2 STATISTICS AND PROBABILITY ACTIVITY!!!

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0 Q Let Us Study You have learned from the previous lesson the different properties of tdistribution that are very important in this lesson. 1. > The total area under the tdistribution curve is equal to 1 or 100%. The area under the tdistribution curve also represents the probability associated with specic sets of tvalues. That means given the tvalue, you can compute for the area or probability with the use of a table. A t value or tstatistic tells us how many standard deviations from the mean is the given score. The set of tvalues are usually written below the horizontal axis of the tdistribution curve. The exact shape of the tdistribution depends on the degrees of freedom. Remember that the lesser the degree of freedom, the lower is its peak and the thicker is its tails. As the degree of freedom increases, the tails become atter, and the peak becomes higher. That means, given the area or the probability, the tvalue depends on the number of degrees of freedom. 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. {L4 [1.3 d. = 35 0.2 Mi The t-Distribution table In finding the areas and percentiles for a t-distribution you need to familiarize yourself with the t-table. Contidence get Intervals .berg 6 0.10.; 1.05 $0.005 0.01 63.657 3.078 5,314 12.706 31.527 1: 2.520 MADLY MERCK303 WHICH 6065 3 9.925 3.182 4.541 5.841 3 1.638 2.353 4.604 4032 1.476 2.015 2571 1.065 2447 - 3.143- 3.707 1.895 2 365 2.948 3 499 7 1.415 7397. 2 306 04 12.B96- 3.355 9 1.383 1.833 2 262 2.821 3.250 3.169 UMMM 30 1372 ON 2.764 2.201 2.718 3.106 11 1.363 1.796 2 179 . 26213 2.650 3.012 13 1.350 1.771 2.150 2.077 2.602 15 1.341 1.753 2131 2.947 Fact: 2.120 - 2.583 2.921 1.740 2.110 2.567 2.098 :17 1.332 PPA 2.552 WILL 2.878 1.328 1,729 2.043 2.539 2.801 19 2.085 2.845 141.725 2.518 2.831 21 1:323 1.721 2.080 45 - 1321 20 1 2.074 2 508 2 4 4 2.819 1.71 2.063 2500 2.807 1.319 O 2.064 MLM 2492 HHHH 2.797 2485 2.787 1.316 1.702 2.060 25 26 2.056 2.779 2.771 27 1.214 1.7:13 2.052 2473 2.763 29 1.699 2.045 2462 2.756 1.311 2.750 2 449 2 731 32 1.309 1.094 2.037 2.728 2.025 2.434 2.719 36 1,305 1.625 2712 2.429 40 1,303 1.684 2.021 1 2.423 2.704 2412 3 2 690 2678 - 1.209 1.676 2.10 2.403 2 660 1.290 1.671 2.030 2.390 - 2.385 7 2651 HALFTIME 1 295 TH 1.667 1.904 2.381 2645 1.294 2.643 1.030 2.374 2.639 1.292 1.064 2.612 1.660 1.984 2.364 2.626 100 1.290 2.586 1.962 2.330 2.581 1010 1.282 1.046 13 2 326- - 17 2.576 Two tails "This value has been rounded to 1 28 h the textbook, One tail This volvo has boon rounded to 1.65 in the tordbock. The value has been rounded to 2.$3 in the toxchock 'This value has been rounded 15 2.58 In the textbook. Source: Adapted from W. H. Bayer, Handbook of Tables for Probably and Stunmiles And od, CAC Prank, Boon Boon, Ho., 1065, Roprinted with ponrissinLooking 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 0.20 0.10 0.04 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, a 0.10 0.05 0.025 0.01 0.005 d.r. Two tails, a 0,20 0.05 0.02 0.01 3.078 14 12.706 31.821 53.657 1.886 4.303 5.965 9.925 1.638 3.182 4.541 5.841 1.533 2.776 3.747 1.601 1.476 2.571 3.365 4.012 1.943 2447 3,143 3.707 1.413 2.365 2.998 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. I "17.151;in EIETII 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 92'"1 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 O.15(1 0.85 =O.15, recalling that the area of the tdistribution is 1). IIIZA%E ANIHI IIEEEE . Till -5-4-3-2-10 2345 85th Percentile Illustrative Examples: 1. Find the 95th percentile of a tdistribution with 12 degrees of freedom. To nd the tvalue, sketch the graph to locate the 95th percentile in the t distribution. From the definition, 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 t-table to find the 95th percentile or the t-value. 0.5 0.5 0.4 0.4 0.3 0.3 0.2 0.2 0.1 95% 5% 0.1 95% 15% 0.0 0.0 -5 -4 -3 -2 -1 0 1 2 3 4 5 -5 -4 -3 -2 -1 0 1 2 3 5 95th percentile = ? 