10: Inference from Small Samples

10: Inference from Small Samples 10.1 Refer to Table 4, Appendix I, indexing df along the left or right margin and tα across the top. a t.05 = 2.015 with 5 df b t.025 = 2.306 with 8 df c t.10 = 1.330 with 18 df c t.025 ≈ 1.96 with 30 df 10.2 The value Pt(>ta )=ais the tabled entry for a particular number of degrees of freedom. a For a two-tailed test with α = .01, the critical value for the rejection region cuts off α 2 = .005 in the two tails of the t distribution shown below, so that t.005 = 3.055 . The null hypothesis H0 will be rejected if t > 3.055 or t <−3.055 (which you can also write as t > 3.055 ). b For a right-tailed test, the critical value that separates the rejection and nonrejection regions for a right tailed test based on a t-statistic will be a value of t (called tα ) such that Pt( >=tα ) α =.05 and df = 16 . That is, t.05 = 1.746 . The null hypothesis H0 will be rejected if t > 1.746 . c For a two-tailed test with α 2 = .025 and df = 25 , H0 will be rejected if t > 2.060 . d For a left-tailed test withα = .01and df = 7 , H0 will be rejected if t < −2.998 . 10.3 a The p-value for a two-tailed test is defined as pP-value =>( t2.43) =2P(t>2.43) so that 1 Pt()>=2.43 p-value 2 Refer to Table 4, Appendix I, with df = 12 . The exact probability, Pt( > 2.43) is unavailable; however, it is evident that t = 2.43 falls between t.025 = 2.179 and t.01 = 2.681. Therefore, the area to the right of t = 2.43 must be between .01 and .025. Since 1 .01 <<p-value .025 2 the p-value can be approximated as .02 < p-value < .05 237 b For a right-tailed test, pP-value =>(t3.21) with df = 16 . Since the value t = 3.21 is larger than t.005 = 2.921 , the area to its right must be less than .005 and you can bound the p-value as p-value < .005 1 c For a two-tailed test, pP-value =>( t1.19) =2P(t>1.19) , so that Pt()>=1.19 p-value . From 2 Table 4 with df = 25 , t = 1.19 is smaller than t.10 = 1.316 so that 1 pp-value >>.10 and -value .20 2 d For a left-tailed test, pP-value =<(t−8.77) =P(t>8.77) with df = 7 . Since the value t = 8.77 is larger than t.005 = 3.499 , the area to its right must be less than .005 and you can bound the p-value as p-value < .005 10.4 a The stem and leaf plot is shown below. Notice the mounded shape of the data, which justifies the normality assumption. Stem-and-Leaf Display: Scores Stem-and-leaf of Scores N = 20 Leaf Unit = 1.0 1 5 7 2 6 2 5 6 578 8 7 123 (4) 7 5567 8 8 244 5 8 669 2 9 13 b Using the formulas given in Chapter 2 or your scientific calculator, calculate ∑ x 1533 x ==i =76.65 n 20 22 2 ()∑ xi ()1533 ∑−xi 119419 − ss2 ==n 20 =100.7658 and =100.7658 =10.0382 n −119 c Small sample confidence intervals are quite similar to their large sample counterparts; however, these intervals must be based on the t-distribution. Thus, the confidence interval for the single population mean described in this exercise will be s xt± α 2 n where tα 2 is a value of t (Table 4) based on df = n −1 degrees of freedom that has area α 2 to its right. For this exercise, nx==20, 76.65, s=10.0382 and t.025 with n −1= 19degrees of freedom is t.025 = 2.093 . Hence the 95% confidence interval is s 10.0382 xt±⇒α 2 76.65 ±2.093 ⇒76.65 ±4.70 n 20 or 71.95 <<µ 81.35 . Intervals constructed using this procedure will enclose µ 95% of the time in repeated sampling. Hence, we are fairly certain that this particular interval encloses µ . 