Let Us Study When performing a probability distribution of sample means using the zdistribution, it is required that the sample size (n) is sufciently large and the standard deviation (6) of the population is already known. But in reallife, it is almost always impossible for us to know the standard deviation (6) of the population from which our sample is drawn. When either of these problems occurs, the solution is to use a different distribution. Fortunately, the way to work out this type of problem was solved in 1908 by William S. Gosset. Math Trivia: \"The Man Behind T\" William Sealy Gosset - Wikipedia en.wikipedia.org "WILLIAM S. GOSSET William Gosset was an English statistician who worked for the brewery of Guiness. He developed different methods for the selection of the best yielding varieties of barley, an important ingredient in making beer. Gosset found big samples tedious, so he was trying to develop a way to extract small samples but still come up with meaningful prediction. He was a curious and productive researcher and published a number of researches that are still relevant today. However, due to the Guiness Company Policy, he was not allowed to sign the research papers with his own name. So, all of his work is under the pen name \"Student\". Later on, a great mathematician theorist, Ronald Fisher, proved rigorously that Gosset's ndings were correct. Fisher was able then to generalize Gosset's original work and introduced the tstatistics and named it as Student's t-distribution. The Student's T Distribution is one of the biggest breakthrough in statistics as it allowed inuence to small samples with an unknown population variance. Student's t - distribution The Student's tdistribution is a probability distribution that is used to estimate population parameters when the sample size is small (i. e. sample size For example, with the given area of 0.05 on the right tail of t distribution, the tvalue is 1.833 with 9 degrees of freedom. But with 35 degrees of freedom, the tvalue is equal to 1.694. 0.4 0.3 d. = 35 0.2 \"J The t-Distribution table In finding the areas and percentiles for a t-distribution you need to familiarize yourself with the t-table. Contidence Intervals : : : 0.10 1.05 #0.005 Two tous, an $0.20- 0.10 0.01 3.078 6.314 12.706 31.62 (3.657 1.638 2.353 3.182 4.541 5.841 82.142 THAT 1.476 2.015 2571 3.365 3.707 1.415 1.895 2 365 2.948 3.499 1.39710 2306 0 3.355 9 1.383 1.833 2 262 2.821 3.250 5 2.764 3.162 1.363 1.796 2.201 2.718 3.106 11 2 179 2.681 3.055 13 1.350 2.150 2.650 4.012 2077 1.341 1.753 2131 2.602 2.047 15 45 0241 18152 120 -367 2.583 2 2.921 2567 2.098 17 1.333 1.740 2.110 19 1.328 1.729 2.093 2.539 2.601 2.528 2.845 TOTARO1 325 1.323 1.721 2.080 2518 2.831 21 1717. 2.074 2.5080 2.819 23 1.319 1.714 2.069 2 500 2.807 BHAM 2 492 6 HH 2.797 :25 1.316 1.708 2.060 2485 2.787 2056 : 24794 3 2.779 26 07 1.214 1.713 2.052 2473 2.771 2048 1 0 157 2.467 430 2.763 29 1.311 1.699 2.045 2462 2.75h 1.69704 8 141 2042 1 2 12457 1 2 750 2.730 32 1.309 1.694 2.037 2.440 SOON2. 728 1.203 1.625 2.025 2.434 2719 36 EXTRA1.304 710 : 2429 3 2712 1,303 1,684 2.021 1 2.423 2 704 202014 WEARE 2 412 HOE 1 2 690 1.200 1.676 2.009 2.403 2 678 2.390 2 660 60 1.290 1.671 2030 1.294 1.604 2.381 70 1.667 2.643 GROWTH 1 293 : 1 065 1.292 1.G64 1.930 2.374 2.639 2.632 2.620 100 1.290 1.060 1.984 2.364 2.586 100 1.282 .G40 1.962 2.330 2.581 : 8 2 326- 1 2.576' "This value has been rounded to 128 h the textbook, One tail Two talls This valuo has boon rounded to 1.65 in the tordbock :The vaud has boon rounded to 2.