\alpha /2}\times\sqrt{\frac{\hat{p}\times(1-\hat{p})}{n}}=0.1625\pm 2.576\times\sqrt{\frac{0.1625\times(1-0.1625)}{603}}=(0.1238, 0.2012)$