Statistical Intervals for a Single Sample�
1
Confidence Interval and its Properties
2
A confidence interval estimate for μ is an interval of the form
l ≤ μ ≤ u,
where the end-points l and u are computed from the sample data.
There is a probability of 1 − α of selecting a sample for which the CI will contain the true value of μ.
The endpoints or bounds l and u are called lower- and upper-confidence limits ,and 1 − α is called the confidence coefficient.
Confidence Interval on the Mean, Variance Known
3
If is the sample mean of a random sample of size n from a normal population with known variance σ2, a 100(1 − α)% CI on μ is given by
where zα/2 is the upper 100α/2 percentage point of the standard normal distribution.
EXAMPLE Metallic Material Transition
4
Ten measurements of impact energy (J) on specimens of A238 steel cut at 60°C are as follows: 64.1, 64.7, 64.5, 64.6, 64.5, 64.3, 64.6, 64.8, 64.2, and 64.3. The impact energy is normally distributed with σ = 1J. Find a 95% CI for μ, the mean impact energy.
The required quantities are zα/2 = z0.025 = 1.96, n = 10, σ = l, and .
The resulting 95% CI is found as follows:
Interpretation: Based on the sample data, a range of highly plausible values for mean impact energy for A238 steel at 60°C is
63.84J ≤ μ ≤ 65.08J
Sample Size for Specified Error on the Mean, Variance Known
5
If is used as an estimate of μ, we can be
100(1 − α)% confident that the error will not exceed a specified amount E when the sample size is
One-Sided Confidence Bounds
6
A 100(1 − α)% upper-confidence bound for μ is
and a 100(1 − α)% lower-confidence bound for μ is
Example One-Sided Confidence Bound
7
The same data for impact testing are used
to construct a lower, one-sided 95% confidence interval for the mean impact energy.
Recall that zα = 1.64, n = 10, σ = l, and .
A 100(1 − α)% lower-confidence bound for μ is
A Large-Sample Confidence Interval for μ
8
When n is large, the quantity
has an approximate standard normal distribution. Consequently,
is a large sample confidence interval for μ, with confidence level of approximately 100(1 − α)%.
Example Mercury Contamination
9
A sample of fish was selected from 53 Florida lakes, and mercury concentration in the muscle tissue was measured . The mercury concentration values were
1.230 | 1.330 | 0.040 | 0.044 | 1.200 | 0.270 |
0.490 | 0.190 | 0.830 | 0.810 | 0.710 | 0.500 |
0.490 | 1.160 | 0.050 | 0.150 | 0.190 | 0.770 |
1.080 | 0.980 | 0.630 | 0.560 | 0.410 | 0.730 |
0.590 | 0.340 | 0.340 | 0.840 | 0.500 | 0.340 |
0.280 | 0.340 | 0.750 | 0.870 | 0.560 | 0.170 |
0.180 | 0.190 | 0.040 | 0.490 | 1.100 | 0.160 |
0.100 | 0.210 | 0.860 | 0.520 | 0.650 | 0.270 |
0.940 | 0.400 | 0.430 | 0.250 | 0.270 | |
Find an approximate 95% CI on μ.
10
The summary statistics for the data are as follows:
Variable | N | Mean | Median | StDev | Minimum | Maximum | Q1 | Q3 |
Concentration | 53 | 0.5250 | 0.4900 | 0.3486 | 0.0400 | 1.3300 | 0.2300 | 0.7900 |
Example Mercury Contamination (continued)
Because n > 40, the assumption of normality is not necessary to use .
The required values are , and z0.025 = 1.96.
The approximate 95% CI on μ is
Interpretation: This interval is fairly wide because there is variability in the mercury concentration measurements. A larger sample size would have produced a shorter interval.
The t distribution
11
Let X1, X2, …, Xn be a random sample from a normal distribution with unknown mean μ and unknown variance σ2. The random variable
has a t distribution with n − 1 degrees of freedom.
