# A researcher surveyed 220 residents of a city about the number of hours theyspend watching news on television each day. The mean

###### Question:

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?

## Answers

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Margin of error of 0.0485 hours.

Step-by-step explanation:

We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:

[tex]\alpha = \frac{1 - 0.95}{2} = 0.025[/tex]

Now, we have to find z in the Ztable as such z has a pvalue of [tex]1 - \alpha[/tex].

That is z with a pvalue of [tex]1 - 0.025 = 0.975[/tex], so Z = 1.96.

Now, find the margin of error M as such

[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]

In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.

In this question:

[tex]\sigma = 0.35, n = 220[/tex]

The margin of error is of:

[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]

[tex]M = 1.96\frac{0.35}{\sqrt{200}}[/tex]

[tex]M = 0.0485[/tex]

Margin of error of 0.0485 hours.

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