Section 6.4 Suppose X 1, X 2, …, X n are observations in a random sample from a N( ,  2 ) distribution. Then, (n – 1)S 2 ———— has a distribution.  2  2 (n – 1) A 100(1 –  )% confidence interval for  can be derived as follows: (n – 1)S 2  2 1–  /2 (n–1)  ————   2  /2 (n–1) = 1 –   2 P (n – 1)S 2 (n – 1)S 2 ————   2  ————— = 1 –   2  /2 (n–1)  2 1–  /2 (n–1) P (n – 1)S 2 (n – 1)S 2 ————    ————— = 1 –   2  /2 (n–1)  2 1–  /2 (n–1) P 1/2

If U and V are independent random variables with respective  2 (r 1 ) and  2 (r 2 ) distributions, then (from Class Exercise 5.2-7, Text Example 5.2-4, or Text Exercise 5.2-2) the random variable has an F = U / r 1 —— V / r 2 The 100(1 –  ) percentile (or the upper 100  percent point) of an f(r 1, r 2 ) distribution is denoted by f  (r 1, r 2 ), that is, Note: If the random sample does not come from a normal distribution, this confidence interval may not be appropriate, because this confidence interval is sensitive to non-normality. In other words, unlike confidence intervals concerning means, this confidence interval is not robust.

Using the data from Text Exercise 6.2-6, we have (36) (36) ————    —————  (36)  (36) 1/2 (36) (36) ————    ————— /2 The 95% confidence interval for  is <  < We are 95% confident that the standard deviation of the number of colonies in 100 milliliters of water in the west basin is between and Find a 95% confidence interval for  with the data of Text Exercise Since this is not in the textbook table, we interpolate [(36–30)/(40–30)](59.34–46.98) [(36–30)/(40–30)](24.43–16.79)

If U and V are independent random variables with respective  2 (r 1 ) and  2 (r 2 ) distributions, then (from Class Exercise 5.2-7, Text Example 5.2-4, or Text Exercise 5.2-2) the random variable has an F = U / r 1 —— V / r 2 The 100(1 –  ) percentile (or the upper 100  percent point) of an f(r 1, r 2 ) distribution is denoted by f  (r 1, r 2 ), that is, f distribution with r 1 numerator degrees of freedom and r 2 denominator degrees of freedom, which we can call an f(r 1, r 2 ) distribution. P[F  f  (r 1, r 2 )] = . Note: If the random sample does not come from a normal distribution, this confidence interval may not be appropriate, because this confidence interval is sensitive to non-normality. In other words, unlike confidence intervals concerning means, this confidence interval is not robust.

Table VII in Appendix B displays some percentiles for various f distributions. It is useful to observe that 1/F has an f(r 2, r 1 ) distribution. Consequently, P[F  f  (r 1, r 2 )] =   P[1/F  1/f  (r 1, r 2 )] =   1 – P[1/F  1/f  (r 1, r 2 )] =   P[1/F  1/f  (r 1, r 2 )] = 1 –   f 1–  (r 2, r 1 ) = 1/f  (r 1, r 2 ). Suppose that X 1, X 2, …, X n are observations in a random sample from a N(  X,  X 2 ) distribution, that Y 1, Y 2, …, Y m are observations in a random sample from a N(  Y,  Y 2 ) distribution, and that the two random samples are observed independently of one another. Then, (m–1)S Y 2 ————  Y 2 (m–1)  X 2 S Y 2 ————————— = —— ——has an distribution. (n–1)S X 2  Y 2 S X 2 ————  X 2 (n–1) A 100(1 –  )% confidence interval for  X /  Y can be derived as follows: f(m – 1, n – 1)

Suppose the random variable F has an f distribution with r 1 numerator degrees of freedom and r 2 denominator degrees of freedom. 2. (a) (b) (c) (d) (e) (f) If r 1 = 5 and r 2 = 10, then P(F < 5.64) =0.99 If r 1 = 5 and r 2 = 10, then P(F > 4.24) =0.025 f 0.05 (4, 8) =3.84 f 0.95 (4, 8) =1/f 0.05 (8, 4) = f 0.01 (8, 4) =14.80 f 0.99 (8, 4) =1/f 0.01 (4, 8) = 1/6.04 = /7.01 = 0.143

Table VII in Appendix B displays some percentiles for various f distributions. It is useful to observe that 1/F has an f(r 2, r 1 ) distribution. Consequently, P[F  f  (r 1, r 2 )] =   P[1/F  1/f  (r 1, r 2 )] =   1 – P[1/F  1/f  (r 1, r 2 )] =   P[1/F  1/f  (r 1, r 2 )] = 1 –   f 1–  (r 2, r 1 ) = 1/f  (r 1, r 2 ). Suppose that X 1, X 2, …, X n are observations in a random sample from a N(  X,  X 2 ) distribution, that Y 1, Y 2, …, Y m are observations in a random sample from a N(  Y,  Y 2 ) distribution, and that the two random samples are observed independently of one another. Then, (m–1)S Y 2 ————  Y 2 (m–1)  X 2 S Y 2 ————————— = —— ——has an distribution. (n–1)S X 2  Y 2 S X 2 ————  X 2 (n–1) A 100(1 –  )% confidence interval for  X /  Y can be derived as follows: f(m – 1, n – 1)

