LESSSIII NO. 7: THE T- DISTRIBUTION The T Distribution [and the associated 1 scores), are used in hypothesis testing when you want to gure out if you should accept or reject the null hypothesis. In general, this distribution is used when you have a small sample size {under 30} or you don't know the population standard deviation. For practical purposes (i.e. in the real world), this is nearly always the case. So. unlike in your elementary statistics class. you'll likely be using it in real life situations more than the normal distribution. If the size of your sample is large enough, the two distributions are practically the same. The T distribution is a family of distributions that look almost identical to the normal distribution curve, only a bit shorter and mm fatter. The tdistribution is used instead of the normal distribution . 1m when you have small samples. The larger the sample 0129. the more the t distribution looks like the normal distribution. lfl+l-il- . . . d\" d -I II I 2 J T Distribution Formula d k Mm w (X- LII \"comm ,..;=. .' t= I . in Where: :2 = Sampie mean u = Population mean s = standard deviation of the sample mean 11 = sample size Table of the Student's r-d'sfributiun The table gives Ii: 1111111014121\" 0211:! 01 \"nonemmydmorm 10 1.002 1.240 2.120 2000 2021 0000 4010 I... 12 1.000 1.241 2.110 2002 2000 0040 0000 10 1.000 1.204 2.101 2502 2020 0010 0002 0'. 0.1 0.05 0.025 001 0005 0.001 00005 19 1335 1.220 1093 3539 2361 3579 3533 v 20 1.025 1.225 2.000 2.520 2045 0552 0050 I 2% is; 'i 1% g; 02210032610 633$ 21 1.020 1.221 2.000 2010 2001 0.022 0010 3 1000 2.050 0.102 *5\" 5.041 10213 12.924 22 1.021 1.212 2.024 2000 2010 0000 0.202 4 1500 2.102 2.220 0.242 4.004 2.120 0.010 23 \"\"9 L7\" 2.000 35\" '5'" 3'35 0.202 25 1.010 1.201 2.000 2405 2.202 0.400 0.220 g mg I'g g; gg gig; 2% 33% 20 1010 1.210 2.000 2420 2.220 0.400 0.202 0 139, 1000 305 2000 0.000 m 5.041 22 1.014 1.200 2.002 2420 2.221 0.421 0000 9 1333 1:333 2202 2321 0250 4.202 \".51 20 1010 1.201 2.040 2402 2200 0.400 0024 10 1.022 1.012 2220 2204 0100 4.144 4.502 29 13\" '59\" 3-\"5 \"52 3-755 3395 3559 00 1.010 1.002 2.042 2.452 2.250 0005 0040 11 1000 1.200 2201 2210 0100 4.000 4.402 12 1:450 1.202 2120 2001 0.005 0.000 4010 g 1% 13:: :% Egg 3% Egg; 3% 10 1:450 1.221 2100 2050 0012 0052 4221 - - 14 1.040 1.201 2145 2024 2022 0.202 4.140 120 1200 1.001 1.000 2000 2012 0.100 0020 10 1041 1.202 2101 2002 2042 0.200 4.020 i 1257' "5\" "9\" 332\" 57\" 399 329' To find a value in the Table of t-distributiori, there is a need to adjust the sample size n by converting it to degrees of freedom off. (if: n - 1, where n = sample size. Use the t-distribution table to find the critical value. Df is the first column and 01 is the first row then find their intersection. 1. 01 = 0.05,df = 3 Critical value: 1.860 2.42 = 0.025,df = 12 Critical value: 2.179 3. 4:1 = 0.01,df =20 Critical value: 1.325 Example 1: Suppose you do a study of acupuncture to determine how effective it is in relieving pain and you suspect that the data you collected do not represent the target population. The population mean is 7.54. You measure sensory rates for 15 subjects with the results given. Use the sample data to construct a 95% confidence level. 8.6 9.4 7.9 6.8 8.3 7.3 9.2 9.6 8.7 11.4 10.3 5.4 8.1 5.5 6.9 Step 1: Find the sample mean and sample standard deviation. Sample mean X = = 8.2267 Sample standard deviation = 1.6722 Step 2: find the degrees of freedom (df = n - 1) df = 14 Step 3: Find the critical value. Confidence level is 95%. (1- a) 100% = 95% 1 - a = 0.95 a = 0.05 Use the t - distribution table and move to the right until the column headed 0.05 with df = 14. Hence, the critical value is 1.761. Step 4: Compute the test statistic t. t= X - H 8.2267-7.54 1.6722 = 1.5904 Vn V15 rty Test statistic Critical value 06S'T 1.761 The value of the test statistic or the computed t-value is less than the critical value 1.761. therefore, the student is wrong in suspecting that the data are not representative of the target population. Identifying Percentiles Using the t-Distribution Table Example 1: The graph of the distribution below has a df = 6 a. If the shaded area is 0.025, what is the area to the left of t? P = (1 - a) 100% = (1 - 0.025) 100% = (0.975) 100% P = 97.5% P = 0.975 area to the left of t b. What does t represent? Hence, t represents the 97.5th percentile c. Find the value of t. To find the value of t, look under the column headed df. Move to the right until the column headed for 0.025 is reached. The result is 2.447.Example 2: le total saea area are curve 15 0.1 wrt at = 18. Wlc peroentrle Is It an?! What 15 Its value? P = (1 -a )100% = (1 E) 100%9'11is divided by 2 because 0.1 is the 2 total shaded area at the left and right P = 0.95 Thus, t1 represents the 95"1 percentile. To find the value, proceed to the right until the "1 0 t1 column headed 0.05. The result is 1.734