7. the t distribution
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The document provides an overview of the t-distribution, detailing its characteristics, formula, and application in statistical analysis with small sample sizes. It explains how to calculate critical values and test statistics while using examples to illustrate comparisons between means in different scenarios. The text concludes with procedures for identifying percentiles and using t-tables effectively.
7. the t distribution
- 2. Learning Competencies The learner will be able to: 1. Illustrate the t-distribution 2. Construct a t-distribution 3. Identify regions under the t-distribution corresponding to different t-values; and 4. Identify percentiles using the t-table.
- 3. The t-distribution, just like the standard normal curve, is bell-shaped and unimodal. It is symmetric about t=0. However, its variance is greater than 1. This makes its wider and flatter in the middle. It has more area in its tails than that of the standard normal curve. Its shapes depends on the sample size n. As the sample size n becomes larger, the t-distribution gets closer to the standard normal distribution.
- 4. Comparison
- 5. Statistical analysis on some studies which cannot be done using the normal distribution can be done using the t-distribution. The t-distribution is used with small samples taken from the population that is approximately normal. The z-statistic in the previous lessons uses the value of population variance while the t-statistic below uses the sample deviation especially when the population variance is unknown. The z-statistic is used when n>30 while t-statistic is used when n<30.
- 6. The t-distribution formula is: 𝑡 = 𝑥 − 𝜇 𝑠 𝑛 where 𝑥 = 𝑠𝑎𝑚𝑝𝑙𝑒 𝑚𝑒𝑎𝑛 𝜇 = 𝑝𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑚𝑒𝑎𝑛 𝑠 = 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑎𝑚𝑝𝑙𝑒 𝑚𝑒𝑎𝑛 𝑛 = 𝑠𝑎𝑚𝑝𝑙𝑒 𝑠𝑖𝑧𝑒s
- 7. To find a value in the Table of t Critical Values, there is a need to adjust the sample size n by converting it to degrees of freedom df. The number of degrees of freedom is equal to the number of the remaining values in a data set that are free to vary after one or more values have been deducted. In the case of the t-distribution, one value is deducted.
- 8. Hence the formula is: 𝑑𝑓 = 𝑛 − 1 where 𝑑𝑓 = 𝑑𝑒𝑔𝑟𝑒𝑒𝑠 𝑜𝑓 𝑓𝑟𝑒𝑒𝑑𝑜𝑚 𝑛 = 𝑠𝑎𝑚𝑝𝑙𝑒 𝑠𝑖𝑧𝑒
- 9. Example1 A student researcher wants to determine whether the mean score in mathematics of the 25 students in Grade 11 Hebrews is significantly different from the average of the school which is 89. The mean and the standard deviation of the scores of the students in Section Hebrews are 5 and 15, respectively. Assume a 95% confidence level.
- 10. Solution Step1. Find the degrees of freedom. 𝑑𝑓 = 𝑛 − 1 = 25 − 1 = 24 Step2. Find the critical value. Use the Table of t Critical Values. Confidence level is 95%. 1 − 𝛼 100% = 95% 1 − 𝛼 = 0.95 𝛼 = 0.05 (area in two tails) 𝛼 2 = 0.025 (area in one tail)
- 11. Look at 24 under the column headed df. Move to the right along the row until reaching the column headed 0.05, area in two tails or 0.025 for area in one tail. The critical value is 2.064.
- 12. Step3. Compute the test statistic t. 𝑡 = 𝑥 − 𝜇 𝑠 𝑛 = 95 − 98 15 25 = 2 The computed value of t is equal to 2 which is smaller than the t-Critical value of 2.064
- 13. The value of the test statistic or computed t value does not fall in the critical region. Therefore, the mean score of Grade 11 section Hebrews in Mathematics is the same with mean score of all the students taking up Grade 11 Mathematics.
- 14. Example 2 A student suspects that the data she collected for research study do not represent the target population. Here are the data she collected. The population mean is 27. Assuming normality in the target population, is the student’s suspicion correct? Use a 90% confidence level. 16 27 34 20 29 30 22 30 37 25 30 42 26 32 35
- 15. Observation x 𝒙 − 𝒙 𝒙 − 𝒙 𝟐 1 16 16-29=-13 169 2 20 20-29=-9 81 3 22 22-29=-7 49 4 25 25-29=-4 16 5 26 26-29=-3 9 6 27 27-29=-2 4 7 29 29-29=0 0 8 30 30-29=1 1 9 30 30-29=1 1 10 32 32-29=3 9 11 34 34-29=5 25 12 30 30-29=1 1 13 37 37-29=8 64 14 42 42-29=13 169 15 35 35-29=6 36 𝑥 = 435 𝑥 − 𝑥 2 = 634 Step1. Find the sample mean and the sample standard deviation. Sample mean 𝑥 = 𝑥 𝑛 = 435 15 = 29 Sample variance 𝑠2 = 𝑥 − 𝑥 2 𝑛 − 1 𝑠2 = 634 15 − 1 𝑠2 = 45.29 Sample standard deviation 𝒔 = 𝑥 − 𝑥 2 𝑛 − 1 𝒔 = 𝟒𝟓. 𝟐𝟗 = 𝟔. 𝟕𝟑
- 16. Step2. Find the degrees of freedom. 𝑑𝑓 = 𝑛 − 1 = 15 − 1 = 14 Step3. Find the critical value. Confidence level is 90%. 1 − 𝛼 100% = 90% 1 − 𝛼 = 0.90 𝛼 = 0.10 𝛼 2 = 0.05
- 17. Use of t-Critical Values Table. In the column headed df, look at 14. Move to the right until the column headed 0.05 for one tail and 0.10 for two tails is reached. The intersection is 1.761. Hence, the critical value is 1.761.
