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T- Test Example of how to perform a t-test

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Step 1 State the null hypothesis Begin with assuming that any observed differences between the two samples means it occurred by chance and is not significant. This assumption is expressed as the NULL HYPOTHESIS. For our example: The mean height of stressed plants is NOT significantly different from the mean height of nonstressed plants.

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Step 2 Establish the level of significance. Alpha =0.05 This communicates the probability that the researcher erred in rejecting the null hypothesis. At 0.05 level of significance, the probability of error in rejecting the null hypothesis is less than 5%.

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Step 3: Calculating the means. Stressed plants 55 65 50 57 59 73 57 54 62 68 Nonstessed Plants 48 65 59 57 51 63 65 58 44 50

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Step 3: Calculating the Means continued X stressed = ΣX÷n 55+65+50+57+59+73+57+54+62+68 / 10= 60 X nonstressed = 48+65+59+57+51+63+65+58+44+50/ 10= 56

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Step 4 Calculating the variance S 2 =Σ(X 1 -X) 2 ÷n-1 Stressed: =(55-60) 2 +(65-60) 2 + (50-60) 2 ……+(68-60) 2 ÷10-1=49.1 Nonstressed: =(48-56) 2 +(65-56) 2 + (59-56) 2 …..+(50-56) 2 ÷10-1=60.7

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Step 5: Calculating t T= X 1 -X 2 ÷√s 2 1 +s 2 2 ÷n T = 60-56÷√49.1+60.7÷10 =4÷√109.8÷10 =4√10.98 =4÷3.1 =1.3

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Step 6: Determining degrees of freedom =(n-1)+(n-1) = (10-1) + 10-1) =18 If you have more than 2 groups you must compare all of your groups. Example group A, B, and C You would compare A vs. B, A vs. C, and B vs. C, so you need to perform 3 t-test.

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Step 7:Determine the significance of the calculated t. At df=18, alpha of 0.05, t= 2.101; the calculated t of 1.3<2.101 and is not significant at the 0.05 level Because the calculated value of t is NOT significant, the null hypothesis is NOT rejected. Because the null hypothesis was NOT rejected, the research hypothesis that stressed plants would have a greater mean height than nonstressed plants was NOT supported.

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