 X 2 S Y 2 f 1–  /2 (m – 1, n – 1)  —— ——  f  /2 (m – 1, n – 1) = 1 –   Y 2 S X 2 P 1  X 2 S Y 2 ———————  —— ——  f  /2 (m – 1, n – 1) = 1 –  f  /2 (n – 1, m – 1)  Y 2 S X 2 P 1 S X 2  X 2 S X 2 ——————— ——  ——  f  /2 (m – 1, n – 1) —— = 1 –  f  /2 (n – 1, m – 1) S Y 2  Y 2 S Y 2 P 1 S X  X S X ——————— —  ——  f  /2 (m – 1, n – 1) — = 1 –  f  /2 (n – 1, m – 1) S Y  Y S Y P 1/2 Note: If the random samples do not come from a normal distribution, this confidence interval may not be appropriate, because this confidence interval, like the confidence interval for , is sensitive to non-normality and not robust.

Do Text Exercise  X ————— ———  ——  f 0.05 (12, 15) ——— f 0.05 (15, 12)0.318  Y /  X —— ———  ——  2.48 ———  Y /2 The 90% confidence interval for  X /  Y is <  X /  Y < We are 90% confident that the ratio of standard deviations  X /  Y of mint weights is between and This confidence interval indicates that there a statistically significant difference in standard deviation, is with a larger standard deviation for the afternoon shift.

P(– < T < 2.262) =0.975 – 0.10 = P(F 1 > 3.89) =0.01 P(F 2 > ) =P(1/ F 2 < 2.59) =0.95 The random sample X 1, X 2, …, X 10 is taken from a N(20, 100) distribution, and the random sample Y 1, Y 2, …, Y 16 is taken from a N(8, 100) distribution. The following random variables are defined: T =F 1 =F 2 = Find each of the following: 4. (a) (b) (c) X – 20 ——— S X /  10 SX2—SY2SX2—SY2 SY2—SX2SY2—SX2

Modify the worksheets in the Excel file Confidence_Intervals (created previously) so that the limits of confidence intervals for the variance and standard deviation are displayed, and the limits of confidence intervals for the ratio of two variances and ratio of two standard deviations are displayed. 5. (a) (1) Modify the worksheet named One Sample in the Excel file named Confidence_Intervals as follows: Enter the additional labels displayed in columns B through H, right justify the labels in column E, and underline the labels in cells F15 and H15.

Enter the following formulas respectively in cells F14 and G14: =IF(AND(n>0,$F$13>0,$F$13<1),CHIINV((1-$F$13)/2,n-1),"-") =IF(AND(n>0,$F$13>0,$F$13<1),CHIINV((1+$F$13)/2,n-1),"-") Copy the formula in H16 to H17. Enter the following formula in cell H16: =IF(AND(n>0,$F$13>0,$F$13<1),SQRT(F16),"-") (2) Enter the following formulas respectively in cells F16 and F17: =IF(AND(n>0,$F$13>0,$F$13<1),(n-1)*Variance/$F$14,"-") =IF(AND(n>0,$F$13>0,$F$13<1),(n-1)*Variance/$G$14,"-") (3) (4) (5) (6) (b)Use the Excel file Confidence_Intervals to obtain the confidence interval in Text Exercise Save the file as Confidence_Intervals (in your personal folder on the college network).

Modify the worksheet named Two Sample in the Excel file named Confidence_Intervals as follows: 5.-continued (c) (1) Enter the additional labels displayed in columns B through H, right justify the labels in column E. Enter the following formulas respectively in cells F16 and G16: =IF(AND(n_1>0,n_2>0,$F$15>0,$F$15<1),FINV((1-$F$15)/2,n_2-1,n_1-1),"-") =IF(AND(n_1>0,n_2>0,$F$15>0,$F$15<1),FINV((1+$F$15)/2,n_2-1,n_1-1),"-") (2)

Enter the following formulas respectively in cells F18 and F19: =IF(AND(n_1>0,n_2>0,$F$15>0,$F$15<1),$G$16*Variance_1/Variance_2,"-") =IF(AND(n_1>0,n_2>0,$F$15>0,$F$15<1),$F$16*Variance_1/Variance_2,"-") (3) (d)Use the Excel file Confidence_Intervals to obtain the two-sided confidence interval with the 95% confidence level in Text Exercise , instead of the one-sided confidence interval mentioned in part (b). (Open Excel file Data_for_Students in order to copy the data of Exercise ) Save the file as Confidence_Intervals (in your personal folder on the college network). Copy the formula in H18 to H19. Enter the following formula in cell H18: =IF(AND(n>0,$F$15>0,$F$15<1),SQRT(F18),"-") (4) (5) (6)