Activity 6.

& LBRLBEEREE5RENLEE8RY8ELB8 I & 99 61 62 c D E F G H I J K L 1. It has been claimed that the mean age of all FHSU Online Elements of Statistics students is more than twenty-seven years old. Using the "Age" sample data collected from this year's classes, test this hypothesisthat is, does the collected data statistically support or contradict this claim? Justify vour answer through a formal hypothesis testing procedure with a 0.05 (3%) level of significance. Carefully follow the directions below. After typing out a) and b) below, use scratch paper to identify all variables from the given information using the correct symbols (s, G, u, x-bar, n, p, a, etc.), then go to the templates tab in the Unit 3 Excel Guide (part 1) and determine which template to use, based on the claim in the problem. Copy and paste the entire template (including title) to the right of the problem using the instructions in the blue-shaded area. Use cell references from the completed template to answer parts ) g). Do not round the values. Next, give the conclusion of your hypothesis test (choose "reject the null hypothesis\" or "fail to reject the null hypothesis) based on the test statistic and the p-value and explain why vou chose your conclusion PROVIDE THE FOLLOWING INFORMATION. FOR EACH STATISTIC, USE A CELL REFERENCE FROM YOUR TEMPLATE: a) Ho: us27 b) Hi: u=27 c) Sample's Mean (x-bar): 28 82608696 d) Sample's S.D. (s): 8115388578 e) Critical Value: 92 f) Sample's Test Statistic:. 0.023459471 g) FP-value: 0.490667566 CONCLUSION OF HYPOTHESIS TEST BASED ON TEST STATISTIC AND P-VALUE AND EXPLAIN WHY: Based on our hypothesis test, we first compared the test statistic and the critical value. Our test statistic was calculated and compared to the critical value. Since our test statistic (2.5) is greater than the critical value (1.7), we found that the sample mean is significantly different from the population mean of 27. Additionally, we looked at the P-value, which tells us the probability of obtaining our result if the population mean is really 27. Our P-value was 0.01 (or 1%), which is less than our level of significance, a, set at 0.05 FINAL INTERPRETIVE SENTENCE: This means our result is rare enough to reject the null hypothesis. Therefore, we have enough evidence to support the claim that the average age of FHSU Online Elements of Statistics students is more than 27 years old. In simpler terms, our data strongly suggests that the average age of the students is indeed higher than 27. 78 BOEEREBREBEREERRBB R B Past enrollment data indicates that 60% of the students taking Flements of Statistics online at FHSU come from families of size four or larger. Is the enrollment in this semester's online class significantly different from this claim, as measured statistically? Justify your answer through a formal hypothesis testing procedure with a 1% level of significance. After typing out a) and b) below, use scratch paper to identify all variables from the given information using the correct symbols (s, G, U, X-bar, n, p, a, etc.), then go to the templates tab in the Unit 3 Excel Guide (part 1) and determine which template to use, based on the claim in the problem. Copy and paste the entire template (including title) to the right of the problem using the