Need help with this stat assignment. Help would be greatly appreciated.?

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? __________________________________________________________

________________________________________________________________