8.2 Confidence intervals: confidence interval, single population  (Page 2/5)

In summary, as a result of the central limit theorem:

• $\overline{X}$ is normally distributed, that is, $\overline{X}$ ~ $\mathrm{N}({\mu }_X,\frac{\sigma }{\sqrt{n}}).$
• When the population standard deviation $\sigma$ is known, we use a Normal distribution to calculate the margin of error.

Calculating the confidence interval:

• Check the assumptions and conditions
• Calculate the sample mean $\overline{x}$ from the sample data. Remember, in this section, we already know the population standard deviation $\sigma$ .
• Find the Z-score that corresponds to the confidence level.
• Calculate the margin of error ME
• Construct the confidence interval
• Write a sentence that interprets the estimate in the context of the situation in the problem. (Explain what the confidence interval means, in the words of the problem.)

We will first examine each step in more detail, and then illustrate the process with some examples.

Finding z for the stated confidence level

This value can be found by using technology or the confidence levels found at the bottom of the t Table in the appendix of the text. Below is the part of the t Table needed to find the z value corresponding to the level of confidence asked for in a problem.

If asked to calculate a 95% confidence interval, replace the z in the confidence formula with 1.960. $\overline{x}$ + z $\frac{\sigma }{\sqrt{n}}$ and $\overline{x}$ - z $\frac{\sigma }{\sqrt{n}}$

The confidence level, $\mathrm{CL}$ , is the area in the middle of the standard normal distribution. $\mathrm{CL}=1-\alpha$ . So $\alpha$ is the area that is split equally between the two tails. Each of the tails contains an area equal to $\frac{\alpha }{2}$ .

The z-score that has an area to the right of $\frac{\mathrm{\alpha }}{2}$ is denoted by $z_{\frac{\alpha }{2}}$

For example, when $\mathrm{CL}=0.95$ then $\alpha =0.05$ and $\frac{\alpha }{2}=0.025$ ; we write $z_{\frac{\alpha }{2}}=z_{.025}$

The area to the right of $z_{.025}$ is 0.025 and the area to the left of $z_{.025}$ is 1-0.025 = 0.975

$z_{\frac{\alpha }{2}}=z_{0.025}=1.96$ , using a calculator, computer or a Standard Normal probability table.

Me: margin of error

• $\mathrm{ME}=z_{\frac{\alpha }{2}}\cdot \frac{\sigma }{\sqrt{n}}$

Constructing the confidence interval

• The confidence interval estimate has the format $(\overline{x}-\mathrm{ME},\overline{x}+\mathrm{ME})$ .

The graph gives a picture of the entire situation.

$\mathrm{CL}+\frac{\alpha }{2}+\frac{\alpha }{2}=\mathrm{CL}+\alpha =1$ .

Writing the interpretation

Changing the confidence level or sample size

Changing the confidence level

Suppose we change the original problem by using a 95% confidence level. Find a 95% confidence interval for the true (population) mean statistics exam score.

To find the confidence interval, you need the sample mean, $\overline{x}$ , and the ME.

• $\overline{x}=68$
• $\mathrm{ME}=z_{\frac{\alpha }{2}}\cdot \left(\frac{\sigma }{\sqrt{n}}\right)$
• $\sigma =3$ ; $n=36$ ; The confidence level is 95% (CL=0.95)

$\mathrm{CL\; =\; 0.95}$ so $\alpha =1-\mathrm{CL}=1-0.95=0.05$

$\frac{\alpha }{2}=0.025z_{\frac{\alpha }{2}}=z_{.025}$

The area to the right of $z_{.025}$ is 0.025 and the area to the left of $z_{.025}$ is 1−0.025=0.975

$z_{\frac{\alpha }{2}}=z_{.025}=1.96$

$\mathrm{ME}=1.96\cdot \left(\frac{3}{\sqrt{36}}\right)=0.98$

$\overline{x}-\mathrm{ME}=68-0.98=67.02$

$\overline{x}+\mathrm{ME}=68+0.98=68.98$

Interpretation

We estimate with 95 % confidence that the true population mean for all statistics exam scores is between 67.02 and 68.98.

Explanation of 95% confidence level

95% of all confidence intervals constructed in this way contain the true value of the population meanstatistics exam score.

Comparing the results

The 90% confidence interval is (67.18, 68.82). The 95% confidence interval is (67.02, 68.98). The 95% confidence interval is wider. If you look at the graphs, because the area 0.95 is larger than the area 0.90, it makes sense that the 95% confidence interval is wider.

Summary: effect of changing the confidence level

• Increasing the confidence level increases the margin of error, making the confidence interval wider.
• Decreasing the confidence level decreases the margin or error, making the confidence interval narrower.