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.

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:

Where:

is the sample's proportion of success, which in this case is the people that regularly lie during surveys,

is the critical value needed to find the tails of distribution related to the confidence level,

is the sample's size.

First we compute the value:

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

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

i think it will be zero as slope has a formula of difference between initial and final distance divided by itz time interval