4.4 Inferences About Variances

Section 4. Inferences About Variances (ATTENDANCE 7) 127

2 2 σ1 We look at hypothesis tests and conﬁdence intervals for σ and 2 . From a “big σ2 picture” point of view, we look at the (large n) one–sample and two–sample variance (standard deviation) problems.

mean variance proportion µ σ2 π one large n, 3.7, 3.8, 3.9, 3.10, 4.6 4.4 6.2 small n, 4.3, 4.6 sample two large n, 3.11 4.4 6.3 small n, 4.3 multiple chapters not 6.2, 6.3 7, 8, 9 done

Exercise 4.6 (Test and Confidence Interval for Variance (or Standard De- viation), σ) We first look at examples of the hypothesis test and confidence interval when S2 is calculated from a random sample of size n from a normal population with variance σ2. The test statistic is

2 2 (n − 1)S χ = 2 σ0

2 with critical value χc where,

H0 H1 reject H0 at level α if 2 2 2 2 2 2 σ ≤ σ0 σ > σ0 χc > χ (n − 1, α) 2 2 2 2 2 2 σ ≥ σ0 σ < σ0 χc < χ (n − 1, 1 − α) 2 2 2 2 2 2 2 2 σ = σ0 σ =6 σ0 χc > χ (n − 1, α/2) or χc < χ (n − 1, 1 − α/2) The conﬁdence intervals are:

(n 1)S2 (n 1)S2 − − • (two–sided) confidence interval, CI: χ2(n 1,α/2) , χ2(n 1,1 α/2) − − − ³ ´ (n 1)S2 − • lower confidence interval, LCI: χ2(n 1,α) , ∞ − ³ ´ (n 1)S2 − • upper confidence interval, UCI: −∞, χ2(n 1,1 α) − − ³ ´ See Lab 7: Test For Variance.

1. Hypothesis Test: Carbon Content. In a random sample of 28 soil samples, the variance in carbon content was found to be s2 = 0.7 parts per million (ppm). Test if the variance is greater than 0.4 at α = 0.05. 128Chapter 4. Inferences About One Or Two Populations: Interval Data (ATTENDANCE 7)

(a) Test Statistic Versus Critical Value, Standardized. i. Statement. The statement of the test is (circle one) 2 2 A. H0 : σ = 0.4 versus H1 : σ > 0.4 2 2 B. H0 : σ = 0.4 versus H1 : σ < 0.4 2 2 C. H0 : σ = 0.4 versus H1 : σ =6 0.4 ii. Test. The standardized test statistic of s = 0.7 is (n − 1)s2 (28 − 1)(0.7) χ2 test statistic = = = σ2 0.4

(circle one) 30.2 / 47.3 / 82.7. The standardized critical value at 0.05 is χ2(n − 1, α) = χ2(28 − 1, 0.05) = (circle one) 39.2 / 40.1 / 43.2 (Use PRGM INVCHI2 ENTER 27 ENTER 0.95 ENTER) iii. Conclusion. Since the test statistic, 47.3, is larger than the critical value, 40.1, we (circle one) accept / reject the null hypothesis that σ = 0.4. (b) P–Value Versus Level of Signiﬁcance, Standardized. i. Statement. The statement of the test is (circle one) 2 2 A. H0 : σ = 0.4 versus H1 : σ > 0.4 2 2 B. H0 : σ = 0.4 versus H1 : σ < 0.4 2 2 C. H0 : σ = 0.4 versus H1 : σ =6 0.4 ii. Test. Since the standardized test statistic is χ2 = 47.3, the p–value, at n − 1 = 28 − 1 = 27 df, is given by

p–value = P (χ2 ≥ 47.3)

which equals (circle one) 0.01 / 0.05 / 0.10. (Use 2nd DISTR 2:χ2cdf(47.3,E99,27).) The level of signiﬁcance is 0.05. iii. Conclusion. Since the p–value, 0.01, is smaller than the level of significance, 0.05, we (circle one) accept / reject the null hypothesis that σ = 0.4. (c) Related Questions. i. Since the χ2 test statistic is (circle one) A. larger than the critical value, χ2(27, 0.05), we will accept that the variance in carbon content exceeds 0.4 mm. B. larger than the level of signiﬁcance, α = 0.05, we will accept that the variance in carbon content exceeds 0.4 mm. Section 4. Inferences About Variances (ATTENDANCE 7) 129

