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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.

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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

AgeGenderFor the following questions, use only the “age” column:
18F
18MPointsAge Frequency Distribution:Class Width9
19M
21FClass LimitsMidpointFreq.Relative FrequencyCumulative Relative Freq
22FLowHigh
23MLimits2182723100.37040.3704
25FFreq.228373370.25930.6296
26MMid.238474370.25930.8889
26FRF248575310.03700.9259
27FCF158676320.07411.0000
28M
31M2Mean34.19*Round to two decimals
31F2Median33.0*Round to one decimal
33M2Sample Standard deviation:11.87*Round to two decimals
35F2Q125.0*Round to one decimal
36M2Q342.0*Round to one decimal
37F
38F
40MOgive:3Ogive:Polygon:
41FPolygon3
42F
43M
44F
47FTotal:25
54M
58M
60F

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.
AgeGenderPoints95% Confidence Interval for Average Age of Online College Students:
18F
21FSample Mean:34.19Note: Calculation cells should list the numbers and operations used to get your answers. Do not put the generic formula and show all calculation steps.
22FSample St. Dev:11.87Normal Distribution
25F1Sample Size:27T-Distribution
26F
27F2Distribution:T-Distribution
31F
35F2Critical Value:2.06*2 decimals
37F
38F2Margin of Error:4.70*2 decimalsCalculation:2.06*11.87/√27
41F1Lower Bound:29.49*2 decimalsCalculation:34.19 – 4.70
42F1Upper Bound:38.88*2 decimalsCalculation:34.19 + 4.70
44F
47FInterpret (context)We are 95% confident that the average age of online college students lies between 29.49 years and 38.88 years.
60F2
18M
19M95% Confidence Interval for Proportion of Male Online College Students:
23M
26M1Sample Size:27
28M1Number of Males:12
31M2Male Proportion:0.4444Female Proportion0.5556*4 decimals
33M
36M2Distribution:Normal Distribution
40M
43M2Critical Value:1.96*2 decimals
54M
58M2Margin of Error:0.1874*4 decimalsCalculation:1.96*√{0.4444*(1-0.4444)/27}
1Lower Bound:0.2570*4 decimalsCalculation:0.4444-0.1874
1Upper Bound:0.6319*4 decimalsCalculation: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
25Total 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
PointsClaim: The average age of online students is 32 years old. Can you prove it is not?
Normal Distribution
1Ho:µ = 32Note: 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
2Ha:µ ≠ 32
Sample mean:34.19
Sample St. Dev:11.87
2Distribution:T-Distribution
Reject Ho
2Test Statistic:0.96*2 decimalsCalculation:(34.19 – 32)/{11.87/√27}Fail to Reject Ho
2p-value:0.3476*4 decimals
1Decision:Fail to Reject Ho
2Interpretation: (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?
1Ho:p = 0.35
2Ha:p ≠ 0.35
Sample Proportion Males0.4444Sample Proportion Females0.5556
2Distribution:Normal Distribution
3Test Statistic:1.03*2 decimalsCalculation:(0.4444-0.35)/√{0.35*(1-0.35)/27}
2p-value:0.3035*4 decimals
1Decision:Fail to Reject Ho
2Interpretation: (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.
25Total Points

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