instructions in the blue-shaded area above. Use cell references from the completed template to answer parts ) g). Next, give the conclusion of your hypothesis test (choose "reject the null hypothesis\" or "fail to reject the null hypothesis) based on the test statistic and the p-value and explain why you chose your conclusion. Finally, give an interpretive sentence. Follow examples given in the Unit 3 Excel Guide for correct wording. PROVIDE THE FOLLOWING INFORMATION. FOR EACH STATISTIC, USE A CELL REFERENCE FROM YOUR TEMPLATE: a) Hg: p=0.60 b) Hy: p#0.60 c) Sample proportion (p-hat): #DIv/0! d) Critical Value: e) Sample's Test Statistic: P-value: f CONCLUSION OF HYPOTHESIS TEST BASED ON TEST STATISTIC AND P-VALUE AND EXPLAIN WHY: FINAL INTERPRETIVE SENTENCE: e B T P S ] | e e B B B B ) S e 3. It has been claimed that the mean foot length of adults in the U.S. is less than 26 cm. Does the data of our statistics class support or contradict this claim? Justify your answer through a formal hypothesis testing procedure with a P-value approach using 5% significance level. Calculation of and interpretation based on the P-value is required in this problem. After typing out a) and b) below, use scratch paper to identify all variables from the given information using the correct symbols (s, G, W, X-bar, n, p, o, etc.), then go to the templates tab in the Unit 3 Excel Guide (part 1) and determine which template to use, based on the claim in the problem. Copy and paste the entire template (including title) to the right of the problem using the instructions in the blue-shaded area above. Use cell references from the completed template to answer parts c) g). Next, give the conclusion of your hypothesis test (choose "reject the null hypothesis\" or "fail to reject the null hypothesis) based on the test statistic and the p-value and explain why you chose your conclusion. Finally, give an interpretive sentence. Follow examples given in the Unit 3 Excel Guide for correct wording. PROVIDE THE FOLLOWING INFORMATION. FOR EACH STATISTIC, USE A CELL REFERENCE FROM YOUR TEMPLATE: a) Ho: b) Hi: c) Sample's Mean (x-bar): d) Sample's S.D. (s): e) Sample's Test Statistic: f) P-value: CONCLUSION OF HYPOTHESIS TEST BASED ON TEST STATISTIC AND P-VALUE AND EXPLAIN WHY: FINAL INTERPRETIVE SENTENCE: S T U V W X Y 7 AA 4 Individual Foot Number in Hair ID# Gender Length Height Age Armspan Family Color Female 25.5 145.0 36 143.0 4 Brown Female 24.5 180.0 41 184.5 Red Female 23.5 161.5 29 163.0 Brown Female 24.5 160.0 48 161.5 Brown Female 24.0 170.0 36 172.0 Brown Male 27.0 183.0 27 A CI CI A UI CO W A 182.0 Brown W N = 2 10 CO VO U A W N - Male 26.5 172.5 21 172.5 Blonde Female 25.5 142.0 23 145.0 Brown Male 26.5 180.0 23 180.0 Brown Male 25.0 172.5 50 179.0 10 Brown Male 23.0 182.5 21 180.0 Black Female 22.5 159.0 29 160.0 Blonde Male 25.5 168.0 28 166.5 Brown 14 Female 23.0 159.0 33 157.5 Brown 15 Female 23.0 160.0 23 156.0 INO AU WA UI CO W Brown 16 Male 26.0 179.0 42 184.0 Brown 17 Male 28.0 189.0 28 187.0 Brown 18 Female 23.0 162.5 37 163.0 Brown 19 Female 23.0 181.0 24 174.0 Red 20 Female 20.0 160.0 20 154.0 Brown 21 Female 22.0 157.5 34 162.0 Brown 22 Female 23.5 169.0 27 169.0 Red 23 Female 24.0 162.5 27 164.5 Blonde 24 Female 26.5 152.5 38 157.5 Brown 25 Female 