STANDARD DEVIATION OF A NORMAL DISTRIBUTION

Sometimes confidence intervals on the population variance or standard deviation are needed.

When the population is modeled by a normal distribution, the tests and intervals described in this section are applicable. The following result provides the basis of constructing these con- fidence intervals.

Let X1, X2, p, Xnbe a random sample from a normal distribution with mean and variance 2, and let S2be the sample variance. Then the random variable

(8-17) has a chi-square (2) distribution with n1 degrees of freedom.

X2 1n12 S2 2

␹2Distribution

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8-3 CONFIDENCE INTERVAL ON THE VARIANCE AND STANDARD DEVIATION OF A NORMAL DISTRIBUTION 267

0 5 10 15 20 25 x

k = 10 k = 5

k = 2 f(x)

Figure 8-8 Proba- bility density functions of several 2distribu- tions.

(a) α,k

α 2 0

f(x) f(x)

x

(b) 2

0

0.05 0.05

0.95, 10

= 3.94

20.05, 10

= 18.31

Figure 8-9 Percentage point of the 2distribution. (a) The percentage point 2,k. (b) The upper percentage point 20.05,1018.31 and the lower percentage point 20.95,103.94.

The probability density function of a 2random variable is

(8-18) where k is the number of degrees of freedom. The mean and variance of the 2distribution are k and 2k, respectively. Several chi-square distributions are shown in Fig. 8-8. Note that the chi-square random variable is nonnegative and that the probability distribution is skewed to the right. However, as k increases, the distribution becomes more symmetric. As the limiting form of the chi-square distribution is the normal distribution.

The percentage points of the 2 distribution are given in Table IV of the Appendix.

Define as the percentage point or value of the chi-square random variable with k degrees of freedom such that the probability that X2exceeds this value is . That is,

This probability is shown as the shaded area in Fig. 8-9(a). To illustrate the use of Table IV, note that the areas are the column headings and the degrees of freedom k are given in the left column. Therefore, the value with 10 degrees of freedom having an area (probability) of 0.05

P1X2 2,k2 冮

2,k

f1u2 du 2,k

kS, f1x2 1

2k21k 22 x1k221ex2 x0

268 CHAPTER 8 STATISTICAL INTERVALS FOR A SINGLE SAMPLE

to the right is This value is often called an upper 5% point of chi-square with 10 degrees of freedom. We may write this as a probability statement as follows:

Conversely, a lower 5% point of chi-square with 10 degrees of freedom would be 20.95,10 3.94 (from Appendix A). Both of these percentage points are shown in Figure 8-9(b).

The construction of the 100(1 )% CI for 2is straightforward. Because

is chi-square with n1 degrees of freedom, we may write

so that

This last equation can be rearranged as

This leads to the following definition of the confidence interval for 2. Pa1n12S2

22,n1

2 1n12S2 212,n1b1 Pa212,n1 1n12S2

2

22,n1b1 P1212,n1X2 2 2,n121

X2 1n12S2 2

P1X2 20.05,102P1X218.3120.05 20.05,1018.31.

If s2is the sample variance from a random sample of n observations from a normal dis- tribution with unknown variance 2, then a 100(1ⴚ ␣)% confidence interval on ␴2is (8-19) where and are the upper and lower 100兾2 percentage points of the chi-square distribution with n1 degrees of freedom, respectively. A confidence interval for␴has lower and upper limits that are the square roots of the correspond- ing limits in Equation 8-19.

212,n1

2 2,n1

1n12s2 22,n1

2 1n12s2 212,n1

Confidence Interval on the Variance

One-Sided Confidence Bounds on the Variance

It is also possible to find a 10011 2% lower confidence bound or upper confidence bound on 2.

The 100(1 )% lower and upper confidence bounds on 2are

(8-20) respectively.

