The Distribution of the Sample Mean

5.4 The Distribution of the Sample Mean

  The importance of the sample mean X springs from its use in drawing conclusions about the population mean m. Some of the most frequently used inferential procedures are based on properties of the sampling distribution of X. A preview of these proper- ties appeared in the calculations and simulation experiments of the previous section,

  where we noted relationships between E(X) and m and also among V(X), s 2 , and n.

  prOpOSITION Let X 1 ,X 2 ,...,X n

  be a random sample from a distribution with mean value m

  and standard deviation s. Then

  1. E(X) 5m X 5m

  2. V(X) 5 s 2 5s X 2 yn and s X 5s yÏn

  In addition, with T o 5X 1 1 ...1X n (the sample total), E(T o ) 5 nm,

  V (T o ) 5 ns 2 , and s T o 5 Ïns.

  Proofs of these results are deferred to the next section. According to Result 1, the sampling (i.e., probability) distribution of X is centered precisely at the mean of the population from which the sample has been selected. Result 2 shows that the X distri- bution becomes more concentrated about m as the sample size n increases. In marked contrast, the distribution of T o becomes more spread out as n increases. Averaging moves probability in toward the middle, whereas totaling spreads probability out

  over a wider and wider range of values. The standard deviation s X 5s yÏn is often

  called the standard error of the mean ; it describes the magnitude of a typical or representative deviation of the sample mean from the population mean.

  5.4 the Distribution of the Sample Mean 231

  ExamplE 5.25 In a notched tensile fatigue test on a titanium specimen, the expected number of cycles to first acoustic emission (used to indicate crack initiation) is m 5 28,000,

  and the standard deviation of the number of cycles is s 5 5000. Let X 1 ,X 2 ,...,X 25

  be a random sample of size 25, where each X i is the number of cycles on a different randomly selected specimen. Then the expected value of the sample mean number of cycles until first emission is E(X) 5 m 5 28,000, and the expected total number of

  cycles for the 25 specimens is E(T 0 ) 5 nm 5 25(28,000) 5 700,000. The standard

  deviation of X (standard error of the mean) and of T 0 are 5000

  s X 5s yÏn 5

  Ï25 s T o 5 Ïns 5 Ï25(5000) 5 25,000

  If the sample size increases to n 5 100, E(X) is unchanged, but s X 5 500, half of its

  previous value (the sample size must be quadrupled to halve the standard deviation of X).

  n

  the case of a normal Population distribution

  The simulation experiment of Example 5.23 indicated that when the population distribution is normal, a histogram of x values for any sample size n is well approxi- mated by a normal curve.

  prOpOSITION Let X 1 ,X 2 ,...,X n

  be a random sample from a normal distribution with mean

  m and standard deviation s. Then for any n, X is normally distributed (with mean m and standard deviation s yÏn), as is T o (with mean nm and standard deviation Ïns).

  We know everything there is to know about the X and T o distributions when the population distribution is normal. In particular, probabilities such as P(a X b) and P(c T o d) can be obtained simply by standardizing. Figure 5.15 illustrates the X part of the proposition.

  X distribution when n 5 10

  X distribution when n 5 4

  Population distribution

  Figure 5.15 A normal population distribution and

  X sampling distributions

  A proof of the result for T o when n 5 2 is possible using the method in Example 5.22, but the details are messy. The general result is usually proved using a theoretical tool called a moment generating function. One of the chapter references can be consulted for more information.

  232 ChapteR 5 Joint probability Distributions and Random Samples

  ExamplE 5.26 The distribution of egg weights (g) of a certain type is normal with mean value 53

  and standard deviation .3 (consistent with data in the article “Evaluation of Egg Quality Traits of Chickens Reared under Backyard System in Western Uttar

  Pradesh” (Indian J. of Poultry Sci., 2009: 261–262) ). Let X 1 ,X 2 ,…,X 12 denote the

  weights of a dozen randomly selected eggs; these X i ’s constitute a random sample of size 12 from the specified normal distribution.

