Confidence Intervals
Confidence Intervals for Variance and Standard Deviation – (section 4)
15
NOTE: You should also read the text. These are just high-level notes.
Use a chi-square distribution to construct a confidence interval for the variance and standard deviation. (chi is the
Greed symbol 𝜒) The chi-square distribution is positively skewed (right skewed, tail to the right). The area under the
chi-square distribution is equal to 1.
Confidence Interval for variance, 𝜎 2 :
(𝑛−1)𝑠2
2
𝜒𝑅
< 𝜎2 <
Confidence Interval for standard deviation, 𝜎: √
(𝑛−1)𝑠2
(𝑛−1)𝑠2
2
𝜒𝑅
𝜒𝐿2
<𝜎<√
(𝑛−1)𝑠2
𝜒𝐿2
𝜒𝑅2 and 𝜒𝐿2 are the critical values for the chi-distribution and can be found from the chi-distribution table (based upon
degrees of freedom, 𝑛 − 1, and the level of confidence. NOTE: When looking up values from this table you will need
to find the AREA TO THE RIGHT (not the left, as in the other types of distributions)! Also, 𝜒𝑅2 and 𝜒𝐿2 are not
symmetric and equal in value as they are in normal distribution.
Note: the left endpoint requires using 𝜒𝑅2 and the right endpoint requires using 𝜒𝐿2
A population has to be normally distributed in order to use the chi-square distribution.
As the degrees of freedom increases, the shape of the chi-square distribution approaches the shape of a normal curve.
(insert graph pg 338)
Confidence Intervals
Confidence Intervals for Variance and Standard Deviation – (section 4)
Chi-Square Distribution Table
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Confidence Intervals
Confidence Intervals for Variance and Standard Deviation – (section 4)
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Examples (these go along with the text, these are the corresponding “try it yourself” problems)
Example 14: Find the critical values 𝜒𝑅2 and 𝜒𝐿2 for a 90% confidence interval when the sample size is 30.
Example 15: You randomly select and weigh 30 samples of an allergy medicine. The sample standard deviation is 1.20
milligrams. Assuming the weights are normally distributed, construct the 90% confidence interval for the population
variance and standard deviation of the medicine weights.
Answers:
Example 14:
42.557, 17.708
90% of the area under the curve lies between 17.708 and 42.557
Example 15: variance (0.98, 2.36); standard deviation (0.99, 1.54); With 90% confidence, you can say that the
population variance is between 0.98 and 2.36 and the population standard deviation is between 0.99 and 1.54.
Homework (solutions posted online)
36. Find the critical values 𝜒𝑅2 and 𝜒𝐿2 for a confidence level 𝑐 = 0.95 and sample size 𝑛 = 20.
Confidence Intervals
Confidence Intervals for Variance and Standard Deviation – (section 4)
18
For the rest of the problems, assume each sample is taken from a normally distributed population and construct the
indicated confidence intervals for (a) the population variance 𝜎 2 and (b) the population standard deviation 𝜎. Interpret
the results.
37. The reserve capacities (in hours) of 18 randomly selected automotive batteries are shown. Use a 99% level of confidence.
1.70
1.60
1.94
1.58
1.74
1.60
1.86
1.72
1.38
1.46
1.64
1.49
1.55
1.70
1.75
0.88
1.77
2.07
38. A magazine includes a report on the energy costs per year for 32-inch liquid crystal display (LCD) televisions. The article states
that 14 randomly selected 32-inch LCD televisions have a sample standard deviation of $3.90. Use a 99% level of confidence.
39. As part of your spring break planning, you randomly select 10 hotels in Cancun, Mexico, and record the room rate for each
hotel. The results are shown in the stem-and-leaf plot below. Use a 98% level of confidence.
6 9
Key: 7|4=74
7 4
8
9 0 9 9
10
11 2
12
13 6 9
14 9
15 0
40. The annual earnings of 14 randomly selected computer software engineers have a sample standard deviation of $3725. Use an
80% level of confidence.
41. The weighting times (in minutes) of a random sample of 22 people at a bank have a sample standard deviation of 3.6 minutes.
Use a 98% level of confidence.