## Confidence Interval

To test for yield strength, consider 40 aluminum test specimens were subjected to a tensile load test. If the standard deviation is 2.5 and the sample mean is 200 MPa, what is the 95% confidence interval for the aluminum’s yield strength?

##
__
__**Hint**

**Hint**

The population’s standard deviation,
$$\sigma$$
, is known, and the sample size is greater than 30. Thus, we need to use the z distribution. We would only use the t distribution if
$$\sigma$$
is unknown and if the sample size is less than 30.

##
__
__**Hint 2**

**Hint 2**

The Confidence Interval for the Mean µ of a Normal Distribution when a standard deviation
$$\sigma$$
is known:

$$$\bar{X}-Z_{a/2}\frac{\sigma}{\sqrt{n}}\leq \mu \leq \bar{X}+Z_{a/2}\frac{\sigma}{\sqrt{n}}$$$

where
$$\bar{X}$$
is the sample mean,
$$\sigma$$
is the standard deviation,
$$n$$
is the sample size, and
$$Z_a$$
corresponds to the appropriate probability under the normal probability curve for a given
$$Z_{var}$$
(standard normal Z score).

The population’s standard deviation,
$$\sigma$$
, is known, and the sample size is greater than 30. Thus, we need to use the z distribution. We would only use the t distribution if
$$\sigma$$
is unknown and if the sample size is less than 30.

The Confidence Interval for the Mean µ of a Normal Distribution when a standard deviation
$$\sigma$$
is known:

$$$\bar{X}-Z_{a/2}\frac{\sigma}{\sqrt{n}}\leq \mu \leq \bar{X}+Z_{a/2}\frac{\sigma}{\sqrt{n}}$$$

where
$$\bar{X}$$
is the sample mean,
$$\sigma$$
is the standard deviation,
$$n$$
is the sample size, and
$$Z_a$$
corresponds to the appropriate probability under the normal probability curve for a given
$$Z_{var}$$
(standard normal Z score).
$$Z_{a/2}$$
is usually derived from confidence interval table (like in the FE exam handbook):

Therefore, the 95% confidence interval is:

$$$200-1.96\frac{2.5}{\sqrt{40}}\leq \mu \leq 200+1.96\frac{2.5}{\sqrt{40}}$$$

$$$200-1.96(0.395)\leq \mu \leq 200+1.96(0.395)$$$

$$$200-0.775\leq \mu \leq 200+0.775$$$

$$$199.22\leq \mu \leq 200.78$$$

Thus, the answer is (199.22, 200.78)

(199.22, 200.78)