All answers should be complete sentences. We need to find the confidence interva
ID: 3124473 • Letter: A
Question
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. (6 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. (6 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. (6 points)
In the week 2 lab, you found the mean and the standard deviation for the HEIGHT variable for both males and females. Use those values for follow these directions to calculate the numbers again.
(From week 2 lab: Calculate descriptive statistics for the variable Height by Gender. Click on Insert and then Pivot Table. Click in the top box and select all the data (including labels) from Height through Gender. Also click on “new worksheet” and then OK. On the right of the new sheet, click on Height and Gender, making sure that Gender is in the Rows box and Height is in the Values box. Click on the down arrow next to Height in the Values box and select Value Field Settings. In the pop up box, click Average then OK. Write these down. Then click on the down arrow next to Height in the Values box again and select Value Field Settings. In the pop up box, click on StdDev then OK. Write these values down.)
You will also need the number of males and the number of females in the dataset. You can either use the same pivot table created above by selecting Count in the Value Field Settings, or you can actually count in the dataset.
Then in Excel (somewhere on the data file or in a blank worksheet), 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? (9 points)
5. Give and interpret the 99% confidence intervals for males and females on the HEIGHT variable. Which is wider and why? (9 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? (12 points)
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? (12 points)
Predicted percentage between 40 and 70 ______________________________ Actual percentage _____________________________________________ Predicted percentage more than 70 miles ________________________________ Actual percentage ___________________________________________ Comparison ____________________________________________________ _______________________________________________________________ Why? __________________________________________________________ ________________________________________________________________
Drive (miles) State Shoe Size Height (inches) Sleep (hours) Gender Car Color TV (hours) Money (dollars) Coin Die1 Die2 Die3 Die4 Die5 Die6 Die7 Die8 20 OR 7 65 4 F green 1 46.00 4 5 5 1 5 2 1 5 1 28 NY 7 67 4 F black 1 25.00 4 6 2 6 5 5 4 3 1 54 NV 11 65 4 F silver 5 54.00 7 6 2 3 6 1 5 3 6 76 PA 11 63 4 M silver 5 13.00 5 4 5 2 3 3 5 3 6 36 TX 7 66 5 M orange 3 7.00 3 2 4 4 1 6 3 1 6 88 SC 6 63 5 F black 3 43.00 6 6 2 4 2 3 5 3 1 6 MI 10 66 6 M black 5 34.00 3 4 4 5 1 4 4 5 5 25 CA 8 69 6 F silver 4 30.00 4 3 5 3 6 1 3 1 4 40 IL 10 68 6 M silver 5 16.00 5 2 2 4 2 1 3 2 4 42 IL 8 65 6 F red 4 4.00 4 3 4 1 3 2 5 5 5 55 OH 9 67 6 F dark blue 4 27.00 2 2 6 5 5 5 3 5 2 71 MI 11 64 6 F silver 2 1.00 2 4 3 5 4 3 1 2 3 76 NY 9 63 6 M silver 4 38.00 2 3 5 2 1 2 3 4 5 78 CA 11 63 6 M blue 2 47.00 2 3 4 4 5 2 5 5 4 4 NY 9 69 7 F blue 5 21.00 5 4 2 1 5 6 5 2 1 20 IL 5 62 7 F black 3 10.00 2 2 4 5 4 5 5 6 3 25 GA 9 63 7 M silver 3 53.00 5 2 4 4 2 5 2 3 4 29 TX 11 70 7 M silver 2 28.00 3 3 3 2 6 6 6 6 5 33 TX 9 75 7 M silver 5 54.00 7 4 6 2 6 4 3 1 4 36 MI 10 61 7 M blue 6 5.00 2 6 6 4 5 4 4 6 4 36 IL 12 61 7 M red 4 14.00 3 2 5 3 6 4 3 2 4 36 OR 9 73 7 M blue 5 23.00 5 2 5 1 4 5 1 1 1 63 MI 5 69 7 F orange 3 9.00 2 6 2 4 4 6 2 2 5 73 FL 12 70 7 F green 4 15.00 3 1 6 4 6 2 3 2 2 73 SC 11 68 7 F red 2 18.00 5 4 2 3 1 6 4 3 3 76 NY 9 65 7 F red 6 7.00 2 1 6 6 5 1 5 2 5 80 FL 9 66 7 F black 5 22.00 2 6 4 4 2 2 4 1 1 33 SC 11 68 8 M blue 1 9.00 4 5 1 5 6 2 5 2 1 54 FL 9 70 8 F green 2 5.00 3 5 4 3 3 3 3 6 3 63 PA 10 66 8 F silver 3 5.00 7 3 1 2 1 6 5 4 2 76 PA 11 70 8 F blue 3 5.00 4 4 4 5 2 4 4 1 3 80 NV 12 69 8 M white 4 3.00 4 2 3 2 5 6 4 6 3 80 SC 10 63 8 F black 2 7.00 7 4 4 1 6 3 2 4 6 94 KY 11 74 8 M red 3 20.00 4 3 5 3 2 6 4 5 3 6 OR 13 75 10 M silver 6 7.00 4 2 1 2 6 4 5 2 2Explanation / Answer
First we have to find confidence interval for hours of sleep a student gets.
The 95% confidence interval for mean is,
Xbar - E < mu < Xbar + E
where Xbar is sample mean.
E is margin of error.
mu is population mean for sleep.
Xbar = 6.6
sd = 1.3547
n = 35
E = tc*sd/sqrt(n)
where tc is critical value for t-distribution.