1.782 Looking at the graph above, we formed one tail only, therefore, in the t-distribution, 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. The t Distribution Confidence intervals 80% 90% 95% 98% 99% One tail, c 0.10 0.05 0.025 0.01 ).005 0.01 d.f. Two tails, c 0.20 0.05 3.078 6314 12.706 31.821 63.657 1.886 2.920 4.303 6.965 9.925 1.638 2.353 3.182 4.541 5.841 1.533 2. 13: 2.776 3.747 4.604 1.476 2.015 2.571 3.365 4.032 1.440 1.943 2.447 3.143 3.70 1.415 1.$95 2.365 2.998 3.49 1.397 1.$60 2.306 2.896 3.355 1.383 1.$33 2.262 2.821 3.250 1.372 1.$12 2.228 2.764 3.169 1.363 2.201 2.718 3.106 1.356- 1.782 2.179 2.681 3.055 1.350 1.771 2.160 2.650 3.017 1.345 1.761 2.145 2.624 2.97 1.341 1.753 2.131 2.602 2.947 1.337 1.746 2.120 2.583 2.921 17 1 373 1 740 110 7 567 2. Find the 2.5th percentile of a t-distribution 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 t-distribution). Remember, one of the properties of the t-distribution is that it is symmetric about zero, meaning the area to the right of the distribution is the same area to the left. 0.5 0.5 0.4 0.4 0.3 0.3 0.2 1. 2 0 . 1 0. 1 0.025 9.975 0.025 0. 0 0. 0- -5 - -1 0 2 3 4 -5 -4 -3 2 -1 2 2.5th percentile or t-value =? -2.145 The table shows that the area of 0.025 to the right of the distribution has a t-value of 2.145. Thus, the t-value with an area to the left of 0.025 must be - 2.145. Therefore, the 2.5th percentile is -2.145. The t Distribution Confidence intervals 80% 90% 95% 98% 99% One tail, ox 0.10 0.05 0.025 0.01 0.005 d.f. Two tails, a 0.20 0.10 0.02 0.01 3.078 6.314 2.706 $1.821 63.657 1.886 2.920 3 3 6.965 9.925 1.638 2.353 3.1 2 4.54 5.841 1.533 2.132 3.747 4.604 1,476 2.015 2.5 1 3.365 4.032 1.440 1.943 3.14 3.707 1.415 1.895 2.3 5 2.998 3.499 1.397 1.860 306 .89 3.355 GEESFEB.NOVAWN- 1.383 1.833 2 2 2 2.821 3.250 1.372 1.812 2.2 8 2.764 3.169 1.363 1.796 2.20 1 2.718 3.106 1.356 1.782 2.179 2.681 3.055 1.350 1.371 2.650 3.012 1.345 1.761 2.145 2.624 2.977 1.341 1.753 2.13 2.602 2.947 1.337 1.746 2.120 2.583 2.921 1.333 1.740 2.110 2.567 2.898 3. Find the t-values that bound in the middle 95% with df = 8. As you can see in the illustration, it consists of two tails. Thus, we are going to use two tail in the t-distribution and we are looking for the two (2) values of t. The area of both tails is 5% or 0.05. The t Distribution 0.5 Confidence 0.4 intervals 80% 90% 95% 98% 99% 5% 0.3 One tail a 0.10 0.05 0.025 0.01 0.005 d.f. Two tails, a 0.20 0.10 0.05 0.02 0.01 0.2 3.078 6.314 31.821 63.657 0. 1.886 2.920 4.303 6.965 .925 95% 1.63 2.353 3.182 4.541 5.841 0.0 1.533 2.132 2. 176 3.747 4.60 -1 0 1 1.476 2.015 2. 71 3.36 4.032 1.440 1.943 3.14 3.707 1.415 1.895 2.998 3.499 1.397 1.860 2.896 3.35 1.383 1.833 2.821 3.250 .372 1.812 2.22 2 76 3.16 1.363 1.796 2.201 2.718 3.10 1.356 1.782 2.179 2.681 3.055Therefore, the tvalues that bound in the middle of 95% are 2.306 and 2.306. EEBDEE D'MW'U' There is also an alternative way in identifying tvalue of the given distribution. 'flJEivEuTll'm [gm-mummi- (11. 63.657 9.925 5 8M 4.604 4.032 3707 Using one tail 33:: distribution, we divide 5% ifs: or 0.05 by 2 to determine aim 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 n.4~ , , , ,_ , m:::=ga:::: 0.2 3 I\" MIIIEEEHIII ll' IBil -5-4-32-1 345 11.8 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 0.10 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. mammal 7 Condence I . intervals mum-mm:- o -au-.Ig~uuhw|um " 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). For 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 (T) or E = Zx/2 (Vm) Where, Zc Or Zx/2 means the critical values or confidence coefficients, is a symbol for standard deviation, and n 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 ?75. Suppose that Martha wants a 95% level of condence, what is the margin of error? Solution: Step 1: Write the given data, n = 160 6 = P75 95% condence level where zC = 1.96 (refer Condence Level of ZC Table) Step 2: Apply the formula and substitute the given data, 0' E = 24V?) 75 E = (1.96)(5.929) >(use three decimal places for partial answer) E = 1 1.62 > round off the nal 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 condence interval. The condence interval can be written in the form of Lower limit