  The total weight of the 12 eggs is T o 5X 1 1…1X 12 ; it is normally distributed with mean value E(T o ) 5 nm 5 12(53) 5 636 and variance V(T o ) 5 ns 2 512(.3) 2 5

  1.08. The probability that the total weight is between 635 and 640 is now obtained by standardizing and referring to Appendix Table A.3:

  1 Ï1.08

  Ï1.08 2

  P (635 , T o , 640) 5

  , Z,

  5 P(−.96 < Z < 3.85)

  5 F(3.85) 2 F(2.96) < 1 2 .1685 5 .8315

  If cartons containing a dozen eggs are repeatedly selected, in the long run slightly more than 83 of the eggs in a carton will weigh in total between 635 g and 640 g. Notice that 635 , T o , 640 is equivalent to 52.9167 , X , 53.3333 (divide each term in the original system of inequalities by 12). Thus P(52.9167 , X , 53.3333) < .8315. This latter probability can also be obtained by standardizing X directly.

  Now consider randomly selecting just four of these eggs. The sample mean

  weight X is then normally distributed with mean value m X 5m5

  53 and stand-

  ard deviation s X 5s yÏn 5 .3yÏ4 5 .15. The probability that the sample mean

  weight exceeds 53.5 g is then

  P(X . 53.5) 5 P Z.

  5 P(Z . 3.33) 5 1 2 F(3.33) 5 1 2 .9996 5 .0004

  Because 53.5 is 3.33 standard deviations (of X ) larger than the mean value 53, it is exceedingly unlikely that the sample mean will exceed 53.5.

  n

  the central Limit theorem

  When the X i ’s are normally distributed, so is X for every sample size n. The deriva- tions in Example 5.21 and simulation experiment of Example 5.24 suggest that even when the population distribution is highly nonnormal, averaging produces a distri- bution more bell-shaped than the one being sampled. A reasonable conjecture is that if n is large, a suitable normal curve will approximate the actual distribution of

  X . The formal statement of this result is the most important theorem of probability.

  THEOrEm the central Limit theorem (cLt)

  be a random sample from a distribution with mean m and variance s 2 . Then if n is sufficiently large, X has approximately a normal dis- tribution with m

  Let X 1 ,X 2 ,...,X n

  5 m and s X 2 X 5s 2 n, and T o also has approximately a normal

  distribution with m

  5 nm, s 2 T o 5 ns . The larger the value of n, the better the

  To

  approximation. Figure 5.16 illustrates the Central Limit Theorem. According to the CLT, when n is

  large and we wish to calculate a probability such as P(a X b), we need only “pretend” that X is normal, standardize it, and use the normal table. The resulting

  5.4 the Distribution of the Sample Mean 233

  X distribution for large n (approximately normal)

  X distribution for small to moderate n

  Population distribution

  m

  Figure 5.16 The Central Limit Theorem illustrated

  answer will be approximately correct. The exact answer could be obtained only by first finding the distribution of X, so the CLT provides a truly impressive shortcut. The proof of the theorem involves much advanced mathematics.

  ExamplE 5.27 The amount of a particular impurity in a batch of a certain chemical product is a random variable with mean value 4.0 g and standard deviation 1.5 g. If 50 batches are independently prepared, what is the (approximate) probability that the sample average amount of impurity X is between 3.5 and 3.8 g? According to the rule of thumb to be stated shortly, n 5 50 is large enough for the CLT to be applicable.

  X then has approximately a normal distribution with mean value m X 5 4.0 and

  s X 5 1.5 yÏ50 5 .2121, so

  P (3.5 X 3.8)

  Z

  5 F(2.94) 2 F(22.36) 5 .1645

  Now consider randomly selecting 100 batches, and let T o represent the total amount of impurity in these batches. Then the mean value and standard deviation of T o are

  100(4) 5 400 and Ï100(1.5) 5 15, respectively, and the CLT implies that T 0 has

  approximately a normal distribution. The probability that this total is at most 425 g is

  P (T 0 425)

  5 P (Z 1.67) 5 F(1.67) 5 .9525 n

  ExamplE 5.28 Let X 5 the number of different people sent text messages during a particular day by

  a randomly selected student at a large university. Suppose the mean value of X is 7

  and the standard deviation is 6 (values very close to those reported in the article “Cell Phone Use and Grade Point Average Among Undergraduate University Students”

