We need to find the confidence interval for the SLEEP variable. To do this, we n
ID: 3055263 • Letter: W
Question
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) 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) 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) Data given for sleep: 4,4,4,4,5,5,6,6,6,6,6,6,6,6,7,7,7,7,7,7,7,7,7,7,7,7,7,8,8,8,8,8,8,8,10 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) Data Given for Drive(miles): 4,6,6,20,20,25,25,28,29,33,33,36,36,36,36,40,42,54,54,55,63,63,71,73,73,76,76,76,76,78,80,80,80,88,94 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
Question: Give and interpret the 95% confidence interval for the hours of sleep a student gets. (6 points)
Solution: The 95% confidence interval for the hours of sleep a student gets is (7.104618 , 8.323953). This implies we can state that the normal number of hours of rest an understudy gets lies in the given interim with a certainty of 95%.
Question: Give and interpret the 99% confidence interval for the hours of sleep a student gets. (6 points)
Solution: The 99% confidence interval for the hours of sleep a student gets is (6.913047 , 8.515525)
This implies we can state that the normal number of hours of rest an understudy gets lies in the given interim with a certainty of 99%.
Question: 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. (6 points)
Solution:
95 % interval is (7.104618 , 8.323953) = Width = 1.22
99 % interval is (6.913047 , 8.515525) = Width = 1.6
The 99% interim is more extensive on the grounds that we can express that the normal lies in this interim with more certainty than that in the 95% interim.
Question: Give and interpret the 95% confidence intervals for males and females on the HEIGHT variable. Which is wider and why? (9 points)
Solution:
The 95% confidence interval for the height of females is (67.27007 , 69.72993) and its width is 2.46.
This means that we can say that the height of females lies in the given interval with a confidence of 95%.
The 95% confidence interval for the height of males is (69.13502 , 72.1591) and its width is 3.02.
This means that we can say that the height of males lies in the given interval with a confidence of 95%.
The width of the confidence interval of males is higher because males have higher standard deviation in heights
Question: Give and interpret the 99% confidence intervals for males and females on the HEIGHT variable. Which is wider and why? (9 points)
Solution:
The 99% confidence interval for the height of females is (66.8836 , 70.1164) and its width is 3.23.
This means that we can say that the height of females lies in the given interval with a confidence of 99%.
The 95% confidence interval for the height of males is (68.6599 , 72.63422) and its width is 3.97.
This means that we can say that the height of males lies in the given interval with a confidence of 99%.
The width of the confidence interval of males is higher because males have higher standard deviation in heights.
Question: DRIVE Variable:
Find the mean and standard deviation of the DRIVE variable. 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. Now determine the percentage of data points in the dataset that fall within this range. How does this compare with your prediction? (12 points)
Solution:
Mean: 52.5429 miles Standard deviation: 26.5733 miles
Predicted percentage: 0.31846
Actual percentage: 0.4286
Comparison The actual percentage is greater which may be because we have assumed the distribution to be normal while there may be some deviations or because of sampling error.
Question: 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? 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? (12 points)
Solution:
Predicted percentage between 40 and 70: 0.4259
Actual percentage: 0.14286
Predicted percentage more than 70 miles: 0.2556
Actual percentage: 0.4
Correlation The anticipated rate for "in the vicinity of 40 and 70 miles" is more prominent than the real rate. The anticipated rate for "in excess of 70 miles" is lower than the real rate.
Why? The distinctions might be on the grounds that we have accepted the dispersion to be typical while there might be a few deviations or on account of inspecting mistake.
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