if you need the additional worksheets I have done I have them just not sure exactly what data you need to solve.
BUSI 230 Age Gender 43 F o What is the upper bound? 33 M o How do we interpret the results, in context of our study? 23 F 31 F (25 points) Part 4 - Hypothesis Testing (Module 6): Based on your sample, you will conduct a 43 M hypothesis test with a = 0.05 to test two of the claims of the above article. 21 M 42 F Using the same excel sheet as last week, answer the following in the "week 6" tab: 22 F . Claim: the average age of online students is 32 years old. Can you prove it is not? 16 M o What is the null hypothesis? 23 F o What is the alternative hypothesis? 36 M o What distribution should be used? 30 M What is the test statistic? 27 F o What is the p-value? 27 F What is the conclusion? 19 F o How do we interpret the results, in context of our study? 52 M Claim: the proportion of males in online classes is 35%. Can you prove it is not? 20 M o What is the null hypothesis? What is the alternative hypothesis? 48 F o What distribution should be used? 20 M o What is the test statistic? 58 M o What is the p-value? 54 M What is the conclusion? 20 F o How do we interpret the results, in context of our study? 20 M Listed below is the title of each part of the project and the module that corresponds with the submission. Module Homework Assignment Module Project: Data Collection Assignment Module 3 Project: Descriptive Statistics Assignment Module 5 Project: Confidence Intervals Assignment Module 6 Project: Hypothesis Tests Assignment Page 3 of 2Note: The goal of the project is to practice conducting a hypothesis test for a mean and proportion with rec about failed assumptions tests. Use primary methods described in text and used on homework. For the following two hypothesis tests, use alpha = .05 Points Claim: The average age of online students is 32 years old. Can you prove it is not? Ho: 32 TWO TALL Ha: 32 Note: Calculation cells should Sample mean: 0 31. 65 217 list the numbers and operations used to get your answers. Do Sample St. Dev: 0 12.81196 not put the generic formula and show all calculation steps. N Distribution: t. DISt X2 . 025 = 36:78 X2.975 = 10,987 Test Statistic: T. 3 *2 decimals Calculation: HNN p-value: . 1919 *4 decimals Decision: Interpretation: N (context) The average ace i online students Claim: The proportion of males in online classes is 35%. Can you prove it is not? 23 Total Ho: NP . 35 HalMeIN = 33, 58 6= 15,02 Ha: 3.5 5217 1783 Sample Proportion Males o Sample Proportion Females 0 N Distribution: TE, DUF Test Statistic: *2 decimals Calculation: - NW o-value *4 decimals Decision: Interpretation: 2 context 25 Total Points 31.65-. 35 31.30 12.81 V 23 2. 67 = 11.72 Rijob 36. 78 7 11.72 hoped 10. 982 5 1172%Busi 230 - November 13, 2021 Page 1 Age Gender For the following questions, use only the "age" column: 16 M 19 Points Age Frequency Distribution: Class Width 20 M 20 M Class Limits Relative Cumulative 20 Low High Midpoint Freq. Frequency Relative Freq Cum Freq 20 M Limits 24 20 10 0.4348 0.4348 10 21 M Freq 34 30 0.2174 0.6522 15 22 F Mid. 44 0.1739 0.8261 19 HNNNN 40 23 F 22 RF 54 50 0.1304 0.9565 23 F 60 1.0000 23 CF 0.0435 27 F 23.0000 1.0000 27 F Mean 31.65 "Round to two decimals 30 M Median 27.0 *Round to one decimal 31 F NNNNN Sample Standard deviation: 12.81 .Round to two decimals 33 M 20.0 *Round to one decimal 36 M Q3 43.0 *Round to one decimal 42 F 43 F 43 M Ogive: Ogive: w w 48 F Polygon 52 M 54 M 58 M Total: 25 728 23 Numbers Q1 = .25X23= 5.75 or Position 6 = 20 Q3 = .75X23 =17.25 or Position 18 = 43 OGIVE 25 20 Class Limits Class Limit Cum Freq Low High Freq 15 10 FREQUENCY un 10 40 50 10 20 30 Age GroupsNote: The goal of the project is to practice making a confidence interval for a mean and proportion with rec assumptions tests and do not make corrections for small sample size. Use primary methods described in te Age Gender Points 95% Confidence Interval for Average Age of Onlir 19 F 20 F Sample Mean: 31.65217 22 F Sample St. Dev: 12.81196 23 F H Sample Size: 23 F 27 F 2 Distribution: Normal Distribution 27 F 31 F 2 Critical Value: 1.96 *2 decimals 42 F 43 F Margin of Error: 5.23 *2 decimals Calculation: 48 F HAN Lower Bound: 26.42 *2 decimals Calculation: 16 M Upper Bound: 36.88 *2 decimals Calculation: 20 M 20 M Interpret 20 M 2 (context) 95% Certain that the population mean is between : 21 M 30 M 95% Confidence Interval for Proportion of Male On 33 M 36 M Sample Size: 23 43 M Number of Males: 12 52 M Male Proportion: 0.5217 Female Proportion 54 M 58 M N Distribution: -Normal Distribution 728 2 Critical Value: 1.96 *2 decimals 403 Male Sum 33.58333 Male Mean Margin of Error: 8.4984 *4 decimals Calculation: 01792 MaleStdev Lower Bound 25.0850 *4 decimals Calculation: Upper Bound: 42.0817 *4 decimals Calculation: Interpret N (context) 95% Certain that the male population mean is betv 25 Total Pointsal data. Do not worry about failed t and used on homework. e College Students: Note: Calculation cells should list the numbers and operations used to get your answers. Do not put the generic formula and show all calculation steps. 