1n12s2 2,n1

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8-3 CONFIDENCE INTERVAL ON THE VARIANCE AND STANDARD DEVIATION OF A NORMAL DISTRIBUTION 269

EXAMPLE 8-6 Detergent Filling

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 s20.0153 (fluid ounce)2. If the variance of fill volume is too large, an unacceptable proportion of bottles will be under- or overfilled. We will assume that the fill volume is approximately normally distributed. A 95% up- per confidence bound is found from Equation 8-26 as follows:

or

211920.0153

10.117 0.0287 1fluid ounce22 2 1n12s2

20.95,˛19

This last expression may be converted into a confidence inter- val on the standard deviation by taking the square root of both sides, resulting in

Practical Interpretation: Therefore, at the 95% level of confi- dence, the data indicate that the process standard deviation could be as large as 0.17 fluid ounce. The process engineer or manager now needs to determine if a standard deviation this large could lead to an operational problem with under-or over filled bottles.

0.17

EXERCISES FOR SECTION 8-3

8-42. Determine the values of the following percentiles:

20.05,10, 20.025,15, 20.01,12, 20.95,20, 20.99,18, 20.995,16, and 20.005,25. 8-43. Determine the 2 percentile that is required to construct each of the following CIs:

(a) Confidence level95%, degrees of freedom24, one-sided (upper)

(b) Confidence level99%, degrees of freedom9, one- sided (lower)

(c) Confidence level90%, degrees of freedom19, two- sided.

8-44. A rivet is to be inserted into a hole. A random sample of n15 parts is selected, and the hole diameter is measured.

The sample standard deviation of the hole diameter measure- ments is s0.008 millimeters. Construct a 99% lower confi- dence bound for 2.

8-45. Consider the situation in Exercise 8-44. Find a 99%

lower confidence bound on the standard deviation.

8-46. The sugar content of the syrup in canned peaches is normally distributed. A random sample of n10 cans yields a sample standard deviation of s4.8 milligrams. Calculate a 95% two-sided confidence interval for .

8-47. The percentage of titanium in an alloy used in aero- space castings is measured in 51 randomly selected parts. The sample standard deviation is s0.37. Construct a 95% two- sided confidence interval for .

8-48. An article in Medicine and Science in Sports and Exercise [“Electrostimulation Training Effects on the Physical Performance of Ice Hockey Players” (2005, Vol. 37, pp.

455–460)] considered the use of electromyostimulation (EMS) as a method to train healthy skeletal muscle. EMS ses- sions consisted of 30 contractions (4-second duration, 85 Hz) and were carried out three times per week for three weeks on 17 ice hockey players. The 10-meter skating performance test showed a standard deviation of 0.09 seconds. Construct a 95%

confidence interval of the standard deviation of the skating performance test.

8-49. An article in Urban Ecosystems, “Urbanization and Warming of Phoenix (Arizona, USA): Impacts, Feedbacks and Mitigation” (2002, Vol. 6, pp. 183–203), mentions that Phoenix is ideal to study the effects of an urban heat island because it has grown from a population of 300,000 to approximately 3 million over the last 50 years and this is a period with a continuous, de- tailed climate record. The 50-year averages of the mean annual temperatures at eight sites in Phoenix are shown below. Check the assumption of normality in the population with a probabil- ity plot. Construct a 95% confidence interval for the standard deviation over the sites of the mean annual temperatures.

Average Mean

Site Temperature (°C)

Sky Harbor Airport 23.3

Phoenix Greenway 21.7

Phoenix Encanto 21.6

Waddell 21.7

Litchfield 21.3

Laveen 20.7

Maricopa 20.9

Harlquahala 20.1

8-50. An article in Cancer Research [“Analyses of Litter- Matched Time-to-Response Data, with Modifications for Recovery of Interlitter Information” (1977, Vol. 37, pp.

3863–3868)] tested the tumorigenesis of a drug. Rats were randomly selected from litters and given the drug. The times of tumor appearance were recorded as follows:

101, 104, 104, 77, 89, 88, 104, 96, 82, 70, 89, 91, 39, 103, 93, 85, 104, 104, 81, 67, 104, 104, 104, 87, 104, 89, 78, 104, 86, 76, 103, 102, 80, 45, 94, 104, 104, 76, 80, 72, 73