Chemistry, Biology and Environmental Science Examples STRONG Lesson Series

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1 Chemistry, Biology and Environmental Science Examples STRONG Lesson Series 1 Examples from Biology, Chemistry and Environmental Science Applications 1. In a forest near Dahlonega, 1500 randomly selected pine trees were tested for traces of Bark Beetle infestation, and 153 of the trees showed such traces. Test at the α = /05 level of significance the hypothesis that more than 10% of the trees have been infested. 2. A recent study in Denver asked 268 people if they prefer drinking bottled or tap water. Of the 268 randomly selected individuals, 61 were male and 114 were female. Test at the.1 level whether there is a gender difference for bottled water given that, in the survey, 17 of 61 males and 22 of 114 females reported that they prefer bottled water. 3. A random sampling of people who work in a downtown area revealed an established population mean of 30 miles for each worker s one-way commute with a standard deviation of The one-way commute mileages shown below were taken from persons under 30 years of age. Test at the.05 level whether commuters under 30 have shorter commutes than the general population of in-town workers A researcher observes 100 Hollywood celebrities to see who deposits their garbage into a garbage can and who litters. Test at the.1 level based upon the observed data shown below whether celebrity littering depends on gender. Deposit Litter Females 18 7 Males A study was done to examine the color in the tail feathers of northern flickers. Some of the birds had one odd feather that was different in color or length from the rest of the tail feathers, presumably because it was regrown after being lost. They measured the yellowness of one odd feather on each of 16 birds and compared it with the yellowness of one typical feather from the same bird. Because these birds were from a hybrid zone between red-shafted flickers and yellow-shafted flickers, there was a lot of variation among birds in color, making a paired analysis more appropriate. Determine if there is a significant difference in tail feather color using α =.05.

2 Bird Typical Odd Feather Bird Typical Odd Feather An experiment was conducted in which two sets of bean plants received different treatments. Group 1 received fertilizer and Group 2 did not. All other conditions were the same. All plants were of the same strain, received the same amount of sunlight, water, and so forth. After one month, we measured the heights of all plants and recorded the data. Determine if the height of the fertilized plants is significantly better (α =.01) than the height of the unfertilized plants. Fertilized Unfertilized Fertilized Unfertilized Professor Takahashi loves radishes and was curious about the biomass of radish seeds and seedlings. He weighed out 3 batches of radish seeds each weighing 1.5 g and grew them under the following conditions. Group 1: Seeds placed on dry paper towels in light. Group 2: Seeds placed on wet paper towels in light. Group 3: Seeds placed on wet paper towels in the dark. He replicated each experiment 10 times. The results are shown below. Group 1 Group 2 Group

3 8. An experiment is conducted on the effect of alcohol on perceptual motor ability. Ten subjects are each tested twice, once after having two drinks and once after having two glasses of water. The two tests were on two different days to give the alcohol a chance to wear off. Half of the subjects were given alcohol first and half were given water first. The scores of the 10 subjects are shown below. Higher scores reflect better performance. Test to see if alcohol had a significant effect. Report the p value and discuss the significance, if any. Water Alcohol A geologist collects several fist-sized specimens of limestone from a particular area. A qualitative assessment of both texture and color is made with the results shown in the table below. Test whether there is evidence of association between color and texture for these limestones at the.01 level of significance. Light Medium Dark Fine Medium Course During a study of two areas in Dahlonega, Georgia a total of 369 Hemlock trees were tested for traces of the Hemlock wooly adelgid infestation. Out of 369, Area A was found with 106 out of 219 infected trees and Area B had 76 out of 150 infected. Test at the.1 level whether there is a more urgent need to treat the infestation at either Area A or B. 11. Humerous bones from the same species of animal have approximately the same lengthto-width ratios. A certain species of squirrel has a mean ratio of 7.5. Suppose that 41 fossil humerous bones were unearthed at a site where these squirrels are known to have flourished. (We assume that all bones are from the same species.) The length-to-width ratios of these bones has sample mean 8.23 and sample standard deviation Can we conclude (at the α =.05) that these bones belong to a different species? 12. A new method for treating trees affected by the Hemlock wooly adelgid infestation has been developed. You have been hired to determine whether the new method is superior to the standard method recommended by the forestry service. You select 22 hemlock trees that have been highly infected and treat 10 hemlocks with the new method, the other 12 with the standard method. All the Hemlocks are then tested 10 weeks later to measure the extent of disease still present. Test at the.01 level. New Method Standard Method AVG SD N A pool of participants was randomly divided into 5 treatment groups. The groups were

