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?
Expand Hint
The population’s standard deviation,
$$\sigma$$
, is known, and the sample size is greater than 30. Thus, let’s use the z distribution. Only use the t distribution if
$$\sigma$$
is unknown and if the sample size is less than 30.
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, let’s use the z distribution. 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)
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