For this part of the project and based on your sample, you will conduct a hypothesis test witha = 0.05 to test two of the claims of the above article.

• 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?

Age	Gender		Points	95% Confidence Interval for Average Age of Online College Students:
17	M
19	M			Sample Mean:	34.22			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.
20	F			Sample St. Dev:	13.05
22	F		1	Sample Size:	18
24	M
25	F		2	Distribution:	T-Distribution
25	F
30	F		2	Critical Value:	2.11	*2 decimals
30	F
32	F		2	Margin of Error:	6.49	*2 decimals	Calculation:	2.11*13.05/18=6.49
35	F		1	Lower Bound:	27.73	*2 decimals	Calculation:	34.22-6.49=27.73
39	M		1	Upper Bound:	40.71	*2 decimals	Calculation:	34.22+6.49=40.71
43	F
43	M			Interpret (context)	We are 95% confident that the mean age of online students is between 27.73 and 40.71 years old. Since 34.22 is within the interval, it is possible that the mean age of the population is 34.22.
45	M		2
52	F
56	F			95% Confidence Interval for Proportion of Male Online College Students:
59	F
			1	Sample Size:	18
			1	Number of Males:	6
			2	Male Proportion:	0.3333	Female Proportion	0.6667	*4 decimals
			2	Distribution:	Normal Distribution
			2	Critical Value:	1.96	*2 decimals
			2	Margin of Error:	0.2178	*4 decimals	Calculation:	1.96*0.3333(1-0.3333)/18=0.2178
			1	Lower Bound:	0.1155	*4 decimals	Calculation:	0.3333-0.2178=0.1155
			1	Upper Bound:	0.5511	*4 decimals	Calculation:	0.3333+0.2178=0.5511
				Interpret (context)	We are 95% confident that the proportion of online students that are male is between 0.1155 and 0.5511. Since 0.3333 is within the interval, it is possible that the proportion of male students in the population is 0.3333.
			2