EXCEL syntax for critical value is,
=TINV(probability, deg_freedom)
where probability = 1 - c = 1 - 0.95 = 0.05
deg_freedom = n-1 = 34
c = confidecnce level = 0.95
tc = 2.0322
E = (2.0322*1.3547) / sqrt(35) =0.4653
lower limit = Xbar - E = 6.6 - 0.4653 = 6.1347
upper limit = Xbar + E = 6.6 + 0.4653 = 7.0653
95% confidence interval for population mean is (6.1347, 7.0653).
Conclusion : We are 95% confident that the population mean for hours of sleep a student gets is in between 6.1347 and 7.0653.
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Similarly find 99% confidence interval for hours of sleep a student gets.
tc = 2.7284
E = (2.7284*1.3547) / sqrt(35) = 0.6248
lower limit = Xbar - E = 6.6 - 0.6248 = 5.9752
upper limit = Xbar + E = 6.6 + 0.6248 = 7.2248
99% confidence interval for population mean is (5.9752, 7.2248).
Conclusion : We are 99% confident that the population mean for hours of sleep a student gets is in between 5.9752 and 7.2248.
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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.
99% (higher) confidence has a wider interval while 95% (lower) confidence has a narrower interval.
length of 95% CI :
length = upper limit - lower limit = 7.0649 - 6.1351 = 0.9237
And length of 99% CI :
length = upper limit - lower limit = 7.2237 - 5.9763 = 1.2475
We see that as confidence level increases length of the confidence interval is also increases.
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Calculate descriptive statistics for the variable Height by Gender.
Maximum error for the female :
E = (tc*sd) / sqrt(n)
tc by using EXCEL syntax is,
tc = 2.1009
E = (2.1009*5.5684) / sqrt(19) = 1.2379
lower limit = avF - E = 66.4737 - 1.2379 = 65.2358
upper limit = avF + E = 66.4737 + 1.2379 = 67.7116
95% confidence interval for population mean is (65.2358, 67.7116).
Conclusion : We are 95% confident that the population mean for average female height is in between 65.2358 and 67.7116.
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Similarly find 99% confidence interval for hours of sleep a student gets.
tc = 2.8784
E = (2.8784*2.5684) / sqrt(19) = 1.6961
lower limit = Xbar - E = 66.4737 - 1.6961 = 64.7776
upper limit = Xbar + E = 66.4737 + 1.6961 = 68.1698
99% confidence interval for population mean is (64.7776, 68.1698).
Conclusion : We are 99% confident that the population mean for average female height is in between 64.7776 and 68.1698.
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99% (higher) confidence has a wider interval while 95% (lower) confidence has a narrower interval.
length of 95% CI :
length = upper limit - lower limit = 69.9990 - 64.7510 = 5.2481
And length of 99% CI :
length = upper limit - lower limit = 71.0027 - 63.7473 = 7.2554
We see that as confidence level increases length of the confidence interval is also increases.
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Similar procedure for confidence interval for average male height is,
Maximum error for the male :
E = (tc*sd) / sqrt(n)
tc by using EXCEL syntax is,
tc = 2.1314
E = (2.1314*4.9244) / sqrt(16) = 2.6240
lower limit = avM - E = 67.375 - 2.6240 = 64.7510
upper limit = avM + E = 67.375 + 2.6240 = 69.9990
95% confidence interval for population mean is (64.7510, 69.9990).
Conclusion : We are 95% confident that the population mean for average male height is in between 64.7510 and 69.9990
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Similarly find 99% confidence interval for height of male student.
tc = 2.9467
E = (2.9467*4.9244) / sqrt(16) = 3.6277
lower limit = Xbar - E = 67.375 - 3.6277 = 63.7473
upper limit = Xbar + E = 67.375 + 3.6277 = 71.0027
99% confidence interval for population mean is (63.7473, 71.0027).
Conclusion : We are 99% confident that the population mean for average male height is in between 63.7473 and 71.0027.
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99% (higher) confidence has a wider interval while 95% (lower) confidence has a narrower interval.
length of 95% CI :
length = upper limit - lower limit = 67.7116 - 65.2358 = 2.4759
And length of 99% CI :
length = upper limit - lower limit = 68.1698 - 64.7776 = 3.3921
We see that as confidence level increases length of the confidence interval is also increases.
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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?
mean = 50.4286
sd = 25.9777
Here we have to find P(X < 40 miles).
By using EXCEL syntax is,
=NORMDIST(40, mean, stdev,TRUE).
where mean = 50.4286
and sd = 25.9777
P(X < 40 miles) = 0.3440
Now determine the percentage of data points in the dataset that fall within this range.
predicted percentage = 0.3440*100 = 34.40%
34.40% of the data points in the data set 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.
count = 15
probability that values are less than 40 = 15 / 35 = 0.4286
actual percentage = 0.4286*100 = 42.86%.
actual precentage > predicted percentage.
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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?
That is here we have to find P(40 < X < 70).
P(40 < X < 70) = P(X <=70) - P(X <= 40)
EXCEL syntax is,
=NORMDIST(x, mean, stdev, cumulative)
P(40 < X < 70) = 0.7744 - 0.3440 = 0.4303
predicted percentage = 43.03%
Actual percentage = 7/35 = 0.2
Actual percentage = 0.2*100 = 20%
Actual percentage < predicted percentage.
Row Labels Count of Height (inches) F 19 M 16 Grand Total 35 total ave 66.8857 av F 66.4737 avM 67.375 total sd 3.7945 sdF 2.5684 sdM 4.9244 total count 35 countF 19 countM 16Related Questions
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