2-3 Assignment Hypothesis Testing Using Minitab

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2-3 Assignment: Hypothesis Testing Using Minitab® By Laura Turgeon December 17, 2023 QSO-620 Six Sigma Quality Management Professor: Thomas Timmins

Problem 1 Jeffrey, as an eight-year-old, established an average time of 16.43 seconds for swimming the 25-yard freestyle, with a standard deviation of 0.8 seconds. His dad, Frank, thought that Jeffrey could swim the 25-yard freestyle faster by using goggles. Frank bought Jeffrey a new pair of expensive goggles and timed Jeffrey for 15 25-yard freestyle swims. For the 15 swims, Jeffrey's average time was 16 seconds. Frank thought that the goggles helped Jeffrey to swim faster than the 16.43 seconds. Conduct a hypothesis test using a preconceived α=0.05. Mean Time: (ẍ) 16 Standard Deviation (Ơ) .8 Sample Size (n): 15 Significance Level (x): .05 Null Hypothesis u=16.43 Alternative Hypothesis u < 16.43 Rule: Reject Null Hypothesis if X > P-Value Minitab Calculations: One-Sample Z Test of μ = 16.43 vs < 16.43 The assumed standard deviation = 0.8 N Mean SE Mean 95% Upper Bound Z P 15 16.000 0.207 16.340 -2.08 0.019 P-Value is .019 Conclusion : Since X (.05) is > than the P-value (.019) the null hypothesis should be rejected. With this information, I am suggesting that Jeffery can in fact swim the 25-meter free-style faster than 16.43 seconds if he does have the new goggles.

Problem 2 A college football coach thought that his players could bench press an average of 275 pounds. It is known that the standard deviation is 55 pounds. Three of his players thought that the average was more than that amount. They asked 30 of their teammates for their estimated maximum lift on the bench press exercise. The data ranged from 205 pounds to 385 pounds. The actual different weights were (frequencies are in parentheses) 205(3); 215(3); 225(1); 241(2); 252(2); 265(2); 275(2); 313(2); 316(5); 338(2); 341(1); 345(2); 368(2); 385(1). Conduct a hypothesis test using a 2.5% level of significance to determine if the bench press average is more than 275 pounds . Mean Weight: (ẍ) 286.2 Standard Deviation (Ơ) 55 Sample Size (n): 30 Significance Level (x): .025 Null Hypothesis: µ = 275 Alternative Hypothesis: u > 275 Rule: Reject Null Hypothesis if P-Value < .025 Minitab Calculations: One-Sample Z Test of μ = 275 vs > 275 The assumed standard deviation = 55 N Mean SE Mean 97.5% Lower Bound Z P 30 286.2 10.0 266.5 1.12 0.132 P-value is .132

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Conclusion : Since the P-value of .132 > X value of .025, we do not reject the Null Hypothesis. The evidence of this problem is simply not conclusive enough to show that the true mean weight lifted by the football team is more than 275 lbs. Problem 3 Statistics students believe that the average score on the first statistics test is 65. A statistics instructor thinks the average score is higher than 65. He samples ten statistics students and obtains the scores 65; 65; 70; 67; 66; 63; 63; 68; 72; 71. He performs a hypothesis test using a 5% level of significance. The data are from a normal distribution. Sample Size (n): 10 Significance Level (X): .05 Mean Score: 67 We will use a t-test in Minitab to calculate the standard deviation since it is not given. This test is a better option when there is a normal distribution among the variables and the population standard deviation is not known. T-test using Minitab Null Hypothesis U = 65 Alternative Hypothesis U > 65 Rule: Reject Null Hypothesis if X > P-Value Minitab Calculations: One-Sample T: C1 Test of μ = 65 vs > 65

Variable N Mean StDev SE Mean 95% Lower Bound T P C1 10 67.00 3.20 1.01 65.15 1.98 0.040 P-Value is 0.04 Conclusion : Since the P-value of 0.4 is < 0.5, we reject the Null Hypothesis. Just as the teacher expected, the average test score is greater than 65. Problem 4 Joon believes that 50% of first-time brides in the United States are younger than their grooms. She performs a hypothesis test to determine if the percentage is the same or different from 50% . Joon samples 100 first-time brides and 53 reply that they are younger than their grooms. For the hypothesis test, she uses a 1% level of significance. Sample size (n): 100 Significance Level (x): .01 53/100 brides were younger than their grooms, while 47/100 brides were older. This can be moved to a percent: 53% were younger; 47% were older Null Hypothesis p=.50 Alternative Hypothesis p≠.50 Minitab Calculations: Test and CI for One Proportion Test of p = 0.5 vs p ≠ 0.5 Sample X N Sample p 99% CI Z-Value P-Value 1 53 100 0.530000 (0.401441, 0.658559) 0.60 0.549 Using the normal approximation.

The P value is calculated at .549. Conclusion: In this proportioned test, we will compare the X and P values. The X value of 0.1 < the P value of 0.549. This means that with a 99% confidence level, we do not reject the null hypothesis. The evidence is not conclusive enough to show that more first-time brides are younger than their soon to be husbands. Problem 5 Suppose the proportion of households that have three telephone numbers is 30%. The telephone company has reason to believe that the proportion of households is less than 30%. Before they start a big advertising campaign, they conduct a hypothesis test. Their marketing people survey 150 households with the result that 43 of the households have three telephone numbers. Sample Size (n): 150 Significance Level (x): .05 (This is assumed as it is not stated.) The test will be: Null Hypothesis p=0.30 Alternative Hypothesis p ≠0.30 43 of the 150 households surveyed do have three phone numbers (43/150, .29) and 107 households do not have three phone numbers. (107/150, .72) 29% do have 3 phone numbers; 72% do not have 3 phone numbers Minitab Calculations: Test and CI for One Proportion Test of p = 0.3 vs p ≠ 0.3

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Sample X N Sample p 99% CI Z-Value P-Value 1 43 150 0.286667 (0.214300, 0.359033) -0.36 0.722 For this one, we have to use estimation: P-Value is .722 The P Value of .722 X (.05) < the p value of .722). Conclusion : Since the P-value is < the p-value, we do not reject the null hypothesis. There is not enough conclusive evidence to show that the proportion of households that have three phone numbers is not 30%.

References Dean, S. & Illowsky, B. (2023). Hypothesis Testing of Single Mean and Single Proportion: Examples. Retrieved from https://web.archive.org/web/20080719213049 /http://cnx.org/content/m17005/latest/

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