Answer the problem below and show complete solutions and box final answers.

1. The graph of the t-distribution with 15 degrees of freedom is shown below.

Find the value of t such that the:

a. shaded area on the left is equal to 0. 05 b. shaded area on the right is 0.1

c. Total shaded area is 0.02

d. Area to the left of t = 0.95

2. Find the 5th percentile of a t-distribution with 28 degrees of freedom. Prop

3. The area to the left of t, is 0.995, what is the value of t?

LESStI NO. 7: THE T- DISTRIBUTION The T Distribution (and the associated t scores), are used in hypothesis testing when you want to gure out it 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 Wm\" fatter. The tdistribution is used instead of the normal distribution . mm when you have small samples. The larger the sample 3129, the ' more the t distribution looks like the normal distribution. -lll-l+--l-l .3 -2 -I II I 2 J T Distribution Formula d k ( - I I Where: i = Sample mean 1.1 = Population mean 5 = standard deviation of the sample mean n = sample size in WEE! Thetlhleghlslhevahuoft_:, who! 01 252.51... )= a .1515 1 deymofurhm 15 1552 1.215 2.120 2555 2.521 5555 4515 2.5 12 1555 1.2210 2.110 2552 2555 5515 5.555 15 1.550 1.251 2.101 2552 2525 5510 55122 a 0.1 0.05 0.025 001 0.005 0001 021005 19 1323 1-73 2393 1539 1351 3579 3533 2 2.0 1.525 1.225 2.055 2.525 2515 5552 5550 l :g is; 'i 1% 63555025: 333% 5555231 5.33" 21 1525 1.221 2.050 2515 2551 5522 5510 3 1555 2.35 5152 11.5111 5.5111 10 213 1232' 22 1521 1.212 2.024 2505 2.515 5505 5.252 4 1553 2.120 2225 5147 4.501 2.125 5.510 23 1319' 1-7\" 2.055 2500 2502 3'55 5.252 25 1.515 1.2151 2.050 2.155 2.252 52150 5.225 g III: 1% g; gm gig; 3% 25 1515 1.202 2.055 2120 2.225 5.1155 5.202 5 1552 {550 2505 2555 5.555 5501 5.011 22 15111 1.205 2.052 2425 2.221 5221 5550 9 1555 1:533 2252 2321 3150 5.292 5.251 25 1515 1.201 2.045 2452 2755 51105 5524 10 1522 1.512 2225 2251 2150 1.1411 4552 39 1-3" 1-599 2-'3'5 7-452 1755 3395 3559 50 1.510 1.552 2.012 2.152 2.250 5555 5515 11 1555 1.255 2201 2215 5105 1.025 21452 110 1505 1.551 2.021 2 525 2.2011 5502 5 551 12 1555 1.252 2125 2551 5.055 5.550 4515 50 1.255 1521 2000 2550 2550 5252 51150 15 1550 1.221 2150 2550 5012 5552 4221 - - 14 135 1.251 2145 2524 2022 5.252 4.150 120 1255 1.555 1.550 2555 2512 5.150 5525 15 1.5111 1.255 2151 2502 25112 5.255 11025 i \"52 L\" "9\"\" 2.525 2525 \"9\" 329' To nd a value in the Table of t-distribution, there is a need to adjust the sample size n by converting it to degrees of freedom or. (if: n - 1', where n = sample size. Use the t-distribution table to find the critical value. Df is the first column and a is the first row then find their intersection. 1 2: 3. 1:1 = o.05,df = 3 Critical value: 1.860 11 = {1.025,df =12 Critical value: 2.179 01 = o.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 n 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 rty V15 Test statistic Critical value 1.590 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: The total shaded area of a ounre is 0.1 with dt = 13. Which percentile is i. and what is its value? P = (1 -e )100% = (1 - %) 100%.911 is divided by 2 because 0.1 is the total shaded area at the left and right P = 0.95 Thus, t1 represents the 95'\" percentile. To find the value, proceed to the right until the "1 0 '1 column headed 0.05. The result is 1.734 Activity: Find the Percentile Find what is asked in each situation using the t-distribution table. 1. Suppose you have a sample size of10, what is the 95th percentile of its corresponding t-distribution. 2. The total shaded area of a curve is 0.02 with df = 19. What is the value of the 99th percentile? 3. Find the 2.5\"1 and 975'\" percentile of a t-distribution with 11 degrees of freedom