• Define the population, sample, and statistic for the study
• Statement of the two claims in the article that were tested in this project
• Null and alternative hypothesis for both tests run for this project (in words)

Mean

• Summary of sample statistics (mean, standard deviation, median, quartiles, sample size)
• Confidence interval, along with interpretation of the confidence interval
• Description of hypothesis test (alpha, test statistic, p-value, conclusion, interpretation)

Proportion

• Summary of sample statistics (sample size, successes, proportion)
• Confidence interval, along with interpretation of the confidence interval
• Description of hypothesis test (alpha, test statistic, p-value, conclusion, interpretation)

Age Gender For the following questions, use only the "age" column: 26 M 32 F Points Age Frequency Distribution: Class Width 34 F 18 F Class Limits Relative Cumulative 49 F Low High Midpoint Freq Frequency Relative Freq 26 M Limits 18 24 21 0.2857 0.2857 45 M req. 25 31 0.1429 0.4286 20 M NNNN Mid 32 38 35 W A UT W 0.2381 0.6667 23 M RF 39 45 12 0.1905 0.8571 18 F CF 46 52 49 0.1429 1.0000 37 F 48 F Mean 32.62 *Round to two decimals 40 F Median 34 *Round to one decimal 35 M NN NN Sample Standard deviation: 11.21 *Round to two decimals 25 M Q 21.5 *Round to one decimal 18 M 03 12.5 *Round to one decimal 44 M 41 F 52 F Ogive: Ogive: W W 34 M Polygon 20 F Relative Frequency Ogive 1.0000 Total: 25 0.9000 0.8000 0.7000 0.6000 Relative Frequency Vertical (Value) Axis 0.3000 0.2000 0.1000 0.0000 1.00 2.00 3.00 4.00 5.00 Upper Class LimitNote: 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 19 F to get your answers. Do not put 26 M 2 Distribution: T-Distribution the generic formula and show all 15 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 HAN Lower Bound: 27.51 *2 decimals Calculation 32.61905 - 5.10442 = 27.51463 48 F Upper Bound: 37.72 *2 decimals Calculation: 32.61905+ 5.10442 = 37.72347 10 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 Number of Males: 10 20 F Male Proportion: .4762 Female Proportion 0.5238 *4 decimals N 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) HAN 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 2 (context) students will lie in between (0.2626, 0.6898) 25 Total PointsNote: The goal ofthe 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. Forthe'folloviingtwohypois mm u's'e'alpha' secs '0 Q. :s 51' Claim: The average age of online students is 32 years old. Can you prove it is not? \":32 NLte: Calculation cells should list the numbers and operations used to get your answers. Do not put the generic formula and show all calculation steps. I-I Sample mean: Sample St. Dev: 11.21372 2 'E *2 decimals {32-61905-32lll11-21372N21l 2 0-8029 *4 decimals 1 m Fail to Reject Ho Since pvalue=0.802927 > (1:0.05, we can't reject the null hypothesis Interpretation: This means that the claim that the average age of online students is 32 years 2 {context} old is supported. I l | l I Claim: The proportion of males in online classes is 35%. Can you prove it is not? I | l | | l N Sample Proportion Males 0.4762 Sample Proportion Females m Distribution: Normal Distribution -EI m *4declmals m Fall to Reject Ho Interpretation: This means that the claim that the proportion of males in online classes is 35% is 2 {context} supported. 25 Total Points N *2 decimals ,3: 10/21: 0.47519 2: {0.47519-0.asj/{vioss'l'u-oasjmj Since pvalue=0.225398 > cr=0.05, we can't reject the null hypothesis