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Section 6.2

Section 6.2. Suppose X 1 , X 2 , …, X n are observations in a random sample from a N ( ,  2 ) distribution. Then,. Suppose X 1 , X 2 , …, X n are observations in a random sample from a N ( ,  2 ) distribution. Then,. X –  ——— has a distribution.  / n. X – 

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Section 6.2

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  1. Section 6.2 Suppose X1 , X2 , …, Xn are observations in a random sample from a N(, 2) distribution. Then,

  2. Suppose X1 , X2 , …, Xn are observations in a random sample from a N(, 2) distribution. Then, X–  ——— has a distribution.  /n X–  ——— has a distribution. S /n N(0, 1) If 2 is known, then a 100(1 – )% confidence interval for  can be derived as follows: X–  – z/2  ———  z/2 = 1 –   /n P P(– X – z/2  /n  –   –X + z/2  /n) = 1 –  P(X – z/2  /n    X + z/2  /n) = 1 –  Note: If the random sample does not come from a normal distribution, then for a sufficiently large sample size the Central Limit Theorem tells us we can still use this confidence interval.

  3. 1. (a) (b) (c) (d) Do Text Exercise 6.2-2. 85  2.576 72 / 8 The 99% confidence interval for  is 77.27 <  < 92.73 85  1.960 72 / 8 The 95% confidence interval for  is 79.12 <  < 90.88 85  1.645 72 / 8 The 90% confidence interval for  is 80.06 <  < 89.94 85  1.282 72 / 8 The 80% confidence interval for  is <  < 81.15 88.84

  4. Section 6.2 If Z and U are independent random variables with respective N(0, 1) and 2(r) distributions, then (from Theorem 5.5-3, Class Exercise 5.5-7, or Text Exercise 5.5-16) the random variable has a Z ———  U / r T = t distribution with r degrees of freedom (df), which we can call a t(r) distribution. The 100(1 – ) percentile (or the upper 100 percent point) of a t(r) distribution is denoted by t(r), that is, Table VI in Appendix B displays some percentiles for various t distributions. P[Tt(r)] =  . Suppose X1 , X2 , …, Xn are observations in a random sample from a N(, 2) distribution. Then,

  5. 2. (a) (b) (c) (d) Suppose X1 , X2 , … , Xn are a random sample from a N( , 2) distribution. Name the distribution for each of the following random variables: X has a distribution (from Corollary 5.5-1). N(, 2/n) (n– 1)S2 ———– 2 has a distribution (from Theorem 5.5-2). 2(n – 1) X–  ———  /n has a distribution (from Theorem 3.6-1 and Corollary 5.5-1). N(0, 1) X–  ———  /n has a distribution (from Theorem 5.5-2 and from Theorem 5.5-3, Class Exercise 5.5-7, or Text Exercise 5.5-16). t(n – 1) X–  ——— = S /n (n– 1)S2 ———– 2 (n– 1)

  6. Suppose X1 , X2 , …, Xn are observations in a random sample from a N(, 2) distribution. Then, X–  ——— has a distribution.  /n X–  ——— has a distribution. S /n N(0, 1) t(n– 1) If 2 is known, then a 100(1 – )% confidence interval for  can be derived as follows: X–  – z/2  ———  z/2 = 1 –   /n P P(– X – z/2  /n  –   –X + z/2  /n) = 1 –  P(X – z/2  /n    X + z/2  /n) = 1 –  Note: If the random sample does not come from a normal distribution, then for a sufficiently large sample size the Central Limit Theorem tells us we can still use this confidence interval.

  7. If 2 is not known, then a 100(1 – )% confidence interval for  can be derived as follows: X–  – t/2(n – 1)  ———  t/2(n – 1) = 1 –  S /n P P(– X – t/2(n – 1) S /n  –   –X + t/2(n – 1) S /n) = 1 –  P(X – t/2(n – 1) S /n    X + t/2(n – 1) S /n) = 1 –  Note: If the random sample does not come from a normal, distribution, then for a sufficiently large sample size a modified version of the Central Limit Theorem tells us we can still use this confidence interval (i.e., the confidence interval is robust). Without a sufficiently large sample size or with a very non-normal distribution, alternative confidence intervals (e.g., nonparametric confidence intervals in Section 6.10) are available. These confidence intervals are the first two listed in the summary displayed on the back page of the text. Note: One-sided confidence intervals are possible but rarely used.

