Hypothesis Testing (Week 6): Based on your sample, you will conduct a hypothesis test

AnswerSheetMath01D03SP21.xlsx

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Week 3

Age	Gender			For the following questions, use only the "age" column:
18	F
18	M		Points	Age Frequency Distribution:				Class Width	9
19	M
21	F			Class Limits		Midpoint	Freq.	Relative Frequency	Cumulative Relative Freq
22	F			Low	High
23	M	Limits	2	18	27	23	10	0.3704	0.3704
25	F	Freq.	2	28	37	33	7	0.2593	0.6296
26	M	Mid.	2	38	47	43	7	0.2593	0.8889
26	F	RF	2	48	57	53	1	0.0370	0.9259
27	F	CF	1	58	67	63	2	0.0741	1.0000
28	M
31	M		2	Mean	34.19	*Round to two decimals
31	F		2	Median	33.0	*Round to one decimal
33	M		2	Sample Standard deviation:			11.87	*Round to two decimals
35	F		2	Q1	25.0	*Round to one decimal
36	M		2	Q3	42.0	*Round to one decimal
37	F
38	F
40	M	Ogive:	3			Ogive:				Polygon:
41	F	Polygon	3
42	F
43	M
44	F
47	F	Total:	25
54	M
58	M
60	F

Ogive

27 37 47 57 67 0.37037037037037035 0.62962962962962954 0.88888888888888884 0.92592592592592582 0.99999999999999989

Age - Upper Class Limits

Cumulative Relative Frequency

Freqency Polygon

Freq. 23 33 43 53 63 10 7 7 1 2

Age - Class Midpoints

Frequency

Week 5

Note: 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
										18	F
	Points	95% Confidence Interval for Average Age of Online College Students:								21	F
										22	F
		Sample Mean:		34.19			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.			25	F
		Sample St. Dev:		11.87					Normal Distribution	26	F
	1	Sample Size:		27					T-Distribution	27	F
										31	F
	2	Distribution:		T-Distribution						35	F
										37	F
	2	Critical Value:		2.06	*2 decimals					38	F
										41	F
	2	Margin of Error:		4.70	*2 decimals	Calculation:	t(α/2) * s/√n = 2.06*(11.87/√27)			42	F
	1	Lower Bound:		29.49	*2 decimals	Calculation:	μ - ME = 34.19 - 4.70			44	F
	1	Upper Bound:		38.88	*2 decimals	Calculation:	μ + ME = 34.19 + 4.70			47	F
										60	F
		Interpret (context)	We can say with 95% confidence that the average age of online college students lies between 29.49 years and 38.88 years.							18	M
	2									19	M
										23	M
		95% Confidence Interval for Proportion of Male Online College Students:								26	M
										28	M
	1	Sample Size:		27						31	M
	1	Number of Males:		12						33	M
	2	Male Proportion:		0.4444	Female Proportion		0.5556	*4 decimals		36	M
										40	M
	2	Distribution:		Normal Distribution						43	M
										54	M
	2	Critical Value:		1.96	*2 decimals					58	M

	2	Margin of Error:		0.1874	*4 decimals	Calculation:	z(α/2) * √{p(1-p)/n} = 1.96√{0.4444*(1-0.4444)/27}
	1	Lower Bound:		0.2570	*4 decimals	Calculation:	μ - ME = 0.4444-0.1874
	1	Upper Bound:		0.6319	*4 decimals	Calculation:	μ + ME = 0.4444+0.1875
		Interpret (context)	We can say with 95% confidence that the proportion of male online college students lies between 0.2570 and 0.6319.
	2
	25	Total Points

Week 6

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?

Normal Distribution

Ho:

µ = 32

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.

T-Distribution

Ha:

µ ≠ 32

Sample mean:

34.19

Sample St. Dev:

11.87

Distribution:

T-Distribution

Reject Ho

Test Statistic:

0.96

*2 decimals

Calculation:

(34.19 - 32)/{11.87/√27}

Fail to Reject Ho

p-value:

0.3476

*4 decimals

Decision:

Fail to Reject Ho

Interpretation: (context)

There is no sufficient evidence to suggest that the mean age of the online students is not equal to 32 years, at α = 0.05 significance.

Claim: The proportion of males in online classes is 35%. Can you prove it is not?

Ho:

p = 0.35

Ha:

p ≠ 0.35

Sample Proportion Males

0.4444

Sample Proportion Females

0.5556

Distribution:

Normal Distribution

Test Statistic:

1.03

*2 decimals

Calculation:

(0.4444-0.35)/√{0.35*(1-0.35)/27}

p-value:

0.3035

*4 decimals

Decision:

Fail to Reject Ho

Interpretation: (context)

There is no sufficient evidence to suggest that the proportion of males in online classes is not 0.35, at α = 0.05 significance.

Total Points