C. smaller than the critical value, χ2(27, 0.05), we will accept that the variance in carbon content exceeds 0.4 mm. D. larger than the p–value we will accept that the variance in carbon content exceeds 0.4 mm. ii. Since the p–value is (circle one) A. larger than the critical value, χ2(27, 0.05), we will accept that the variance in carbon content exceeds 0.4 mm. B. smaller than the level of signiﬁcance, α = 0.05, we will accept that the variance in carbon content exceeds 0.4 mm. C. smaller than the critical value, χ2(27, 0.05), we will accept that the variance in carbon content exceeds 0.4 mm. D. larger than the test statistic we will reject that the variance in carbon content is equal to 0.4 mm. iii. Population, Parameter, Sample and Statistic. Match the statistical items with the appropriate parts of this soil example.

terms carbon example (i) population (i) variance in carbon content, of 28 samples, s2 (ii) sample (ii) variance in carbon content, of all samples, σ2 (iii) statistic (iii) carbon content, of all samples (iv) parameter (iv) carbon content, of 28 samples terms (i) (ii) (iii) (iv) carbon example 2. Hypothesis Test: Fish Lengths. In a random sample of 18 ﬁsh lengths, the standard deviation in lengths was found to be s = 12 cm. Test if the standard deviation is less than 13 cm at α = 0.05. (a) Test Statistic Versus Critical Value, Standardized. i. Statement. The statement of the test is (circle one)

A. H0 : σ = 13 versus H1 : σ > 13

B. H0 : σ = 13 versus H1 : σ < 13

C. H0 : σ = 13 versus H1 : σ =6 13 ii. Test. The standardized test statistic of s = 12 is (n − 1)s2 (18 − 1)(12)2 χ2 test statistic = = = σ2 (13)2

(circle one) 14.5 / 60.1 / 82.7. The standardized critical value at α = 0.05 is χ2(n − 1, 1 − α) = χ2(18 − 1, 0.95) = (circle one) 8.7 / 40.1 / 43.2 (Use PRGM INVCHI2 ENTER 17 ENTER 0.05 ENTER) 130Chapter 4. Inferences About One Or Two Populations: Interval Data (ATTENDANCE 7)

iii. Conclusion. Since the test statistic, 14.5, is larger than the critical value, 8.7, we (circle one) accept / reject the null hypothesis that σ = 13. (b) P–Value Versus Level of Signiﬁcance, Standardized. i. Statement. The statement of the test is (circle one)

A. H0 : σ = 13 versus H1 : σ > 13

B. H0 : σ = 13 versus H1 : σ < 13

C. H0 : σ = 13 versus H1 : σ =6 13 ii. Test. Since the standardized test statistic is χ2 = 14.5, the p–value, at n − 1 = 18 − 1 = 17 df, is given by

p–value = P (χ2 ≤ 14.5)

which equals (circle one) 0.00 / 0.05 / 0.37. (Use 2nd DISTR 2:χ2cdf(−E99,14.5,17).) The level of significance is 0.05. iii. Conclusion. Since the p–value, 0.37, is larger than the level of significance, 0.05, we (circle one) accept / reject the null hypothesis that σ = 13. 3. Confidence Interval: Car Door and Jamb. In a random sample of 28 cars, the variance in the distance between door and jamb was found to be s2 = 0.7. Let σ2 = 0.4. Calculate a 95% CI using the formula, (n − 1)S2 (n − 1)S2 , Ãχ2(n − 1, α/2) χ2(n − 1, 1 − α/2)! (a) The critical value χ2(n − 1, α/2) = χ2(28 − 1, 0.05/2) = (circle one) 8.7 / 40.1 / 43.2 (Use PRGM INVCHI2 ENTER 27 ENTER 0.975 ENTER) (b) The critical value χ2(n − 1, 1 − α/2) = χ2(28 − 1, 1 − 0.05/2) = is (circle one) 14.6 / 40.1 / 43.2 (Use PRGM INVCHI2 ENTER 27 ENTER 0.025 ENTER) (c) So, the CI is (n 1)s2 (n 1)s2 (28 1)0.7 (28 1)0.7 − − − − χ2(n 1,α/2) , χ2(n 1,1 α/2) = 43.2 , 14.6 = − − − (circle³ one) (0.6, 1.6) / (0´ .5,³1.2) / (0.44, 1.29)´ . (d) Since the 95% CI (0.44, 1.29) excludes 0.4, this indicates that the variance in the distance between door and jamb (circle one) is / is not 0.4 mm. 4. Confidence Interval: Machine Parts. In a random sample of 18 machine parts, the variance in lengths was found to be s2 = 144. Let σ = 92. Calculate a 90% CI. Section 4. Inferences About Variances (ATTENDANCE 7) 131