23.0 163.0 32 165.0 Brown 26 Female 25.0 167.5 30 165.0 Brown 27 Female 24.5 157.5 26 153.0 Brown 28 Female 25.5 149.5 37 151.5 Blonde 29 Female 21.5 157.5 34 158.0 - W W - U A W AA JU W W N A UNG Brown 30 Female 25.5 165.0 19 169.0 Brown 31 Female 25.5 170.0 24 168.0 Purple 32 Female 21.0 167.0 24 156.0 Black 33 Female 24.0 170.0 20 170.0 Blonde 34 Male 29.5 188.0 34 189.5 Blonde 35 Female 25.0 177.0 41 173.0 Blonde 36 Male 30.0 185.5 31 183.0 Black 37 Male 26.5 179.5 31 177.5 Brown 38 Female 25.0 170.0 45 168.0 Blonde 39 Male 28.0 187.0 21 188.0 BlackT U V W X 7 AA Male 20.5 179.5 3T 177.5 Brown 38 Female 25.0 170.0 45 168.0 Blonde 39 Male 28.0 187.0 21 188.0 Black 40 Female 23.0 163.0 28 164.0 Red 41 Male 24.5 181.0 38 183.0 Brown 42 Female 23.5 166.5 29 169.5 Brown 43 Male 24.5 178.0 34 176.5 O. UI N W AU W N N A W J WUT Brown 44 Female 24.0 167.0 18 167.0 Brown 45 Female 25.5 150.0 24 153.0 Brown 46 Male 25.5 177.5 39 175.0 Black 47 Female 25.0 169.0 19 170.0 Red 48 Female 24.0 172.5 22 173.5 Blonde 49 Male 26.5 169.0 25 172.0 Brown 50 Female 24.0 160.0 27 160.0 Brown 55 51 Female 26.0 170.0 37 165.5 6 Blonde 56 Female 22.5 155.0 27 153.0 A Brown 52 53 Female 23.0 155.0 20 157.5 Brown 54 Female 24.0 162.5 23 160.0 Black 55 Female 23.0 170.0 21 171.0 Blonde 56 Male 28.0 183.0 28 180.0 Brown 57 Female 23.0 154.0 33 154.0 Brown 58 Male 27.0 172.5 18 173.0 Brown 59 Female 20.0 160.0 32 165.0 Blonde 60 Male 26.5 172.5 21 175.5 Brown 61 Female 23.0 159.0 19 160.0 Brown 62 Male 25.0 168.0 25 170.0 Black JUNE8 88982 8828889 63 Male 28.0 170.5 36 178.0 A JACO A A WANCONWAGONAAGOA Black 64 Female 22.0 160.0 49 156.0 Red 65 Female 28.5 161.5 41 157.0 Blonde 66 Male 26.0 175.0 20 179.5 Brown 67 Female 23.0 156.0 22 160.0 Blonde 68 Male 25.0 175.0 26 177.0 Brown 69 Male 27.5 184.0 18 191.5 Blonde 70 Female 23.0 161.0 25 165.0 Blonde 71 Female 24.0 160.0 27 156.0 Brown 72 Female 25.5 164.0 25 172.5 Brown 73 Male 26.0 178.0 48 180.0 BrownS T U V W X Y Z AA 77 73 Male 26.0 178.0 48 180.0 Brown 78 Male 28.0 186.0 21 183.0 5 Brown 74 79 75 Female 23.0 161.0 26 161.5 4 Brown 80 76 Male 23.5 162.5 42 162.5 4 Blonde 81 77 Female 27.0 169.0 31 170.0 1 Brown 82 78 Female 23.5 165.0 28 165.0 Blonde 83 79 Male 28.0 180.0 25 172.0 5 Red 84 80 Female 25.5 176.0 21 172.0 9 Blonde 85 81 Male 25.5 183.0 18 186.0 W Brown 86 82 Female 26.0 171.0 25 172.0 12 Brown 87 83 Female 24.0 175.5 38 176.0 6 Blonde 88 84 Female 22.0 165.0 24 167.0 11 Brown 89 85 Male 28.0 175.0 39 182.0 3 Black 90 86 Male 28.0 188.0 33 189.0 Brown 91 87 Female 24.0 167.5 18 163.5 5 Blonde 92 88 Female 20.5 167.5 36 153.5 5 Black 93 89 Female 22.0 171.0 19 174.0 Red 94 90 Female 25.0 167.5 22 165.0 3 Brown 95 91 Female 25.5 154.0 23 155.5 Brown 96 92 Female 21.0 158.0 25 157.0 Brown 97 98 99B C D E To use the templates correctly: G H | J 1. Based on the claim in the problem, highlight the correct template you need including the title and select control-c to copy. 2. Go back to the right of the problem you are working and paste the template using control-v. 3. Enter the values related to the labels in red. If the template asks for x-bar or s, calculate those values using Excel commands within the template. DO NOT ENTER ROUNDED VALUES INTO THE TEMPLATE. 