The Confidence Interval on Mean, Variance Unknown
12
If and s are the mean and standard deviation of a random sample from a normal distribution with unknown variance σ2, a 100(1 − α)% confidence interval on μ is given by
where tα/2,n−1 the upper 100α/2 percentage point of the t distribution with n − 1 degrees of freedom.
One-sided confidence bounds on the mean are found by replacing tα/2,n-1 with t α,n-1.
Example Alloy Adhesion �
13
Construct a 95% CI on μ to the following data.
The sample mean is and sample standard deviation is s = 3.55.
Since n = 22, we have n − 1 =21 degrees of freedom for t, so t0.025,21 = 2.080.
The resulting CI is
Interpretation: The CI is fairly wide because there is a lot of variability in the measurements. A larger sample size would have led to a shorter interval.
19.8 | 10.1 | 14.9 | 7.5 | 15.4 | 15.4 |
15.4 | 18.5 | 7.9 | 12.7 | 11.9 | 11.4 |
11.4 | 14.1 | 17.6 | 16.7 | 15.8 | |
19.5 | 8.8 | 13.6 | 11.9 | 11.4 | |
14
Let X1, X2, …, Xn be a random sample from a normal distribution with mean μ and variance σ2, and let S2 be the sample variance. Then the random variable
has a chi-square (χ2) distribution with n − 1 degrees of freedom.
Confidence Interval on the Variance and Standard Deviation
15
If s2 is the sample variance from a random sample of n observations from a normal distribution with unknown variance σ2, then a 100(1 – α)% confidence interval on σ2 is
where and are the upper and lower 100α/2 percentage points of the chi-square distribution with
n – 1 degrees of freedom, respectively.
A confidence interval for σ has lower and upper limits that are the square roots of the corresponding limits.
16
One-Sided Confidence Bounds
The 100(1 – α)% lower and upper confidence bounds on σ2 are
Example Detergent Filling
17
An automatic filling machine is used to fill bottles with liquid detergent. A random sample of 20 bottles results in a sample variance of fill volume of s2 = 0.01532. Assume that the fill volume is approximately normal. Compute a 95% upper confidence bound.
A 95% upper confidence bound is found as follows:
A confidence interval on the standard deviation σ can be obtained by taking the square root on both sides, resulting in
Approximate Confidence Interval on a Binomial Proportion �
18
If is the proportion of observations in a random sample of size n, an approximate 100(1 − α)% confidence interval on the proportion p of the population is
where zα/2 is the upper α/2 percentage point of the standard normal distribution.
Example 8 Crankshaft Bearings
19
In a random sample of 85 automobile engine crankshaft bearings, 10 have a surface finish that is rougher than the specifications allow. Construct a 95% two-sided confidence interval for p.
A point estimate of the proportion of bearings in the population that exceeds the roughness specification is .
A 95% two-sided confidence interval for p is computed as
Interpretation: This is a wide CI. Although the sample size does not appear to be small (n = 85), the value of is fairly small, which leads to a large standard error for contributing to the wide CI.
Approximate One-Sided Confidence Bounds on a Binomial Proportion
20
The approximate 100(1 − α)% lower and upper confidence bounds are
respectively.
Tolerance and Prediction Intervals�
21
Prediction Interval for Future Observation
The prediction interval for Xn+1 will always be longer than the confidence interval for μ.
A 100 (1 − α)% prediction interval (PI) on a single future observation from a normal distribution is given by
Example Alloy Adhesion
22
The load at failure for n = 22 specimens was observed, and found that and s = 3.55. The 95% confidence interval on μ was 12.14 ≤ μ ≤ 15.28. Plan to test a 23rd specimen.
A 95% prediction interval on the load at failure for this specimen is
Interpretation: The prediction interval is considerably longer than the CI. This is because the CI is an estimate of a parameter, but the PI is an interval estimate of a single future observation.