• 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?
  • What distribution should be used?
  • 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?
  • What is the null hypothesis?
  • What is the alternative hypothesis?
  • What distribution should be used?
  • 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?

Note: The goal of the project is to practice conducting a hypothesis test for a mean and proportion with real data. Do not worry 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? 1 Ho: N Ha: Note: Calculation cells should list the numbers and operations used to Sample mean: 32.61905 get your answers. Do not put the Sample St. Dev: 11.21372 generic formula and show all calculation steps 2 Distribution: Test Statistic *2 decimals Calculation: H NN p-value: *4 decimals Decision: Interpretation: (context) Claim: The proportion of males in online classes is 35%. Can you prove it is not? Ho: Ha Sample Proportion Males 0.4762 Sample Proportion Females 0.5238 2 Distribution: Test Statistic: *2 decimals |Calculation: - NW p-value: *4 decimals Decision: Interpretation: N (context) 25 Total PointsNote: The goal of the project is to practice making a confidence interval for a mean and proportion with real data. Do not worry about failed assumptions tests and do not make corrections for small sample size. Use primary methods described in text and used on homework. Age Gender Points 95% Confidence Interval for Average Age of Online College Students 26 M 32 F Sample Mean: 32.61905 34 F Sample St. Dev: 11.21372 Note: Calculation cells should list 18 F 1 Sample Size: 21 the numbers and operations used 49 F to get your answers. Do not put 26 M 2 Distribution: T-Distribution the generic formula and show all 45 M calculation steps 20 M 2 Critical Value: 2.09 *2 decimals 23 M 18 F Margin of Error: 5.1 *2 decimals Calculation: 20 * s = 2.086 *11.21372 / 21 = 5 37 F Lower Bound 27.51 *2 decimals Calculation: 32.61905 - 5.10442 = 27.51463 18 F Upper Bound: 37.72 *2 decimals Calculation: 32.61905+ 5.10442 = 37.72347 40 F 35 M Interpret We are 95% confident that the true average age of all online college students will lie in 25 M 2 (context) between (27.51, 37.72). 18 M 44 M 95% Confidence Interval for Proportion of Male Online College Students 41 F 52 F Sample Size: 21 34 M N H P Number of Males: 10 20 F Male Proportion 0.4762 Female Proportion 0.5238 *4 decimals 2 Distribution: Normal Distribution 2 Critical Value: 1.96 *2 decimals Margin of Error: 0.2136 *4 decimals Calculation: 1.96 * 21/0.4762 * (1-0.4762) Lower Bound: 0.2626 *4 decimals Calculation: 0.47619 - 0.21361 = 0.26258 Upper Bound: 0.6898 *4 decimals Calculation: 0.47619 + 0.21361 = 0.68980 Interpret We are 95% confident that the true proportion of males among all online college students 2 (context) will lie in between (0.2626, 0.6898) 25