II. Probability and Statistics for Reliability
B. Statistical inference
2. Statistical interval estimates (Evaluate)
Compute confidence intervals, tolerance intervals, etc., and draw conclusions from the results.
When using a sample to estimate a parameter, we can calculate where the true (unknown) value may reside.
Additional References
Point and Interval Estimates (article)
Confidence Interval for a Proportion – Normal Approximation (article)
Confidence Interval for Variance (article)
Statistical Confidence (article)
Quick Quiz
1-39. Identify which of the following statements concerning statistical inference is false.
(A) The confidence interval is a range of values that may include the true value of a population parameter.
(B) The confidence interval normally encompasses the statistical tolerance limits of the population parameter.
(C) Estimation is the process of analyzing a sample result to predict the value of the population parameter.
(D) The point estimate is a single value used to estimate the population parameter.
(B) The confidence interval normally encompasses the statistical tolerance limits of the population parameter.
Confidence intervals apply to the limits of the statistical parameter being estimated such as the mean or variance. Tolerance intervals provide a set of limits for the future individual values, not the mean or variance. Tolerance intervals do not apply to estimates of population parameters.
1-42. What sort of mathematical models are used for statistical inference?
(A) exponential
(B) inferential
(C) deterministic
(D) probabilistic
(D) probabilistic
We are dealing with variation and using a sample to estimate parameters for a population. We base our ability to estimate confidence intervals and hypothesis testing on the tenets of probability.
I got a different result on the Proportion example. I have:
p=.1533
z=1.645
n=150
Is this correct?
So far so good. p = 23/150 = 0.153333…
To get the confidence interval multiple the z alpha/2 which is 1.645 in this case, as you list, times the SQRT term.
Not done yet with calculations.
I’ve done this problem 4 times and always get a different result.
The square root term is:
SQRT[(.1533 * .8467 )/ 150] = .0294
So CI = .1533 ± (1.645) *(.0294)
CI is 0.1049 < p < .2017
But the example shows .1236 < p < .1824
What am I doing wrong?
Hi Raye, got me again… you are right and I just confirmed your calculations – this is the example at about minute 22 in the video on the confidence interval about a proportion.
Cheers,
Fred