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July 21, 2017UncategorizedBy classhelp24@gmail.com [email if you need fresh paper done]

I need help with my Stats homework. Below is the content.?

MATH 221 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)

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. (5
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? (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)
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? __________________________________________________________
________________________________________________________________

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