(a) The critical value χ2(n − 1, α/2) = χ2(28 − 1, 0.10/2) = (circle one) 8.7 / 27.6 / 43.2 (Use PRGM INVCHI2 ENTER 17 ENTER 0.95 ENTER) (b) The critical value χ2(n − 1, 1 − α/2) = χ2(28 − 1, 1 − 0.10/2) = is (circle one) 8.7 / 40.1 / 43.2 (Use PRGM INVCHI2 ENTER 17 ENTER 0.05 ENTER) (c) So, the CI is (n 1)s2 (n 1)s2 (18 1)122 (18 1)122 − − − − χ2(n 1,α/2) , χ2(n 1,1 α/2) = 27.6 , 8.7 = − − − (circle³ one) (70, 100) / (84´ .6³, 291.4) / (88.6, ´281.4). (d) Since the 95% CI (88.6, 281.4) includes 92, this indicates that the variance in lengths (circle one) is / is not 92 mm.

Exercise 4.7 (Hypothesis Test and Conﬁdence Interval of σ1 ) We now look at σ2 2 2 examples of the hypothesis test and conﬁdence interval when S1 and S2 are calculated from two independent random samples of size n1 and n2 from two normal populations 2 2 with variances σ1 and σ2. The test statistic is

2 S1 F = 2 S2 with critical value Fc where,

H0 H1 reject H0 at level α if 2 2 σ1 σ1 2 ≤ δ0 2 > δ0 Fc > δ0F (n1 − 1, n2 − 1, α) σ2 σ2 2 2 σ1 σ1 2 ≥ δ0 2 < δ0 Fc < δ0F (n1 − 1, n2 − 1, 1 − α) σ2 σ2 2 2 σ1 σ1 2 = δ0 2 =6 δ0 Fc > δ0F (n1 − 1, n2 − 1, α/2) or Fc < δ0F (n1 − 1, n2 − 1, 1 − α/2) σ2 σ2 where δ0 is a nonnegative number (usually δ0 = 1). The conﬁdence intervals are:

2 2 S1 1 S1 1 • (two–sided) confidence interval, CI: 2 , 2 S F (n1 1,n2 1,α/2) S F (n1 1,n2 1,1 α/2) 2 − − 2 − − − ³ ´ 2 S1 1 • lower confidence interval, LCI: 2 , ∞ S F (n1 1,n2 1,α) 2 − − ³ ´ 2 S1 1 • upper confidence interval, UCI: −∞, 2 S F (n1 1,n2 1,1 α) 2 − − − ³ ´ See Lab 7: Tests and CIs For σ1 . σ2

1. Hypothesis Test: Plasma Levels Again. The average and standard deviation of the plasma levels for males and females are given below. 132Chapter 4. Inferences About One Or Two Populations: Interval Data (ATTENDANCE 7)

males (1) females (2) y¯ 3.259 1.413 s2 0.16 0.13 n 9 6

2 2 Test if σ1 > σ2 at α = 0.05. (a) Test Statistic Versus Critical Value. i. Statement. The statement of the test is (circle one) 2 2 2 2 A. H0 : σ1 = σ2 versus H1 : σ1 > σ2 2 2 2 2 B. H0 : σ1 = σ2 versus H1 : σ1 < σ2 2 2 2 2 C. H0 : σ1 = σ2 versus H1 : σ1 =6 σ2 2 2 where, notice, if σ1 = σ2, then 2 σ1 2 = δ0 = (circle one) −1 / 0 / 1 σ2 2 2 and if σ1 > σ2, then 2 σ1 2 > δ0 = (circle one) −1 / 0 / 1. σ2 ii. Test. The test statistic is 2 s1 0.16 F test statistic = 2 = = s2 0.13