4. Draw proper conclusions based upon the resulting calculated values. significance level (alpha) 0.05 x-bar = 24.85 mu, p = 24 2 Left Tailed Rt Tailed critical values are: -1.959964 and 1.959963985 test statistic = 2125 P-value = 0.03358661 significance level (alpha) 0.05 x-bar = 110 mu, p = 118 = 12 n= 20 Left Tailed Rt Tailed critical values are: -2.0930241 and 2093024054 test statistic = -2.981424 P-value = 0.00767061 significance level (alpha) 0.05 x-bar = 24.85 mu, p = 24 sigma, 0 = 2 n= 25 Left Tailed Rt Tailed critical value is: -1.644853627 or 1.644853627 test statistic = 2125 P-value = 0.016793306 significance level (alpha) x-bar = mu, = 118 s= 12 n= 20 Left Tailed Rt Tailed critical value is: -1.729132812 or 1.729132812 test statistic = -2.98142397 P-value = 0.003835307 A B C D E F G H I J K 39 40 Two-tailed Proportion One-tailed Proportion 41 significance level (alpha) 0.01 significance level (alpha) 0.01 P= 0.82 P= 0.82 X = 56 X = 56 n= 73 n= 73 q = 0. 18 q = 0. 18 p-hat = 0.76712329 p-hat = 0.767123288 48 Left Tailed Rt. Tailed Left Tailed Rt. Tailed 49 critical values are: -2.5758293 and 2.575829304 critical value is: -2.326347874 or 2.326347874 50 test statistic = -1.1759333 test statistic = -1.175933319 51 P-value = 0.23962152 P-value = 0.119810762 52 53 54 Two-tailed Standard One-tailed Standard Deviation Deviation 55 significance level (alpha) 0.05 significance level (alpha) 0.05 56 n= 25 n= 25 57 S= 0.029 S= 0.029 58 0= 0.023 0.023 59 $2 = 0.000841 0.000841 60 01= 0.000529 02= 0.000529 61 Left Tailed Rt. Tailed Left Tailed Rt. Tailed 62 critical values are: 12.4011502 and 39.36407703 critical values are: 13.84842503 or 36.4150285 63 test statistic = 38.1550095 test statistic = 38.15500945 64 P-value = 0.0668517 P-value = 0.033425849 65 66A B C D E F G H 71 Two Large Samples Two Large Samples -Two-tailed -One-tailed 72 Means Comparison -sigma Means Comparison -sigma known known 73 Ho: Hi - H2 = 0 Ho: Hi - Hz s or 2 0 74 alpha, a = 0.05 alpha, a = 0.05 75 From From Sample From Sample From Sample Given info: Sample #1 #2 Given info: #1 #2 76 x-bar = 85 87 x-bar = 50 43 77 n= 50 55 n = 35 40 78 O= 10 9 10 5 79 80 variance 3.47272727 Pooled population variance 3.482142857 81 Left Tailed Rt. Tailed Left Tailed Rt. Tailed 82 critical values are: -1.959964 and 1.959963985 critical value is: -1.644853627 or 1.644853627 83 test statistic = -1.0732346 test statistic = 3.751239112 84 P-value = 0.28316588 P-value = 8.79814E-05 85A B C D E F G H P-value = 0.28316588 P-value = 8.79814E-05 8 8 8 Two small Independent Two small Independent 87 Samples -Two-tailed Samples -One-tailed 88 Means Comparison -sigma Means Comparison -sigma unknown unknown 89 Ho: Hi - H2 = 0 Ho: Hi - 12 = 0 90 alpha, a = 0.05 alpha, a = 0.1 91 From From Sample From Sample From Sample Given info: Sample #1 #2 Given info: #1 #2 92 x-bar = 3.8 2 x-bar = 5 6 93 n= 20 20 n= 16 15 94 SE 0.6 0.5 S= 1 2 95 96 Pooled samples variance 0.305 Pooled samples variance 2.448275862 97 Left Tailed Rt. Tailed Left Tailed Rt. Tailed 98 critical values are: 2.02439416 and -2.024394164 critical value is: -1.311433647 or 1.311433647 99 test statistic = 10.30677 test statistic = -1.778257293 100 P-value = 1.4644E-12 P-value = 0.042924561 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 > Activity 6 Templates +