Age Gender For the following questions] use only the "age" column: 31 F 43 F Points Age Frequency Distribution: Class Width 8 25 M 19 F Class Limits Relative Cumulative 42 F Low High Midpoint Freq. Frequency Relative Freq 22 M Limits 1 15 22 19 6 0.316 0.316 17 M Freq. 1 23 30 27 2 0.105 0.421 50 F Mid. 1 31 38 35 6 0.316 0.737 49 M RF 1 39 46 43 3 0.158 0.395 38 M CF 1 47 54 51 2 0.105 1.000 34 M 21 F 2 Mean 31.32 *Round to two decimals 42 M 2 Median 31.0 *Round to one decimal 37 F 2 Sample Standard deviation: 10.70 \"Round to two decimals 17 F 2 (11 22.0 *Round to one decimal 22 F 2 Q3 42.0 *Round to one decimal 23 F 31 M 32 F 23;)" : ogw' Age of Students in Class Pelvgon' Age of Students in Each Class Limit 3 1.200 7 a 1.000 a 5 g 0.500 g; 5 Total: 25 _ E \"m g 4 E om E 3 '3' E 2 g 0.200 g 1 0.05; fail reject Ho. There is not enough evidence to conclude Interpretation: that the avg age of online students is NOT 32 years old @ a=0.05 17 18 19 Claim: The proportion of males in online classes is 35% 20 21 Ho: p =0.35 22 Ha: p #0.35 23 24 Sample Proportion Males 0.4211 Sample Proportion Females 0.5789 25 26 2 Distribution: Normal Distribution 27 28 Test Statistic: 0.65 *2 decimals Calculation: (P-Po) / V[(Po(1-Po) / n] Also in TI84 - 1-PropZTest p-value: 0.5156 *4 decimals Po: 0.35, x:8, n:19, prop:#Po 30 Since P>0.05, we fail to reject null hypothesis. There is not sufficient 31 3 Interpretation: evidence to conclude the proportion of males in online classes is NOT 35%. 32 33 25 Total Points 34 35 36 Week 3 Week 5 Week 6 (+