$3 in the texthock, "This value has been rounded 15 2.58 in the textbook. Source: Adapted from W. H. Bayer, Handbook of Tables for Probabily and Sturtles and id, CAC Prom, Boon Paton, Ro, 1086, Aoprinted with ponrissionLooking at the table below, the first column in the left-side is the degree of freedom, the first-three rows in the top is the area under the distribution while the rest of the entries in the body are the values of t (t-values). The I Distribution Ofe usa a 0.20 0.10 0.05 Area of the distribution Degree of Freedom t-values For instance, what is the value of t, given a degree of freedom 6 and an area of 0.05 (one-tailed) to the right of the distribution? In finding the t- value, look at the area to the right of the distribution which is 0.05 (one- tailed) and the degree of freedom 6, intersect the two points to determine the t-value. As shown, the intersection is at 1.943, therefore, the area 0.05 to the right of the distribution has a t-value of 1.943. The / Distribution Confidence Intervals 80% 90% 95% 98% 99% One tail, c 0.10 0.05 0.025 0.01 0.005 d.f. Two tails, a 0.20 0.0 0.02 0.01 3.078 14 12.70 31.821 53.657 1.886 303 5.965 9.925 1.638 3.182 .341 3.841 1.533 2.776 3.747 4.601 1.476 2 571 3.365 4.032 1,943 2447 3,143 3.707 1.415 L.NYS 2.365 2.99% 3.499 1.397 1.860 2.306 2.896 3.355 1.383 1.833 2.262 2.821 3.250Another example, what is the value of t, with an area of 10% (twotailed) of the distribution with a sample size of 15? In this case, locate 10% or 0.1 in the twotailed of the area, and intersect this point to the degree of freedom 14 (d.f. = n 1 =15 1= 14). The tvalue corresponds to the area 0.1 in the distribution is 1.761. l iraraamiitru Condence intervals 80% 90% 95% 93% _m-m-m-m- L \"0"\"; Jmmmm E\". l 6. 12 706 3| ll 63.65? 2 2. ' 9.915 3 2. 5 R41 4 2. 4. 6M 5 2. 4. 012 I5 1. ' 3. T0? '7 l. ' 3.499 3 l. ' 3.355 9 1. 3.250 1. ' 3.169 1. 3. [06 3.055 3.012 2.917 2.94? 2.92: Identifying Percentiles Using the t-table Percentiles represent the number of scores that fell below a given value. For instance, a student with a summative test score of 50 is at 92nd percentile, this means that their score is higher than 92 percent of the other scores. Percentiles expresses the comparison of one value to the other. In tdistribution, percentile is the value that is less than the probability in the given percentage. 0-4 Percentile is also known as tvalue. 0.3 For example, the 85th percentile of the tdistribution is the tvalue whose left tail probability is 85% or 0.1 0.85 and whose righttail probability is 15% or 0.15(1 0.85 ' -3 -2 =0.15, recalling that the area of the tdistribution is 1). 0.5 85th Percentile Illustrative Examples: 1. Find the 95"h percentile of a tdistribution with 12 degrees of freedom. To find the tvalue, sketch the graph to locate the 95th percentile in the t distribution. From the denition, 95th percentile refers to the value of t that has an area of 95% to the left of the distribution, thus the area to the right of the 95th percentile is 5% or 0.05 (1 0.95 = 0.05, recalling the total area of the given distribution which is equal to 1). Use the ttable to nd the 95th percentile or the tvalue. '-5-43-2-1u1-3>4 5 95th percentile = ? 1.782 Looking at the graph above, we formed one tail only, therefore, in the tdistribution, we use one tailed with an area 0.05 to the right. As shown in table, the 95th percentile is 1.782. This means that the t value of 1.782 has an area of 95% (0.95) to the left of it or has an area of 5% (0.05) to its right. ' HiEfiEiilu L 2. Find the 2.5th percentile of a tdistribution with sample size 15. The 2.5th percentile is the value of the variable t that has an area of 2.5% or 0.025 to the left, hence, the area to the right of 2.5th percentile is 97.5% or 0.975 (1 0.025, again, 1 is the total area of the tdistribution). Remember, one of the properties of the tdistribution is that it is symmetric about zero, meaning the area to the right of the distribution is the same area to the left. 2.5\": percentile or t-value =? The table shows that the area of 0.025 to the right of the distribution has a tValue of 2.145. Thus, the tValue with an area to the left of 0.025 must be 2.145. Therefore, the 2.5th percentile is 2. 