(circle one) 1.03 / 1.23 / 2.27. The critical value at α = 0.05, with n1 − 1 = 9 − 1 = 8 and n2 − 1 = 6 − 1 = 5 degrees of freedom, is F (n1 − 1, n2 − 1, α) = F (8, 5, 0.05) = (circle one) 3.22 / 4.15 / 4.82 (Use PRGM INVF ENTER 8 ENTER 5 ENTER 0.95 ENTER) iii. Conclusion. Since the test statistic, 1.23, is less than the critical value, 2 2 4.82, we (circle one) accept / reject the null hypothesis that σ1 = σ2. (b) P–Value Versus Level of Signiﬁcance. i. Statement. The statement of the test is (circle one) 2 2 2 2 A. H0 : σ1 = σ2 versus H1 : σ1 > σ2 2 2 2 2 B. H0 : σ1 = σ2 versus H1 : σ1 < σ2 2 2 2 2 C. H0 : σ1 = σ2 versus H1 : σ1 =6 σ2 ii. Test. Since the test statistic is F = 1.23, with n1 − 1 = 9 − 1 = 8 and n2 − 1 = 6 − 1 = 5 degrees of freedom, the p–value is given by p–value = P (F ≥ 1.23)

which equals (circle one) 0.14 / 0.35 / 0.43. (Use 2nd DISTR 9:F cdf(1.23,E99,8,5).) The level of signiﬁcance is 0.05. Section 4. Inferences About Variances (ATTENDANCE 7) 133

iii. Conclusion. Since the p–value, 0.43, is larger than the level of signiﬁ- cance, 0.05, we (circle one) accept / reject the null hypothesis that 2 2 σ1 = σ2. (c) Related Questions. Match the statistical terms with the appropriate parts of the plasma example. terms plasma example (i) population (i) nine male, six female plasma levels 2 s1 (ii) sample (ii) F = 2 s2 (iii) statistic (iii) all 17–year–olds plasma levels 2 σ1 (iv) parameter (iv) F = 2 σ2 terms (i) (ii) (iii) (iv) plasma example 2. Hypothesis Test: Plasma Levels Again. The average and standard deviation of the plasma levels for males and females is given below. males (1) females (2) y¯ 2.159 1.713 s2 0.16 0.15 n 9 13

2 2 Test if σ1 =6 σ2 at α = 0.05. (a) Test Statistic Versus Critical Value. i. Statement. The statement of the test is (circle one) 2 2 2 2 A. H0 : σ1 = σ2 versus H1 : σ1 > σ2 2 2 2 2 B. H0 : σ1 = σ2 versus H1 : σ1 < σ2 2 2 2 2 C. H0 : σ1 = σ2 versus H1 : σ1 =6 σ2 ii. Test. The test statistic is 2 s1 0.16 F test statistic = 2 = = s2 0.15

(circle one) 1.07 / 1.23 / 2.27. 0.5 Since this is two–sided, the critical value at α = 2 = 0.025, with n1 − 1 = 9 − 1 = 8 and n2 − 1 = 13 − 1 = 12 degrees of freedom, is F (n1 − 1, n2 − 1, α/2) = F (8, 12, 0.025) = (circle one) 3.51 / 4.15 / 4.82 (Use PRGM INVF ENTER 8 ENTER 12 ENTER 0.975 ENTER) and F (n1 − 1, n2 − 1, 1 − α/2) = F (8, 12, 0.975) = (circle one) 0.148 / 0.238 / 2.825 (Use PRGM INVF ENTER 8 ENTER 12 ENTER 0.025 ENTER) 134Chapter 4. Inferences About One Or Two Populations: Interval Data (ATTENDANCE 7)

iii. Conclusion. Since the test statistic, 1.07, is in between the two critical values, (0.238, 3.51), we (circle one) accept / reject the null 2 2 hypothesis that σ1 = σ2. (b) P–Value Versus Level of Significance. i. Statement. The statement of the test is (circle one) 2 2 2 2 A. H0 : σ1 = σ2 versus H1 : σ1 > σ2 2 2 2 2 B. H0 : σ1 = σ2 versus H1 : σ1 < σ2 2 2 2 2 C. H0 : σ1 = σ2 versus H1 : σ1 =6 σ2 ii. Test. Since the test statistic is F = 1.07, the p–value is given by p–value = 2 × P r(F ≥ 1.07) which equals (circle one) 0.14 / 0.44 / 0.88. (Use 2nd DISTR 9:F cdf(1.07,E99,8,12).) The level of significance is 0.05. iii. Conclusion. Since the p–value, 0.88, is larger than the level of significance, 0.05, we (circle one) accept / reject the null hypothesis that 2 2 σ1 = σ2.

4.5 Statistical Models

Not covered.