  (College Student J., 2011: 544–551) . Among 100 randomly selected such students, how likely is it that the sample mean number of different people texted exceeds 5? Notice that the distribution being sampled is discrete, but the CLT is applicable whether the variable of interest is discrete or continuous. Also, although the fact that the stand- ard deviation of this nonnegative variable is quite large relative to the mean value sug- gests that its distribution is positively skewed, the large sample size implies that X does

  have approximately a normal distribution. Using m X 5 7 and s X 5 .6,

  P (X . 5)

  Note: The cited article stated that text messaging frequency was negatively corre- lated with GPA.

  n

  234 ChapteR 5 Joint probability Distributions and Random Samples

  The CLT provides insight into why many random variables have probability dis- tributions that are approximately normal. For example, the measurement error in a scientific experiment can be thought of as the sum of a number of underlying per- turbations and errors of small magnitude.

  A practical difficulty in applying the CLT is in knowing when n is sufficiently large. The problem is that the accuracy of the approximation for a particular n depends on the shape of the original underlying distribution being sampled. If the underlying distribution is close to a normal density curve, then the approximation will be good even for a small n, whereas if it is far from being normal, then a large n will be required.

  rule of thumb The Central Limit Theorem can generally be used if n . 30.

  There are population distributions for which even an n of 40 or 50 does not suffice, but such distributions are rarely encountered in practice. On the other hand, the rule of thumb is often conservative; for many population distributions, an n much less than 30 would suffice. For example, in the case of a uniform population distribution, the CLT gives a good approximation for n 12.

  ExamplE 5.29 Consider the distribution shown in Figure 5.17 for the amount purchased (rounded to the nearest dollar) by a randomly selected customer at a particular gas station

  (a similar distribution for purchases in Britain (in £) appeared in the article “Data

  Mining for Fun and Profit,” Statistical Science, 2000: 111–131 ; there were big

  spikes at the values, 10, 15, 20, 25, and 30). The distribution is obviously quite non-normal.

  We asked Minitab to select 1000 different samples, each consisting of n 5 15 observations, and calculate the value of the sample mean X for each one. Figure 5.18 is a histogram of the resulting 1000 values; this is the approximate

  Probability

  Purchase amount

  Figure 5.17 Probability distribution of

  X 5 amount of gasoline purchased ()

  5.4 the Distribution of the Sample Mean 235

  Figure 5.18 Approximate sampling distribution of the sample mean amount purchased when n 5 15 and the population distribution is as shown in Figure 5.17

  sampling distribution of X under the specified circumstances. This distribution is clearly approximately normal even though the sample size is actually much smaller than 30, our rule-of-thumb cutoff for invoking the Central Limit Theorem. As further evidence for normality, Figure 5.19 shows a normal probability plot of the 1000 x values; the linear pattern is very prominent. It is typically not non- normality in the central part of the population distribution that causes the CLT to fail, but instead very substantial skewness.

  Figure 5.19 Normal probability plot from Minitab of the 1000 x values based on samples of size n 5 15

  n

  other Applications of the central Limit theorem

  The CLT can be used to justify the normal approximation to the binomial distribu- tion discussed in Chapter 4. Recall that a binomial variable X is the number of suc- cesses in a binomial experiment consisting of n independent successfailure trials

  with p 5 P(S) for any particular trial. Define a new rv X 1 by

  1 if the 1st trial results in a success

  X 1 5

  0 if the 1st trial results in a failure

  and define X 2 ,X 3 ,...,X n analogously for the other n 2 1 trials. Each X i indicates

  whether or not there is a success on the corresponding trial.

  236 ChapteR 5 Joint probability Distributions and Random Samples

  Because the trials are independent and P(S) is constant from trial to trial, the

  X i ’s are iid (a random sample from a Bernoulli distribution). The CLT then implies that if n is sufficiently large, both the sum and the average of the X i ’s have approxi- mately normal distributions. When the X i ’s are summed, a 1 is added for every S that

  occurs and a 0 for every F, so X 1 1...1X n 5 X. The sample mean of the X i ’s is

  X n, the sample proportion of successes. That is, both X and Xn are approximately normal when n is large. The necessary sample size for this approximation depends on the value of p: When p is close to .5, the distribution of each X i is reasonably symmetric (see Figure 5.20), whereas the distribution is quite skewed when p is near

  0 or 1. Using the approximation only if both np 10 and n(1 2 p) 10 ensures that n is large enough to overcome any skewness in the underlying Bernoulli distribution.

