what other data is missing? Data Values for Height Std Dev: Females: 2.662, Male
ID: 3205999 • Letter: W
Question
what other data is missing?
Data Values for Height Std Dev: Females: 2.662, Males: 3.080
calculate the maximum error for the females and the maximum error for the males. To find the maximum error for the females, type =CONFIDENCE.T(0.05,stdev,#), using the females’ height standard deviation for “stdev” in the formula and the number of females in your sample for the “#”. Then you can use a calculator to add and subtract this maximum error from the average female height for the 95% confidence interval. Do this again with 0.01 as the alpha in the beginning of the formula to find the 99% confidence interval.
Find these same two intervals for the male data by using the same formula, but using the males’ standard deviation for “stdev” and the number of males in your sample for the “#”.
4. Give and interpret the 95% confidence intervals for males and females on the HEIGHT variable. Which is wider and why? (7 points)
5. Give and interpret the 99% confidence intervals for males and females on the HEIGHT variable. Which is wider and why? (7 points)
6. 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? (10 points)
Drive Variable: 36,71,42,76,76,20,28,55,33,36,40,4,25,25,63,94,36,54,73,80,76,78,88,20,25,29,36,63,80,80,88,36,73,94,6
Mean ______________ Standard deviation ____________________
Predicted percentage ______________________________
Actual percentage _____________________________
Comparison
7. 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? (11 points)
Predicted percentage between 40 and 70 ______________________________
Actual percentage
Predicted percentage more than 70 miles ________________________________
Actual percentage ___________________________________________
Comparison ____________________________________________________
_______________________________________________________________
Why?
Explanation / Answer
mean height and number of observation is mission to calculate the 95% and 99% confidence interval
(6)
(7)
Predicted percentage between 40 and 70 =74.44-31.85=42.59
Actual percentage=100*5/35=14.29
Predicted percentage more than 70 miles=100-74.44=25.56
Actual percentage =100*14/35=40
Comparison:predicted value is not matching with the actual value
Why: we predicted using normal approximation and in real sense the data is not following normal distribution. that is why predicted is not matiching with actual.
s.n. x 1 4 2 6 3 20 4 20 5 25 6 25 7 25 8 28 9 29 10 33 11 36 12 36 13 36 14 36 15 36 16 40 17 42 18 54 19 55 20 63 21 63 22 71 23 73 24 73 25 76 26 76 27 76 28 78 29 80 30 80 31 80 32 88 33 88 34 94 35 94 mean= 52.5428571 stdev= 26.5732538 predicted%= 31.8459615 actual%=100*16/35= 45.7142857Related Questions
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