2 10.5 a Using the formulas given in Chapter 2, calculate ∑ xi = 70.5 and ∑=xi 499.27 . Then ∑ x 70.5 x ==i =7.05 n 10 238 22 2 ()∑ xi ()70.5 ∑−xi 499.27 − ss2 ==n 10 =.249444 and =.4994 n −19 b With df =−n 1=9, the appropriate value of t is t.01 = 2.821(from Table 4) and the 99% upper one- sided confidence bound is s .249444 xt+⇒.01 7.05 +2.821 ⇒7.05 +.446 n 10 or µ < 7.496 . Intervals constructed using this procedure will enclose µ 99% of the time in repeated sampling. Hence, we are fairly certain that this particular interval encloses µ . c The hypothesis to be tested is H:0aµ = 7.5 versus H:µ < 7.5 and the test statistic is x −−µ 7.05 7.5 t == =−2.849 sn .249444 10 The rejection region with α = .01and n −1= 9degrees of freedom is located in the lower tail of the t- distribution and is found from Table 4 as tt< −=.01 −2.821 . Since the observed value of the test statistic falls in the rejection region, H0 is rejected and we conclude that µ is less than 7.5. d Notice that the 99% upper one-sided confidence bound for µ does not include the value µ = 7.5 . This would confirm the results of the hypothesis test in part c, in which we concluded that µ is less than 7.5. 2 10.6 a Using the formulas given in Chapter 2, calculate ∑ xi = 12.55 and ∑=xi 13.3253 . Then ∑ x 12.55 x ==i =.896 n 14 2 2 2 ()∑ xi ()12.55 ∑−xi 13.3253 − s ==n 14 =.3995 n −113 With df =−n 11=3, the appropriate value of t is t.025 = 2.16 (from Table 4) and the 95% confidence interval is s .3995 xt±⇒.025 .896 ±2.16 ⇒.896 ±.231 n 14 or .665 <<µ 1.127 . Intervals constructed using this procedure will enclose µ 95% of the time in repeated sampling. Hence, we are fairly certain that this particular interval encloses µ . ()∑ x 2 ()4.9 2 ∑−x2 i 6.0058 − ∑ x 4.9 i b Calculate x ==i =1.225 and s ==n 4 =.0332 n 4 n −13 and the 95% confidence interval is s .0332 xt±⇒.025 1.225 ±3.182 ⇒1.225 ±.053 or 1.172 <<µ 1.278 . n 4 The interval is narrower than the interval in part a, even though the sample size is smaller, because the data is so much less variable. c For white tuna in water, 239 ()∑ x 22()10.24 ∑−x2 i 13.235 − ∑ x 10.24 i x ==i =1.28 and s ==n 8 =.1351 n 8 n −17 and the 95% confidence interval is s .1351 xt±⇒.025 1.28 ±2.365 ⇒1.28 ±.113 or 1.167 <<µ 1.393 . n 8 For light tuna in oil, ()∑ x 2 ()12.62 2 ∑−x2 i 19.0828 − ∑ x 12.62 i x ==i =1.147 and s ==n 11 =.6785 n 11 n −110 and the 95% confidence interval is s .6785 xt±⇒.025 1.147 ±2.228 ⇒1.147 ±.456 or .691 <<µ 1.603 . n 11 1.4 1.3 1.2 t s 1.1 Co 1.0 0.9 0.8 Light Oil Light Water White Oil White Water The plot of the four treatment means shows substantial differences in variability. The cost of light tuna in water appears to be the lowest, and quite difference from either of the white tuna varieties. 10.7 Similar to previous exercises. The hypothesis to be tested is H:0aµ = 5 versus H:µ < 5 ∑ x 29.6 Calculate x ==i =4.933 n 6 22 2 ()∑ xi ()29.6 ∑−xi 146.12 − ss2 ==n 6 =.01867 and =.1366 n −15 The test statistic is x −−µ 4.933 5 t == =−1.195 sn .1366 6 The critical value of t with α = .05 and n −1= 5degrees of freedom is t.05 = 2.015 and the rejection region is t <−2.015 . Since the observed value does not fall in the rejection region, H0 is not rejected. There is no evidence to indicate that the dissolved oxygen content is less than 5 parts per million. ∑ x 608 10.8 Calculate x ==i =60.8 n 10 240 22 2 ()∑ xi ()608 ∑−xi 37538 − ss2 ==n 10 =63.5111 and =7.9694 n −19 The 95% confidence interval based on df = 9 is s 7.9694 xt±⇒.025 60.8 ±2.262 ⇒60.8 ±5.701 n 10 or 55.099 <<µ 66.501 . 10.9 a Similar to previous exercises. The hypothesis to be tested is H0a: µ =<100 versus H : µ 100 ∑ x 1797.095 Calculate x ==i =89.85475 n 20 22 2 ()∑ xi ()1797.095 ∑−xi 165,697.7081− ss2 ==n 20 =222.1150605 and =14.9035 n −119 The test statistic is x −µ 89.85475 −100 t == =−3.044 sn 14.9035 20 The critical value of t with α = .01and n −11= 9degrees of freedom is t.01 = 2.539 and the rejection region is t <−2.539 .

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