Project Part 3 – Confidence Intervals

Based on the class sample, you will create a 95% confidence interval for the mean age and the proportion of males in the population of all online college students.

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.

Using the same excel sheet as last week, answer the following in the “week 5” tab:

- For the average age, form a 95% confidence interval:

o What distribution should be used?

o What is the critical value?

o What is the error bound?

o What is the lower bound?

o What is the upper bound?

o How do we interpret the results, in context of our study?

- For the proportion of males, form a 95% confidence interval:

o What distribution should be used?

o What is the critical value?

o What is the error bound?

o What is the lower bound?

o What is the upper bound?

o How do we interpret the results, in context of our study?

__Project Part 3 – Confidence Intervals__

Based on the class sample, you will create a 95% confidence interval for the mean age and the proportion of males in the population of all online college students.

__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.

Using the same excel sheet as last week, __answer the following in the “week 5” tab:__

· For the average age, form a 95% confidence interval:

. What distribution should be used?

. What is the critical value?

. What is the error bound?

. What is the lower bound?

. What is the upper bound?

. How do we interpret the results, in context of our study?

· For the proportion of males, form a 95% confidence interval:

. What distribution should be used?

. What is the critical value?

. What is the error bound?

. What is the lower bound?

. What is the upper bound?

. How do we interpret the results, in context of our study?

## 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 | ||||||||||||

Ogive27 37 47 57 67 0.37037037037037035 0.62962962962962954 0.88888888888888884 0.92592592592592582 0.99999999999999989

Age – Upper Class Limits

Cumulative Relative Frequency

Freqency PolygonFreq. 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 | Points | 95% Confidence Interval for Average Age of Online College Students: | ||||||||||

18 | F | ||||||||||||

21 | 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. | |||||||||

22 | F | Sample St. Dev: | 11.87 | Normal Distribution | |||||||||

25 | F | 1 | Sample Size: | 27 | T-Distribution | ||||||||

26 | F | ||||||||||||

27 | F | 2 | Distribution: | T-Distribution | |||||||||

31 | F | ||||||||||||

35 | F | 2 | Critical Value: | 2.06 | *2 decimals | ||||||||

37 | F | ||||||||||||

38 | F | 2 | Margin of Error: | 4.70 | *2 decimals | Calculation: | 2.06*11.87/√27 | ||||||

41 | F | 1 | Lower Bound: | 29.49 | *2 decimals | Calculation: | 34.19 – 4.70 | ||||||

42 | F | 1 | Upper Bound: | 38.88 | *2 decimals | Calculation: | 34.19 + 4.70 | ||||||

44 | F | ||||||||||||

47 | F | Interpret (context) | We are 95% confident that the average age of online college students lies between 29.49 years and 38.88 years. | ||||||||||

60 | F | 2 | |||||||||||

18 | M | ||||||||||||

19 | M | 95% Confidence Interval for Proportion of Male Online College Students: | |||||||||||

23 | M | ||||||||||||

26 | M | 1 | Sample Size: | 27 | |||||||||

28 | M | 1 | Number of Males: | 12 | |||||||||

31 | M | 2 | Male Proportion: | 0.4444 | Female Proportion | 0.5556 | *4 decimals | ||||||

33 | M | ||||||||||||

36 | M | 2 | Distribution: | Normal Distribution | |||||||||

40 | M | ||||||||||||

43 | M | 2 | Critical Value: | 1.96 | *2 decimals | ||||||||

54 | M | ||||||||||||

58 | M | 2 | Margin of Error: | 0.1874 | *4 decimals | Calculation: | 1.96*√{0.4444*(1-0.4444)/27} | ||||||

1 | Lower Bound: | 0.2570 | *4 decimals | Calculation: | 0.4444-0.1874 | ||||||||

1 | Upper Bound: | 0.6319 | *4 decimals | Calculation: | 0.4444+0.1875 | ||||||||

Interpret (context) | We are 95% confident that the proportion of male online college students lie 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 | ||||||||||||

1 | 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 | ||||||||

2 | Ha: | µ ≠ 32 | ||||||||||

Sample mean: | 34.19 | |||||||||||

Sample St. Dev: | 11.87 | |||||||||||

2 | Distribution: | T-Distribution | ||||||||||

Reject Ho | ||||||||||||

2 | Test Statistic: | 0.96 | *2 decimals | Calculation: | (34.19 – 32)/{11.87/√27} | Fail to Reject Ho | ||||||

2 | p-value: | 0.3476 | *4 decimals | |||||||||

1 | Decision: | Fail to Reject Ho | ||||||||||

2 | 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? | ||||||||||||

1 | Ho: | p = 0.35 | ||||||||||

2 | Ha: | p ≠ 0.35 | ||||||||||

Sample Proportion Males | 0.4444 | Sample Proportion Females | 0.5556 | |||||||||

2 | Distribution: | Normal Distribution | ||||||||||

3 | Test Statistic: | 1.03 | *2 decimals | Calculation: | (0.4444-0.35)/√{0.35*(1-0.35)/27} | |||||||

2 | p-value: | 0.3035 | *4 decimals | |||||||||

1 | Decision: | Fail to Reject Ho | ||||||||||

2 | 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. | ||||||||||

25 | Total Points |