  8. 3. (a) The mean and standard deviation for a random sample of 25 heart rates of teenagers are respectively found to be 73.438 beats per minute and 6.815 beats per minute. Find and interpret a 99% confidence interval for the mean heart rate of teenagers. 73.438  (2.797) (6.815) /25 The 99% confidence interval for  is 69.626 <  < 77.250 We are 99% confident that the mean heart rate of teenagers is between 69.626 and 77.250 beats per minute.

  9. (b) (c) How would each of a 90% confidence interval and a 95% confidence interval be different from the one found in part (a)? The 90% confidence interval and the 95% confidence interval will each have shorter length than the 99% confidence interval. How would the confidence interval in part (a) most likely have been different if it had been based on a simple random sample of 40 teenagers? The larger sample size would tend to decrease the length of the confidence interval.

  10. 4. (a) (1) (2) (3) (4) Create the Excel file Confidence_Intervals which displays the limits of a confidence interval for the mean, based on the Student’s t distribution. Create a worksheet named One Sample in an Excel file named Confidence_Intervals as follows: In Sheet1, select cells A1:A500, and color these cells with a light color such as yellow. Change the name of Sheet1 to One Sample. Color the cell F7 with a light color such as yellow. Enter the labels displayed in columns B through G, and right justify the labels in column E.

  11. (5) (6) Select cells A1:A500. From the main menu, select the Formulas tab, select the option Name Manager, and click the New button in the dialog box which appears. (7) Type the new range name Data, click the OK button to return to the Name Manager dialog box, and click the Close button to close the dialog box. (8) Select cells E1:F4, and in the Formulas tab select the option Create from Selection; then, with Left column checked in the dialog box that appears, click OK. (9) Enter the following formulas respectively in cells F1:F4: =COUNT(Data) =IF(COUNT(Data)>0,AVERAGE(Data),"-") =IF(COUNT(Data)>1,VAR(Data),"-") =IF(COUNT(Data)>1,STDEV(Data),"-") (10) Enter the following formulas respectively in cells F8:F10: =IF(AND(n>0,$F$7>0,$F$7<1),TINV(1-$F$7,n-1),"-") =IF(AND(n>0,$F$7>0,$F$7<1),Mean-$F$8*Standard_Deviation/SQRT(n),"-") =IF(AND(n>0,$F$7>0,$F$7<1),Mean+$F$8*Standard_Deviation/SQRT(n),"-") (11) Save the file as Confidence_Intervals (in your personal folder on the college network).

  12. (b) Use the Excel file Confidence_Intervals to obtain the confidence interval in Example 6.2-5 of the textbook. (Open Excel file Data_for_Students in order to copy the data of Example 6.2-5.)

  13. 5. (a) Let Y1 < Y2 < … < Yn be the order statistics for a random sample of size n taken from a U(0, ) distribution. Find the p.d.f. for Yn . Yn is the largest order statistic, and its p.d.f. is gn(y) = n [F(y)]n–1f(y) for 0 < y <  where f(x) and F(x) are respectively the p.d.f and d.f. for the distribution from the random sample is taken. 0 if x 0 1 — for 0 < x <   f(x) = F(x) = x /  if 0 < x  1 if  < x nyn–1 —— for 0 < y <  n gn(y) =

  14. (b) Find the value of c for which we have P(c < Yn < ) = 1 –  .   nyn–1 ——dy = n cn 1 – — n yn — = n P(cYn) = c y = c P(cYn) = 1 –   c = 1/n

  15. 5.-continued (c) Derive a 100(1 – )% confidence interval for . P(1/nYn) = 1 –  Yn P 1/n— 1 = 1 –    P 1 —–1/n = 1 –  Yn P (Yn–1/nYn) = 1 –  A 100(1 – )% confidence interval for  is given by (yn , –1/nyn) .

  16. (d) Suppose a random sample of size 10 is taken from a U(0, ) distribution, and the maximum value in the sample is observed to be 12.3. Find a 95% confidence interval for . (yn , –1/nyn)  (12.3 , 0.05–1/1012.3)  We are 95% confident that  is between 12.3 and 16.6.

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