I really need help with questions 4 and 5 please Drive (miles) 33 28 77 71 54 81
ID: 3219298 • Letter: I
Question
I really need help with questions 4 and 5 please
Drive (miles)
33
28
77
71
54
81
38
29
6
78
63
36
63
36
86
76
42
83
20
94
39
88
25
80
42
71
73
55
25
76
40
4
36
54
6
Sleep (hours)
8
4
8
6
8
8
8
7
7
9
8
7
8
5
8
7
7
8
7
8
10
5
7
8
8
8
7
8
6
4
6
7
7
5
10
MATH 399N Statistics for Decision Making
Week 6 iLab
Name:
Statistical Concepts:
Data Simulation
Confidence Intervals
Normal Probabilities
Short Answer Writing Assignment
All answers should be complete sentences.
We need to find the confidence interval for the SLEEP variable. To do this, we need to find the mean and then find the maximum error. Then we can use a calculator to find the interval, (x – E, x + E). First, find the mean. Under that column, in cell E37, type =AVERAGE(E2:E36). Under that in cell E38, type =STDEV(E2:E36). Now we can find the maximum error of the confidence interval. To find the maximum error, we use the “confidence” formula. In cell E39, type =CONFIDENCE.NORM(0.05,E38,35). The 0.05 is based on the confidence level of 95%, the E38 is the standard deviation, and 35 is the number in our sample. You then need to calculate the confidence interval by using a calculator to subtract the maximum error from the mean (x-E) and add it to the mean (x+E).
1.) Give and interpret the 95% confidence interval for the hours of sleep a student gets. (5 points)
The 95% confidence interval for the hours of sleep a student gets is (6.180, 7.098). This means that we are 95% confident that the mean number of hours of sleep a student gets is between 6.180 hours and 7.098 hours.
Then, you can go down to cell E40 and type =CONFIDENCE.NORM(0.01,E38,35) to find the maximum error for a 99% confidence interval. Again, you would need to use a calculator to subtract this and add this to the mean to find the actual confidence interval.
2.) Give and interpret the 99% confidence interval for the hours of sleep a student gets. (5 points)
The 99% confidence interval is (6.024, 7.254). This means that we are 99% confident that the mean number of hours of sleep a student gets is between 6.024 hours and 7.254 hours.
3.) Compare the 95% and 99% confidence intervals for the hours of sleep a student gets. Explain the difference between these intervals and why this difference occurs. (10 points)
The 99% confidence interval for the hours of sleep a student gets is wider than the 95% confidence interval. When the confidence level increases, the confidence interval is wider because the margin of error is greater.
4.) Find the mean and standard deviation of the DRIVE variable by using =AVERAGE(A2:A36) and =STDEV(A2:A36). Assuming that this variable is normally distributed, what percentage of data would you predict would be less than 40 miles? This would be based on the calculated probability. Use the formula =NORM.DIST(40, mean, stdev,TRUE). Now determine the percentage of data points in the dataset that fall within this range. To find the actual percentage in the dataset, sort the DRIVE variable and count how many of the data points are less than 40 out of the total 35 data points. That is the actual percentage. How does this compare with your prediction? (15 points)
Mean= Standard deviation=
Predicted percentage:
Actual percentage:
Comparison:
5.) What percentage of data would you predict would be between 40 and 70 and what percentage would you predict would be more than 70 miles? Subtract the probabilities found through =NORM.DIST(70, mean, stdev, TRUE) and =NORM.DIST(40, mean, stdev, TRUE) for the “between” probability. To get the probability of over 70, use the same =NORM.DIST(70, mean, stdev, TRUE) and then subtract the result from 1 to get “more than”. Now determine the percentage of data points in the dataset that fall within this range, using same strategy as above for counting data points in the data set. How do each of these compare with your prediction and why is there a difference? (15 points)
Predicted percentage between 40 and 70= ______________________________
Actual percentage =_____________________________________________
Predicted percentage more than 70 miles= ________________________________
Actual percentage ___________________________________________
Comparison ____________________________________________________
Why? __________________________________________________________
Explanation / Answer
a). Mean = 51.657 miles
Standard deviation =25.801 miles
Predicted percentage = 0.3257 or 32.57%
Actual percentage = 0.40 or 40%
Comparison: The predicted percentage of 32.57 is lessthan the actual percentage, but is pretty close.Keeping in mind that we only had 35 data points ,that is understandable.As the amount of data increases,the actual percentage of 40% would get closer to the predicted percentage.
b). Predicted percentage between 40 and 70= 0.4357 or 43.57%
Actual percentage =0.229 or 22.9%
Predicted percentage more than 70 miles= 0.2386 or 23.86%
Actual percentage=0.371 or 37.1%
Comparison:In this case,The predicted was a little of.About 5% on the between 40 and 70 case and 2% on the greaterthen 70 case.However they were still pretty close.
Why?: The reason is that we had only 35 data points,as the amount of data increases, the actual percentage should get closer to the predicted percentage.