A researcher surveyed 220 residents of a city about the number of hours they spend watching news on television each day. The mean of the sample was

1.8 with a standard deviation of 0.35.

The researcher can be 95% confident that the mean number of hours all the

residents of the city are watching news on television is 1.8 with what margin

of error?

Margin of error of 0.0485 hours.

Step-by-step explanation:

We have that to find our level, that is the subtraction of 1 by the confidence interval divided by 2. So:

Now, we have to find z in the Ztable as such z has a pvalue of .

That is z with a pvalue of , so Z = 1.96.

Now, find the margin of error M as such

In which is the standard deviation of the population and n is the size of the sample.

In this question:

The margin of error is of:

Margin of error of 0.0485 hours.

96 39 cause 135 -96=39 im smart (i used a calculater)