145. 3. Find the tvalues that bound in the middle 95% with (if = 8. As you can see in the illustration, it consists of two tails. Thus, we are going to use two tail in the tdistribution and we are looking for the two (2) values of t. The area of both tails is 5% or 0.05. ai'illfulm Condence . E iIIIIIHII ,IIIWERH: Ild Therefore, the tvalues that bound in the middle of 95% are 2.306 and 2.306. 6136556 l-'I'JUUI There is also an alternative way in identifying tvalue of the given distribution. ' Wham [-mmmm MIMI-Dim 31.821 (1.1. 63.657 9.925 5. \"I 4.604 4.032 3707 Using one tail 33:: distribution, we divide 5% 3:: or 0.05 by 2 to determine 3.I06 3.055 3.0l2 2977 the area of each tail. As shown, the area to the right of the distribution is 0.025. Use this area to determine the value of t. The tvalue with an area of 0.025 to the right of the distribution is 2.306. Since it is symmetrical, hence, the tvalue with an area of 0.025 to the left of the distribution is 2.306. 4. What is the area to the right of 1.8 under a tdistribution with 10 degrees of freedom? In the illustration the tvalue of 1.8 is somewhere between 1 and 2, and we are going to nd the area to the right of it. \"IIIIIIIIII 0.47 7777 , w:::=gi:::: 0.2 1' L IIImIII MII''iI -54-3-21 345 1.8 0.1 So, looking at the table, you need to focus on the 10 degrees of freedom line. You will observe that the tvalue of 1.8 cannot be found in this row but you observe that 1.8 is between 1.372 and 1.812. The table tells you that the area to the right of 1.372 is O. 10 and the area to the right of 1.812 is 0.05. You gure out earlier that our tValue of 1.8 falls in between two values 1.372 and 0.10 and it tends to reason then, that the area to the right of 1.8 must be between those two values 0.10 and 0.05. ml'lilllw'l 7 Condence inwn'als 30% _m dJ. -uo-uqnuh.w~m So, using the table you found that the area to the right of 1.8 under the tdistribution with 10 degrees of freedom lies somewhere between 0.10 and 0.05. What is then the area to the left of 1.8? Since the area under the entire curve is 1, the area to the left of 1.8 is equal to 1 minus the area to the right of 1.8. So, based on the table the area to the left of 1.8 under the t distribution with 10 degrees of freedom must lie somewhere between 0.90 and 0.95 (1 0.10 = 0.90 and 1 0.05 = 0.95). a Let Us Practice .' Independent Practice: \"BLACKPINK in your Area\" Use the ttable to identify the area under the tdistribution given the t value and the number of degrees of freedom. Draw a tdistribution for each number then choose from BLACK or PINK to shade the required area. Example: Area to the right of 1.746 under a tdistribution with 16 degrees of freedom. Let Us Try! .\\ Hello, I hope you are having a good day. Before proceeding to Lesson 3, let us check what you know by answering the following. Direction: Choose the letter of the correct answer and write the correct letter on the separate sheet of paper. 1. Which of the following is the formula for the margin of error for a sample mean? 5 A. E = Z0 (E) c. E = 20%) t 6 B. E = Zc() D. E = Zc() 2. Given: n=60, 6 =0.7, level of condence is 96%. What is the margin of error of the given data? A. 0.118 C. 0.185 B. 0.151 D. 0.158 3. Calculate the sample size of the given data: E = 5, 6 = 20, with a 95% condence level. A. 60 C. 62 B. 61 D. 63 O a Let Us Study We are enjoying many things at present because of the past. Gadgets, foods, cars, medicines, computers and other things came up into reality because of research which plays an important role in our life. Computing for the length of the condence interval is an important part of the research. As much as possible, we should use a high level of condence in estimating our parameters to give us correct information that helps us build stronger tomorrow. 