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IB Biology / IHS Just My Cup of “t”! (test, that is!) Or: How to conduct an unpaired two-tailed t-test Adapted from Jennifer Lockwood, Newbury Park High School, NP, CA Introduction What is statistics? Statistics is the science of collecting, organizing, analyzing, and interpreting data in order to make decisions. There are many different types of statistical tests. A t-test examines the difference between the means of two sets of data.* See fig. 1. This is done in order to determine if any observed difference is due to chance alone or if the difference between the means is the result of some other factor; in other words, if that difference is significant. Fig. 1 In science, researchers are often looking to see if experimental group data are significantly different from control group data. (If the data are significantly different, the researchers may propose that their treatment, the independent variable, within the experimental group caused the observed difference in the data.) How does a researcher determine if results are significant? By convention, scientists agree that if there is less than a 5% chance of getting the observed difference by chance, they conclude that they have found a statistically significant difference between the two groups being tested. Two-tailed (or, two-sample) t-tests are used when the researcher wants to find out if the results would be interesting in either direction (i.e. if the treatment had a positive or negative effect). And in an unpaired two-tailed t-test, there is no requirement that the two groups be of equal sizes. The two-sample t-test is one of the most commonly used hypothesis tests. How to conduct an unpaired two-tailed (two-sample) t-test 1) Define the null hypothesis. 2) Calculate the t-test statistic (tcalc). It's time-consuming to calculate so in this handout we often calculate it for you. 3) Specify the degrees of freedom (df) using this formula: df = (n1 + n2) - 2, where n1 and n2 are the number of observations in each data set. For example, if an experimental group has 12 participants and a control group has 10 participants, then df = (12+10) – 2 = 20. 4) Look up the t critical value (tcrit) in a t-distribution table for the particular df at P = 0.05. 5) If tcalc < tcrit, we accept the null hypothesis. If tcalc > tcrit, we reject our null hypothesis. Let’s look at the example in Table 1. Two different classes take the same Biology test. The calculated means, sample standard deviations (SD), and sample sizes are shown at the right. _______________________________________ *The data should be normally distributed with a sample size of at least 10. Table 1. Per. 2 & 4 Bio test results out of 62 possible points Period 2 Period 4 17 students 15 students Mean: 53.2 points Mean: 53.9 points Sample SD: 4.4 Sample SD: 5.5 Working through the example: 1) The null hypothesis: there is no difference between period 2's and 4’s mean test scores except that which is due to random chance. 2) The t-test statistic (tcalc) in this example is 0.3997. We calculated it for you to save you time. In "real life" one consults a software program or online calculator.** 3) We determine the degrees of freedom (df) by applying the above formula: df = (n1 + n2) – 2 = (17 + 15) – 2 = 32 – 2 = 30 Table 2. tcrit values for a t 4) Now we consult a table of t-critical values. In the row for 30 df, tcrit is 2.04 at P = 0.05. Note, P ranges from 0 (not likely) to 1 (certain). Table 2 only shows a few of the values for P: 0.05, 0.025, 0.01 & 0.005. 5) Recall, tcalc was 0.3997, which is < tcrit (2.04) at P = 0.05. When tcalc < tcrit, we must accept our null hypothesis: there is no difference between period 2's and 4’s test scores except that which is due to random chance. **If you must see the t-test formula, it is: 2 Let's practice some more! 1. An experiment is performed to discover the effect of using different amounts of fertilizer on the growth of bean plants. After germination, the beans are planted in sterilized soil to which fertilizer is added. As you can see in Table 3, there are ten trials of each of 3 levels of fertilizer treatment and a control. The height of each of the bean plants is measured in centimeters ( + 0.5 cm) 25 days after planting. Table 3. The effect of the amount of fertilizer on bean plant height Group A: no Group B: 0.001% Group C: 0.01% Group D: 0.1% Trial Number fertilizer fertilizer fertilizer fertilizer 1 2 3 4 5 6 7 8 9 10 10 7 8 7 8 10 9 8 7 9 8 8 7 10 10 7 8 8 9 9 7 10 8 10 8 9 9 8 6 7 12 13 15 10 10 15 10 10 14 10 a) Find the Mean and Standard Deviation of each set of data. (**In Excel: to calculate mean =AVERAGE(data range) to calculate sample SD =STDEV(data range). Treatment Group Group Group Group A: B: C: D: Table 4. Statistics generated from Table 3 data Sample t-Test score when treatment is Mean (cm) Standard compared to control (to save time, we'll just use 2 of 4 combo's) Deviation (cm) no fertilizer 0.001% fertilizer 0.01% fertilizer 0.1% fertilizer 0.20 4.61 b) Discuss the variability of the bean plant data using the sample standard deviation values. c) Compare the means of each group of data. d) Is there a significant difference in bean plant height between the Group B fertilizer treatment and the control? 3 e) Explain how you figured out the answer to (d) using the 5 steps on page 1 of this handout. f) Is there a significant difference in bean plant height between the Group D fertilizer treatment and the control? g) Explain how you figured out the answer to (f) using the 5 steps on page 1 of this handout. Table 5. Resting heart rates (bpm). Use for problem 2. 2. Resting heart rates (beats/min) are measured in 20 non-runners and 20 marathon-running athletes. The data are shown in Table 5 at the right. On average, do male marathon runners have a significantly different resting heart rate than untrained men? Show ALL steps 1-5 from page 1 of this handout. Non-runner males 72 75 68 61 82 78 79 77 71 75 74 79 81 83 77 74 70 69 78 77 Male Marathon runners 62 48 47 51 55 54 60 48 57 56 55 50 59 50 55 58 61 60 49 53 3. Scientists determined the average size of salmon that spawned in two different streams. 38 salmon were sampled for one stream and 22 for the other stream. The value of “tcalc” was found to be 1.29. Is the average size of the salmon in each stream significantly different? Explain. (Note: to use Table 2, use the degrees of freedom row closest to the d.f. for this problem.) 4