  Figure 5.20 Two Bernoulli distributions: (a) p 5 .4 (reasonably symmetric); (b) p 5 .1 (very skewed)

  Consider n independent Poisson rv’s X 1 ,…, X n , each having mean value mn. It can be shown that X 5 X 1 1…1X n has a Poisson distribution with mean value m

  (because in general a sum of independent Poisson rv’s has a Poisson distribution). The CLT then implies that a Poisson rv with sufficiently large m has approximately

  a normal distribution. A common rule of thumb for this is m . 20.

  Lastly, recall from Section 4.5 that X has a lognormal distribution if ln(X) has

  a normal distribution. Let X 1 ,X 2 ,...,X n

  be a random sample from a distribution

  for which only positive values are possible [P(X i . 0) 5 1]. Then if n is sufficiently

  large, the product Y 5 X 1 X 2 ?...?X n has approximately a lognormal distribution.

  To verify this, note that ln(Y) 5 ln(X 1 ) 1 ln(X 2 ) 1 . . . 1 ln(X n )

  Since ln(Y) is a sum of independent and identically distributed rv’s [the ln(X i )’s], it is approximately normal when n is large, so Y itself has approximately a lognormal dis- tribution. As an example of the applicability of this result, Bury (Statistical Models in Applied Science, Wiley, p. 590) argues that the damage process in plastic flow and crack propagation is a multiplicative process, so that variables such as percentage elongation and rupture strength have approximately lognormal distributions.

  EXERCISES Section 5.4 (46–57)

  46. Young’s modulus is a quantitative measure of stiffness of

  Stamping Processes: A Stochastic Analysis” (Intl. J.

  an elastic material. Suppose that for aluminum alloy

  of Advanced Manuf. Tech., 2010: 117–134) ).

  sheets of a particular type, its mean value and standard

  a. If X is the sample mean Young’s modulus for a ran-

  deviation are 70 GPa and 1.6 GPa, respectively (values

  dom sample of n 5 16 sheets, where is the sampling

  given in the article “Influence of Material Properties

  distribution of X centered, and what is the standard

  Variability on Springback and Thinning in Sheet

  deviation of the X distribution?

  5.4 the Distribution of the Sample Mean 237

  b. Answer the questions posed in part (a) for a sample

  a. Is it plausible that X is normally distributed?

  size of n 5 64 sheets.

  b. For a random sample of 50 such pairs, what is the

  c. For which of the two random samples, the one of part

  (approximate) probability that the sample mean

  (a) or the one of part (b), is X more likely to be

  courtship time is between 100 min and 125 min?

  within 1 GPa of 70 GPa? Explain your reasoning.

  c. For a random sample of 50 such pairs, what is the

  47. Refer to Exercise 46. Suppose the distribution is normal

  (approximate) probability that the total courtship

  (the cited article makes that assumption and even

  time exceeds 150 hr?

  includes the corresponding normal density curve).

  d. Could the probability requested in (b) be calculated

  a. Calculate P (69 X 71) when n 5 16.

  from the given information if the sample size were

  b. How likely is it that the sample mean diameter

  15 rather than 50? Explain.

  exceeds 71 when n 5 25?

  51. The time taken by a randomly selected applicant for a

  48. The National Health Statistics Reports dated Oct. 22, 2008 ,

  mortgage to fill out a certain form has a normal distribu-

  stated that for a sample size of 277 18-year-old American

  tion with mean value 10 min and standard deviation

  males, the sample mean waist circumference was 86.3 cm. A

  2 min. If five individuals fill out a form on one day and

  somewhat complicated method was used to estimate various

  six on another, what is the probability that the sample

  population percentiles, resulting in the following values:

  average amount of time taken on each day is at most

  52. The lifetime of a certain type of battery is normally dis- tributed with mean value 10 hours and standard deviation

  a. Is it plausible that the waist size distribution is at

  1 hour. There are four batteries in a package. What life-

  least approximately normal? Explain your reasoning.

  time value is such that the total lifetime of all batteries in

  If your answer is no, conjecture the shape of the

  a package exceeds that value for only 5 of all packages?

  population distribution.