1.96 * 2.67 31.65 - 5.23 31.65 + 5.23 26.42 and 36.88. line College Students: 0.4783 *4 decimals 1.96 * 4.3359 33 - 8.498364 33.58333 + 8.498364 veen 25.0850 and 42.0817Busi 230 - November 13, 2021 lage 1 Age Gender For the following questions, use only the "age" column: 16|M 19 F Points Age Frequency Distribution: Class Width 20 M 20 M Class Limits Relative Cumulative 20 F Low High Midpoint Freq Frequency Relative Freq Cum Freq 20 M Limits 15 24 20 10 0.4348 0.4348 10 21 M Freq 25 34 30 0.2174 0.6522 15 22 F Mid. 0.8261 19 HNNNN 35 40 0.1739 23 F RF 45 54 50 0.1304 0.9565 22 23 F CF 55 60 0.0435 1.0000 23 27 F 23.0000 1.0000 27 F Mean 31.65 *Round to two decimals 30 M Median 27.0 *Round to one decimal 31 F NNNNN Sample Standard deviation: 12.81 *Round to two decimals 33 M Q1 20.0 *Round to one decimal 36 M Q3 43.0 *Round to one decimal 42 F 43 F 43 M Ogive: Ogive: w w 48 F Polygon 52 M 54 M 58 M Total: 25 728 23 Numbers Q1 = .25X23= 5.75 or Position 6 = 20 Q3 = .75X23 =17.25 or Position 18 = 43 OGIVE 25 20 Class Limits Class Limit Cum Freq Low High Freq 15 10 FREQUENCY 10 10 20 30 40 Age GroupsMary Weigandt - Prof Andrew Past 2 Busi 230 - November 13, 2021 Frequency Polygon Class Limits Freq. Low High Midpoint 15 24 20 10 25 34 30 44 40 54 50 55 64 60 23.0000 Frequency Polygon 12 FREQUENCY 20 30 40 50 60 MIDPOINTNote: The goal of the project is to practice making a confidence interval for a mean and proportion with assumptions tests and do not make corrections for small sample size. Use primary methods described in Age Gender Points 95% Confidence Interval for Average Age of Onlit 19 F 20 F Sample Mean: 31.65217 22 F Sample St. Dev: 12.81196 23 F Sample Size: 23 23 F 27 F 2 Distribution: Normal Distribution 27 F 31 F 2 Critical Value: 1.96 *2 decimals 42 F 43 F N Margin of Error: 5.23 *2 decimals Calculation: 48 F 1 Lower Bound: 26.42 *2 decimals Calculation: 16 M Upper Bound: 36.88 *2 decimals Calculation: 20 M 20 M Interpret 20 M 2 (context) 95% Certain that the population mean is between 21 M 30 M 95% Confidence Interval for Proportion of Male Or 33 M 36 M Sample Size: 23 Total Sample = 23 for Male 43 M Number of Males: 12 52 M 2 Male Proportion: 0.5217 Female Proportion 54 M 58 M 2 Distribution: Normal Distribution 728 2 Critical Value: 1.96 *2 decimals 403 Male Sum 33.58333 Male Mean Margin of Error: 8.4984 *4 decimals Calculation: 15.01792 MaleStdev HEN Lower Bound: 25.0850 *4 decimals Calculation: Upper Bound: 42.0817 *4 decimals Calculation: Interpret 2 (context) 95% Certain that the male population mean is bet 25 Total Pointscontext of our study ? (25 points) Part 4 - Hypothesis Testing (Module 6): Based on your sample, you will conduct a hypothesis test with a = 0.05 to test two of the claims of the above article. Using the same excel sheet as last week, answer the following in the "week 6" tab: Claim: the average age of online students is 32 years old. Can you prove it is not? What is the null hypothesis? What is the alternative hypothesis? O What distribution should be used? o What is the test statistic? What is the p-value? What is the conclusion? How do we interpret the results, in context of our study? . Claim: the proportion of males in online classes is 35%. Can you prove it is not? o What is the null hypothesis? O What is the alternative hypothesis? What distribution should be used? O What is the test statistic? O What is the p-value? O What is the conclusion? How do we interpret the results, in context of our study? Listed below is the title of each part of the project and the module that corresponds with the submission. Module Homework Assignment Module 1 Project: Data Collection Assignment Module 3 Project: Descriptive Statistics Assignment Module 5 Project: Confidence Intervals Assignment Module 6 Project: Hypothesis Tests Assignment
