by samara
Last Updated August 13, 2019 23:19 PM

Given the value of alpha (confidence level), how I can compute the he confidence coefficient? The value of the confidence coefficient is given for 90%, 95% and 99%. My question is how to find the value of the confidence coefficient for 92% as example. Source of the picture is : link.

You can find the confidence coefficient from tables of the standard normal distribution. I don't know what style of table you may have at hand, but for a 92% confidence interval, you want the z-value that cuts 4% of the probability from the upper tail of the standard normal distribution. (You may find a graph at the top of the table with a shaded area that gives an idea what probabilities are to be found in the table.)

Look in the body of the table for the *nearest* relevant value, then get the z-value from the margins of the table. By symmetry, the negative of that z-value will cut probability 4% from the *lower* tail, leaving 92% of the probability in the center. The answer is $z = 1.75$

If you're having trouble figuring out how the printed table works, remember that the confidence coefficient for a 95% confidence interval is $1.96.$ Look in the margins of the table for $1.96$ and then see how the corresponding value in the table is related to .9750 or .0250.$

Alternatively, you can use software. Software can be a little more accurate because you can get the exact value to several decimal places. (Using a normal table, you will find probabilities to four places in the body of the table, but you may not find an *exact* match to cut 4% from the upper tail; then without interpolation, you can get only two places behind the decimal point from the margins of the table.)

In R statistical software, the computation is as shown below. (To get 4% above, you want 96% below.) You get 1.7507, rounding to 4 places.

```
qnorm(.96)
[1] 1.750686
```

Either way, $Z_{\alpha/2} = Z_{.08/2} = Z_{.04} = 1.76$ so that a 92% confidence interval is of the form $\bar X \pm 1.76\frac{\sigma}{\sqrt{n}.}$

In the figure below, the area (probability) under the standard normal curve between the dotted vertical lines, at $\pm 1.75,$ is $0.92.$

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