  53. Rockwell hardness of pins of a certain type is known to

  b. Suppose that the population mean waist size is

  85 cm and that the population standard deviation is

  have a mean value of 50 and a standard deviation of 1.2.

  15 cm. How likely is it that a random sample of 277

  a. If the distribution is normal, what is the probability

  individuals will result in a sample mean waist size

  that the sample mean hardness for a random sample

  of at least 86.3 cm?

  of 9 pins is at least 51?

  c. Referring back to (b), suppose now that the popu-

  b. Without assuming population normality, what is the

  lation mean waist size in 82 cm. Now what is the

  (approximate) probability that the sample mean

  (approximate) probability that the sample mean

  hardness for a random sample of 40 pins is at least

  will be at least 86.3 cm? In light of this calcula-

  tion, do you think that 82 cm is a reasonable value

  54. Suppose the sediment density (gcm) of a randomly

  for m?

  selected specimen from a certain region is normally distrib-

  49. There are 40 students in an elementary statistics class.

  uted with mean 2.65 and standard deviation .85 (suggested

  On the basis of years of experience, the instructor knows

  in “Modeling Sediment and Water Column Interactions

  that the time needed to grade a randomly chosen first

  for Hydrophobic Pollutants,” Water Research, 1984:

  examination paper is a random variable with an expected

  1169–1174).

  value of 6 min and a standard deviation of 6 min.

  a. If a random sample of 25 specimens is selected, what

  a. If grading times are independent and the instructor

  is the probability that the sample average sediment

  begins grading at 6:50 p.m. and grades continuously,

  density is at most 3.00? Between 2.65 and 3.00?

  what is the (approximate) probability that he is

  b. How large a sample size would be required to ensure

  through grading before the 11:00 p.m. TV news

  that the first probability in part (a) is at least .99?

  begins?

  55. The number of parking tickets issued in a certain city on

  b. If the sports report begins at 11:10, what is the prob-

  any given weekday has a Poisson distribution with

  ability that he misses part of the report if he waits

  parameter m 5 50.

  until grading is done before turning on the TV?

  a. Calculate the approximate probability that between

  50. Let X denote the courtship time for a randomly selected

  35 and 70 tickets are given out on a particular day.

  female–male pair of mating scorpion flies (time from the

  b. Calculate the approximate probability that the total

  beginning of interaction until mating). Suppose the mean

  number of tickets given out during a 5-day week is

  value of X is 120 min and the standard deviation of X is

  between 225 and 275.

  110 min (suggested by data in the article “Should I Stay

  c. Use software to obtain the exact probabilities in (a)

  or Should I Go? Condition- and Status-Dependent

  and (b) and compare to their approximations.

  Courtship Decisions in the Scorpion Fly Panorpa

  56. A binary communication channel transmits a sequence

  Cognate” (Animal Behavior, 2009: 491–497) ).

  of “bits” (0s and 1s). Suppose that for any particular bit

  238 ChapteR 5 Joint probability Distributions and Random Samples

  transmitted, there is a 10 chance of a transmission error

  in the first transmission is within 50 of the number of

  (a 0 becoming a 1 or a 1 becoming a 0). Assume that bit

  errors in the second?

  errors occur independently of one another.

  57. Suppose the distribution of the time X (in hours) spent by

  a. Consider transmitting 1000 bits. What is the approx-

  students at a certain university on a particular project is

  imate probability that at most 125 transmission

  gamma with parameters a 5 50 and b 5 2. Because a is

  errors occur?

  large, it can be shown that X has approximately a normal

  b. Suppose the same 1000-bit message is sent two

  distribution. Use this fact to compute the approximate

  different times independently of one another. What is

  probability that a randomly selected student spends at

  the approximate probability that the number of errors

  most 125 hours on the project.