- 18. Step4. Compute the test statistic t. 𝑡 = 𝑥 − 𝜇 𝑠 𝑛 𝑡 = 29 − 27 6.73 15 𝑡 = 2 6.73 15 𝑡 = 1.151 The value of test statistic or computed t- value is less than the tabular value of 1.761. Therefore, the student is wrong in suspecting that the data are not representative of the target population.
- 19. Example 3 A sample of size n=20 is a simple random sample selected from a normally distributed population. Find the value of t-such that the shaded area to the left of t is 0.05.
- 20. Solution a. Find the degrees of freedom df. 𝑑𝑓 = 𝑛 − 1 = 20 − 1 = 19
- 21. Since the t-distribution is symmetric about 0, then the area to the right of t is 0.05 and to the left of t is also 0.05. In the Table of t-Critical Values, move down the first column headed df until the df until df=19. Move to the right along this row reaching the column headed 0.05 (area in one tail) or 0.10 (area in two tails). −𝑡0.05= −1.729 Since we are interested to the area at the left of t, hence, the negative value.
- 22. Example 4 Suppose you have a sample of size n=12 from a normal distribution. Find the critical value 𝑡𝛼 2 that corresponds to a 95% confidence level.
- 23. Solution a. Find the degrees of freedom df 𝑑𝑓 = 𝑛 − 1 = 12 − 1 = 11 b. Confidence level is 95% 1 − 𝛼 100% = 95% 1 − 𝛼 = 0.95 𝛼 = 0.05 𝛼 2 = 0.025
- 24. In the Table of t-Critical Values, move down the first column headed df until df=11, then move to the right along this row until the column headed 0.05 for two tails is reached.
- 25. Example 5 For a t-distribution with 25 degrees of freedom, find the values of t-such that the area to the right of t is 0.05.
- 26. Solution a. The degrees of freedom df=25. Using the Table of Critical Values of t, move down the column headed df until df=25. Move to the right along this row until reaching the column headed 0.05 for the area in two tails. 𝑡𝛼 2 = 𝑡0.05 = 1.708
- 27. Example 6 For a t-distribution with 14 degrees of freedom, find the value of t such that the area between –t and t is 0.90.
- 28. Solution: a. Find the degrees of freedom df=14. b. 1 − 𝛼 100% = 90% 1 − 𝛼 = 0.90 𝛼 = 0.10 𝛼 2 = 0.05 Using the Table of Critical Values of t, move down the column headed df until df=14. Move to the right along this row until reaching the column headed 0.10 for the area in two tails. 𝑡𝛼 2 = 𝑡0.05 = 1.761
- 30. Example 7 The graph of a distribution with df=25 is shown below. a. If the shaded area on the right is 0.05, what is the area to the left of 𝑡1? b. What does 𝑡1 represent? c. Find the value of 𝑡1.
- 31. Solution a. 1 − 𝛼 100% = ρ 1 − 0.05 = 𝜌 0.95 100% = 𝜌 𝜌 = 95% 𝜌 = 0.95 𝑎𝑟𝑒𝑎 𝑡𝑜 𝑡ℎ𝑒 𝑙𝑒𝑓𝑡 𝑜𝑓𝑡1 b. Hence 𝑡1 represents the 95th percentile 𝑡0.95 c. To find the value of 𝑡1, look at 18 under the column headed df. Proceed to the right until the column headed 0.05 for one tail is reached. The result is 1.753.
- 32. Example 8 The graph of a t-distribution with 18 df is shown below. a. If the total shaded area of the curve is 0.02, what is the area to the left of 𝑡1? b. What does 𝑡1 represent? c. What is the value of 𝑡1?
- 33. Solution a. 1 − 𝛼 100% = ρ 1 − 0.02 2 100% = 𝜌 1 − 0.01 = 𝜌 𝜌 = 0.99 b. 𝑡1 represents the 99th percentile 𝑡0.99 c. To find the value of 𝑡1, look at 18 under column headed df. Proceed to the right until the column headed 0.01 for one tail is reached. The result is 2.552.