# Statistics question

**Question description**

*One can calculate the 95% confidence interval for the mean with the population standard deviation known. This will give us an upper and a lower confidence limit. What happens if we decide to calculate the 99% confidence interval? Describe how the increase in the confidence level has changed the width of the confidence interval. Do the same for the confidence interval set at 90%. Do an example and give real values for the intervals in your post. *

The formula for a confidence interval is:

x̄ +/-

z_{α/2 }∙**σ/ sqrt(n) **

Use that formula with the numbers plugged into it.

x̄ is the sample mean and n is the sample size (no. of samples collected).

The population standard deviation is denoted by the greek letter sigma

**σ**:

(1-α)*100% is the level of

confidence.

For example, in a 95% confidence interval

1- α =.95 Solving for α

gives

α =.05(Also, see Table 8.1 on

page 359, under 95% confidence). Thus:

α/2 = .05/2 =.025z

_{α/2}denotes the z

value on the bell-shaped curve whose right-tail area is α/2 .

For example, on page

292 (under "what is this z value?"), the z value there is denoted by z

_{.10}since

the right-tail area of that z value is .10.

If you are doing a 95% confidence interval for the mean using formula 8.1 on

page 358, then

z_{α/2 }= z_{.025}

= 1.96

by looking up the z-value on table 8.1 on page 359 (under confidence level

95%). This means that the area to the right of z=1.96 is .025.

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