4.6 Predicting Future Observations

Exercise 4.8 (Prediction Interval For The Mean of m Future Observations, Normal) Three prediction intervals for the average of m future unknown observations are: • (two–sided) prediction interval, PI: 1 1 2 1 1 2 Y − t(n − 1, α/2) m + n σ , Y + t(n − 1, α/2) m + n σ µ r³ ´ r³ ´ ¶ 1 1 2 • lower prediction bound, LPB: Y − t(n − 1, α) m + n σ , ∞ µ r³ ´ ¶ 1 1 2 • upper prediction bound, UPB: −∞, Y + t(n − 1, α) m + n σ µ r³ ´ ¶ If n is large (n > 30, say), and the prediction is about one individual, the t critical values can be replaced by z critical values and the resulting prediction intervals are said to be normal limits. From a “big picture” point of view, we look at more of the details of the large n and small n one–sample mean problems. Section 6. Predicting Future Observations (ATTENDANCE 7) 135

mean variance proportion µ σ2 π one large n, 3.7, 3.8, 3.9, 3.10, 4.6 4.4 6.2 small n, 4.3, 4.6 sample two large n, 3.11 4.4 6.3 small n, 4.3 multiple chapters not 6.2, 6.3 7, 8, 9 done 1. Average Touch–Sensitivity. In a random sample of 32 totally blind people taken from a normally distributed population, the average touch–sensitivity is found to be y¯ = 0.013 with a standard deviation of σ = 0.003. (a) The 95% lower prediction bound, LPB, for the mean of m = 10 future observations of touch–sensitivity is given by 1 1 2 1 1 2 y¯ − t(n − 1, α) m + n σ , ∞ ≈ 0.013 − z(0.05) 10 + 32 0.003 , ∞ = µ r ¶ µ r ¶ (circle one) (0.009³, ∞) /´(0.010, ∞) / (0.011, ∞). ³ ´ (Notice that t(n − 1, α) is replaced by z(0.05) since n = 32 > 30 and, also, for z(0.05), type 2nd DISTR 3:invNorm(0.95) ENTER.) (b) The 93% UPB, for the mean of m = 15 future observations of touch– sensitivity is given by 1 1 2 1 1 2 −∞, y¯ + z(α) m + n σ = −∞, 0.013 + z(0.07) 15 + 32 0.003 = µ r ¶ µ r ¶ (circle one) (−∞³, 0.012)´ / (−∞, 0.013) / (−∞, 0.014)³ . ´ (2nd ENTRY will retrieve the last typed calculator command.) (c) The 98% PI, for the mean of m = 12 future observation of touch–sensitivity is given by 1 1 2 1 1 2 y¯ § z(α/2) m + n σ = 0.013 § z(0.01) 12 + 32 0.003 = r r (circle one) (0³ .009, ´0.013) / (0.010, 0.014)³ / (0.011´ , 0.015). (For z(0.02/2) = z(0.01), type 2nd DISTR 3:invNorm(0.99) ENTER.) (d) (Review.) The 98% confidence interval (not a prediction interval and for no future observations) of touch–sensitivity is given by 1 2 1 2 y¯ + z(α/2) n σ , y¯ − z(α/2) n σ = µ r³ ´ r³ ´ ¶ 1 2 1 2 0.013 + z(0.01) 32 0.003 , 0.013 + z(0.01) 32 0.003 = µ r r ¶ (circle one) (0.009³, 0.´012) / (0.010, 0.013) / ³(0.´012, 0.014). (Type STAT TESTS ZInterval... Stats 0.013 0.003 32 0.98 Calculate.) (e) True / False Prediction Intervals are always wider (larger, broader) than confidence intervals because they estimate both the population average as well as the uncertainty of future unknown observations, whereas confidence intervals estimate just the population average. 136Chapter 4. Inferences About One Or Two Populations: Interval Data (ATTENDANCE 7)

2. Average Ph Levels in Soil. In a random sample of 28 soil samples taken from a normally distributed population, the average Ph level is found to be y¯ = 10.55 with a standard deviation of σ = 3.01. The 95% prediction interval, PI, for m = 5 future observations of the Ph levels is given by 1 1 2 1 1 2 y¯ − t(n − 1, α/2) m + n σ , y¯ + t(n − 1, α/2) m + n σ = µ r³ ´ r³ ´ ¶ 1 1 2 1 1 2 = 10.55 − t(27, 0.05/2) 5 + 28 3.01 , 10.55 + t(27, 0.05/2) 5 + 28 3.01 = µ r r ¶ (circle one) (7.55, 13.54) ³/ (8.146´ , 13.954) / (8.146, 14.954)³. ´ (For t(27, 0.025), type PRGM INVT 27 ENTER 0.975 ENTER.) 3. Log–Normal Distribution: Soil–Water Fluxes. Y is log–normal if ln Y is normal. Consider the following sample of ten soil–water fluxes. 0.306, 0.363, 0.437, 0.787, 0.899, 1.272, 1.424, 1.634, 1.682, 5.128 Assume the soil–water flux has a log–normal distribution; in other words, the natural logarithm (ln) of the soil–water flux has a normal distribution. Calculate a 95% PI for this data. (a) True / False Since the soil–water flux is nonnormal (in fact, it is log– normal) we are not able to calculate a 95% PI for the data as given, since one of the assumptions of calculating a PI is that the data follows from a normal distribution. (b) Use your calculator and transform the data by taking the ln of these ten measurements (and so transform the data into that which follows from a normal distribution). This gives (circle one) i. 0.306, 0.363, 0.437, 0.787, 0.899, 1.272, 1.424, 1.634, 1.682, 5.128 ii. -1.185, -1.103, -0.829, -0.239, -0.106, 0.241, 0.354, 0.491, 0.520, 1.635