4. Compute for the length of the confidence interval with a sample size of 50 and has a sample mean of 20 with standard deviation 18 having a 98% confidence level. A. [ 25.93 , 14.07 ] C. [ 24.30 , 13.57 ] B. [ 26.85 , 13.15 ] D. [ 24.26, 15.74 ] 5. A random sample of 150 stores from Abrizo Mall have mean average daily income of P25,000 and a standard deviation of P12,000. Find the 95% confidence interval for the mean average daily income of all stores in Abrizo Mall. A. [ P26,800 , P23,200] C. [ P26,900.60 , P23,099.40] B. [ P26,750.50 , P23,249.50] D. [ P26,920.40, P23,079.60] 6. How does the level of confidence affect the sample size? (Consider that other factors are constant A. The higher the level of confidence, the smaller the sample size required B. The lower the level of confidence, the larger the sample size required. C. The lower the level of confidence, the smaller the sample size required D. The sample size remains constant in every level of confidence. 7. A researcher wants his/ her estimate with a minimum margin of error. Which of the following is suggested? A. Reduce the sample size. C. Increase the sample size. B. Use a sample size of 30. D. Use a sample size of less than 25. 8. What is the sample size needed if the margin of error is 30 and the standard deviation of 45 with a 98% confidence level? A. 10 C. 12 B. 11 D. 13For you to be guided in this lesson, these are the following topics to be discussed: 1. Confidence Level vs Confidence Interval 2. Computing Margin of Error 3. Computing Confidence Interval 4. Computing Length of Confidence Interval 5. Computing Appropriate Sample Size Using the Length of the Confidence Interval 6. Factors Affecting Sample Size Determination CONFIDENCE LEVEL VS CONFIDENCE INTERVAL Confidence level is the likelihood measure of the confidence interval that is represented by a percentage that refers to all possible samples that can be estimated to contain the true population parameter. However, the level of confidence can be any quantity or number, but the most common values are shown in the table. Table 1. Confidence Level of Zc Table Confidence 99% 98% 96% 95% 92% 90% 85% 80% 70% Level 0.99 0.98 0.96 0.95 0.92 0.90 0.85 0.80 0.70 Zc 2.58 2.33 2.05 1.96 1.75 1.645 1.44 1.28 1.04 Confidence intervals (also called the interval estimates) are intervals which contain the actual values for our estimates, they are one way to represent how "good" an estimate is. They are important reminder of the limitations of the estimates. We must get the confidence interval when we estimate of something, for example the mean with known and unknown standard deviation. We compute for it to know how close we are to the actual value of parameter. However, this approximate may or may not contain the correct or true parameter value. COMPUTING MARGIN OF ERROR Margin of error is the range of values above and below the given statistical number or sample in a confidence interval. To compute for the margin of error, use the formula E = Zc() or E = Zx/2 (Vm) Where, Zc Or Zx/2 means the critical values or confidence coefficients, is a symbol for standard deviation, and as the sample sizeEXAMPLE 1 Martha owns a shoe store. She used 160 pairs of shoes as her samples with a price standard deviation of P75. Suppose that Martha wants a 95% level of confidence, what is the margin of error? Solution: Step 1: Write the given data, n = 160 8 = P75 95% confidence level where zo = 1.96 (refer Confidence Level of Zc Table) Step 2: Apply the formula and substitute the given data, E = Zc( 7) E = (1.96)(160 E = (1.96)(5.929) - (use three decimal places for partial answer) E = 11.62 - round off the final answer in two decimal places) COMPUTING THE CONFIDENCE INTERVAL Since you are already familiar on how to compute for the margin of error, you are now ready to compute for the confidence interval. The confidence interval can be written in the form of Lower limit