4 administered different daily doses of vitamin C over a 12-month period: none, 250 mg, 500 mg, 1000 mg and 2000 mg. The data in the table represents the number of cold and flu viruses reported by the participant. At the.05 level, analyze the data using the correct statistical procedure. 0mg 250mg 500mg 1000mg 2000mg Solutions to Biology, Chemistry and Environmental Science Applications 1. In a forest near Dahlonega... pine trees were tested for traces of Bark Beetle infestation... Solution. z-test for population proportion (one-sample) H a : p >.1 Verification: Both 1500(.10) = 150 > 10 and 1500(.90) = 1350 > 10, OK to proceed. Plus 4 ˆp = 155 (z =.3954) p =.3462 >.05 = α 1504 Fail to reject H A recent study in Denver asked 268 people if they prefer drinking bottled or tap water. Solution. z-test for population proportion (two-samples) H a : p M > p F Verification: Both groups have 10+ successes (17 and 22) and 10+ failures (44 and 92), OK to proceed. Plus 4 ˆp ˆp M = 18 and ˆp 63 F = 23 (z = 1.33) p =.1837 > Fail to reject H A random sampling of people... each worker s one-way commute... Solution. t-test, not z-test! H a : µ < 30 Verification: Sample size: n = 20 > 15, no histogram or stemplot needed, but n < 40, so check boxplot for outliers. OK to proceed. (t =.7753) p =.2239 >.05

5 Fail to reject Ho. 4. A researcher observes 100 Hollywood celebrities... Solution. χ 2 Test of Independence H a : Littering habits are dependent upon gender. Verification: No low cell counts. OK to proceed. (χ 2 = 2.0) p =.1573 >.1 = α Fail to reject. 5. A study was done to examine the color in the tail feathers of northern flickers... Solution. Dependent Samples (Matched Pairs) t-test H a : µ g 0 Verification: Sample size: n = 16 15, no histogram or stemplot needed, but n < 40, so check boxplot for outliers. None are present. OK to proceed. (t = 4.065) p =.2239 >.05 Fail to reject H 0 6. An experiment was conducted in which two sets of bean plants received different treatments... Solution. Dependent Samples (Matched Pairs) t-test H a : µ F > µ U Verification: Sample size: n = 32 15, no histogram or stemplot needed, but n < 40, so check boxplot for outliers. None are present in either data set. OK to proceed. (t = 6.16) p = <.05 Reject H 0 7. Professor Takahashi loves radishes... Solution. ANOVA H 0 : µ 1 = µ 2 = µ 3, H a : At least one group mean is different Verification: N = n 1 + n 2 + n 3 = 30 > 20 with largest-to-smallest group size ratio of 1 : 1, which is less than 2:1 as required. OK to proceed. (F = 15.50) p = <.05 = α Reject the null, proceed to post hoc testing, Tukey s HSD. Formula: HSD ij = q MSB n ij HSD = (HSD is same for all 30

6 three groups since group sizes are identical.) Only one pairwise difference significant: x 2 x 3 = = > = HSD An experiment is conducted on the effect of alcohol on perceptual motor ability... Solution. Dependent Samples (Matched Pairs) t-test H 0 : µ g = 0, H a : µ g > 0 Verification: n = 10 < 15 must check both a boxplot (no outliers) and histogram. Fails verification. Histogram gives no evidence data were drawn from a normal distribution. NOT OK to proceed. 9. A geologist collects several fist-sized specimens of limestone from a particular area... Solution. χ 2 Test of Independence H 0 : Texture is independent of color, H a : Texture depends upon color Verification: Lowest cell count (of 9) is 6.4 > 5. No low cell counts in Expected matrix B at all (could have as many 2). OK to proceed. (χ 2 = 17.73) p = <.01 Reject H During a study... a total of 369 Hemlock trees were tested... Solution. z-test for population proportion (two-samples) H a : p A p B Verification: Well over 10 successes and 10 failures in both samples. OK to proceed. Plus 4 ˆp: ˆp A = 107 and ˆp 221 B = 77 (z =.3954) p =.3462 >.05 = α 152 Fail to reject H 0. (z =, 425) p =.6705 >.1 Fail to reject H Humerous bones... Solution. t-test (z-test not used in these situations in modern practice) H a : µ 7.5 Verification: Sample size: n = 41 > 40, no graphical checks needed, robustness of t statistic guarantees that, regardless of violations of assumptions, it is OK to proceed. (t = 4.173) p = <.05

7 Reject H 0 which provides evidence that the fossils are not of the same species. 12. A new method for treating... the Hemlock wooly adelgid infestation... Solution. Independent Samples t-test H a : µ New < µ Std Verification: Sample size: n = 22 < 40, so we must check both samples for outliers using boxplots. However, the data are not available for scrutiny. We can conduct the test (and will), but results may be spurious. Sort of OK to proceed. (t = ) p =.0667 <.1 Reject H 0 which provides evidence that the new treatment works better. 13. A pool of participants... were administered different daily doses of vitamin C... Solution. ANOVA (by-hand) H 0 : µ 0 = µ 250 = µ 500 = µ 1k = µ 2k, H a : At least one group mean is different Verification: N = n 1 +n 2 +n 3 +n 4 +n 5 = 20 which is enough, thanks to the robustness of the F -statistic, to guarantee accurate p-values even if assumptions are violated. The largest-to-smallest group size ratio of 1 : 1, which is less than 2:1 as required. OK to proceed. F 6.8 = 3.93 > 3.06 = F α 1.73 Reject the null, proceed to post hoc testing, Tukey s HSD. Formula: HSD ij = q MSB n ij HSD = 4.37 groups since group sizes are identical.) We find two pairwise differences significant: x 1 x 5 = 3 > 2.87 and x 2 x 5 = 3 > (HSD is same for all three