(Type STAT EDIT, then type ten fluxes into L1 and define L2 as ln(L1).) (c) The average (mean) of the ten ln–fluxes is y¯ = −0.013 and the standard deviation is s = (circle one) 0.853 / 0.925 / 1.253. (d) The 95% prediction interval, PI, for the mean of m = 5 future observations of the ln–fluxes is given by 1 1 2 1 1 2 y¯ − t(n − 1, α/2) m + n σ , y¯ + t(n − 1, α/2) m + n σ = µ r³ ´ r³ ´ ¶ 1 1 2 1 1 2 = −0.013 − t(9, 0.05/2) 5 + 10 0.853 , −0.013 + t(9, 0.05/2) 5 + 10 0.853 = µ r r ¶ (circle one) (−1.06, 1.04)³/ (−0.´681, 0.855) / (−0.581, 0.955).³ ´ (e) The 95% prediction interval, PI, for m = 5 future observations of the original soil–water fluxes (not the ln–fluxes!) is obtained by introducing e· 1.06 1.04 in the following way (e− , e ) = (circle one) (0.343, 2.840) / (0.458, 2.128) / (0.558, 2.228). Section 7. The Assumptions for t, χ2 and F Test Procedures (ATTENDANCE 7)137

4.7 The Assumptions for t, χ2 and F Test Proce- dures

Exercise 4.9 (Violating the Assumptions of the Statistical Model.) All statistical models thus far involve one or more of the following three assumptions (sometimes called the analysis of variance (ANOVA) assumptions):

1. The populations have normal distributions.

2. The population variances are equal but unknown.

3. The data are independent random samples from the populations.

1. Normality. A distribution is nonnormal if it is skewed. A distribution is also nonnormal if it either heavy–tailed or light–tailed.

normal

heavy-tailed light-tailed

Figure 4.1 (Light–Tailed and Heavy–Tailed Distributions)

Use your calculator to draw a box–plot of the 28 Ph levels of soil data:

4.3 5 5.9 6.5 7.6 7.7 7.7 8.2 8.3 9.5 10.4 10.4 10.5 10.8 11.5 12 12 12.3 12.6 12.6 13 13.1 13.2 13.5 13.6 14.1 14.1 15.1

The box–plot indicates the data is (circle one) skewed right / symmetric / skewed left and (circle one) heavy–tailed / normal / light–tailed (where heavy–tailed distributions are indicated by extreme outliers).

2. Normality Again. Log–normal data is an example where a nonnormal distribution can be transformed to a normal distribution (and so allow statistical inference which requires normality to proceed). It is (circle one) always / sometimes possible to do this to nonnormal data.

3. Equal Variance. A violation of the assumption of equal variance most detri- mentally eﬀects calculation of tests and conﬁdence intervals for (circle one) 138Chapter 4. Inferences About One Or Two Populations: Interval Data (ATTENDANCE 7)

s (a) θ = µ, where σˆθˆ = √n and we assume normality and a small random sample 1 1 (b) θ = µ − µ , where σˆˆ = S + 1 2 θ p n1 n2 and we assume normality, smallq independent random samples 2 2 and unknown σ1 = σ2

2 2 S1 S2 (c) θ = µ − µ , where σˆˆ = + 1 2 θ n1 n2 and we assume normalityr, small independent random samples 2 2 and unknown σ1 =6 σ2 SD (d) θ = µD = µ1 − µ2, where σˆθˆ = √n and we assume normality and small dependent (paired) random samples

4. Independence. True / False Independence is violated if the sample has not been collected properly. A systematic dependence of error occurs in time series.