Upcoming SlideShare
×

Like this presentation? Why not share!

# Week 7 quiz_review_mini_tab_2011

## on Aug 15, 2011

• 25,309 views

Week 7 Lecture for help on the Week 7 Quiz.

Week 7 Lecture for help on the Week 7 Quiz.

### Views

Total Views
25,309
Views on SlideShare
25,309
Embed Views
0

Likes
2
907
0

No embeds

### Report content

• Comment goes here.
Are you sure you want to

## Week 7 quiz_review_mini_tab_2011Presentation Transcript

• Week 7 LectureStatistics for Decision Making
B. Heard
(This material can not be copied or posted without the author’s consent. Students may download one copy for personal use.)
• Standard Normal Distribution
The “standard” normal distribution is a normal distribution with mean zero and where the standard deviation (and variance) equals one.
The Total Area under the curve is one (1) or 100% (This is true for all normal distributions regardless of the mean and standard deviation).
Week 7 Lecture
• Using Minitab for Normal Distribution calculations.
Use Calc >> Probability Distributions >> Normal
Examples Follow
Week 7 Lecture
• Example
The average fish in Happy Lake weighs 2 pounds with a standard deviation of 0.5 pounds. If Bob catches a fish that weighs 3.2 pounds. What could you say about the catch?
Week 7 Lecture
• Week 7 Lecture
Since this is the Cumulative Distribution Function, it “fills” from left to right. Therefore, you could say his catch was in the “Top 1 %”
• Example
The average fish in Happy Lake weighs 2 pounds with a standard deviation of 0.5 pounds. If Bob catches a fish that weighs 1.35 pounds. What could you say about the catch?
Week 7 Lecture
• Week 7 Lecture
Since this is the Cumulative Distribution Function, it “fills” from left to right. Therefore, you could say his catch was in the “Bottom 10 %”
• Other types of questions
If you have a normal distribution with a mu = 100 and sigma = 15, what number corresponds to a z = -2
Week 7 Lecture
-2 = (x – 100)/15
Multiply both sides by 15 to get
-30 = x – 100
Add 100 to each side to get
70 = x
So “70” is my answer, I just did a little Algebra.
• Another type of question
Say we take 120 samples of size 81 each from a distribution we know is normal. Calculate the standard deviation of the sample means if we know the population variance is 25.
Week 7 Lecture
The Central Limit Theorem tells us the variance is the Population variance divided by the Sample Size. We can just take the square root to get the standard deviation.
Week 7 Lecture
Variance = 25/81 or 0.309
Standard Deviation = Square Root(25/81) = 5/9 = 0.556
• Finding z scores
Example
The area to the left of the “z” is 0.6262. What z score corresponds to this area.
Use Calc >> Probability Distributions >> Normal
(Set Mean = 0 and Standard Deviation to 1 and use “INVERSE Cumulative Probability”
Week 7 Lecture
• Week 7 Lecture
Answer is 0.322 rounded to three decimals. Remember the distribution fills from left to right.
• Another type of question
In a normal distribution with mu = 40 and sigma = 10 find P(32 < x < 44)
Easy, but this takes a couple of steps.
Using Calc >> Probability Distributions >> Normal find the probabililties that x < 32 and x < 44 using the Cumulative Probability option.
Week 7 Lecture
• Continued
Week 7 Lecture
Get results for
Both 32 and then 44.
Week 7 Lecture
Subtract
0.655422 – 0.211855
To get
0.443567
Or 0.444 rounded to
three decimals
P(32 < x < 44) = 0.444 based on
the given mean and std deviation.
• Confidence Intervals and Examples
Charts follow
Week 7 Lecture
• Interpreting Confidence Intervals
If you have a 90% confidence interval of (15.5, 23.7) for a population mean, it simply means “There is a 90% chance that the population mean is contained in the interval (15.5, 23.7)
It’s really that simple.
Week 7 Lecture
• Finding Confidence Intervals
A luxury car company wants to estimate the true mean cost of its competitor’s automobiles. It randomly samples 180 of its competitors sticker prices. The mean cost is \$65,000 with a standard deviation of \$3200. Find a 95% confidence interval for the true mean cost of the competitor’s automobiles. Write a statement about the interval.
Week 7 Lecture
• It randomly samples 180 of its competitors sticker prices. The mean cost is \$65,000 with a standard deviation of \$3200. Find a 95% confidence interval…
Use Stat >> Basic Statistics >> 1 sample Z
Make sure to click Options and set to 95%
Week 7 Lecture
• Week 7 Lecture
Week 7 Lecture
Confidence Interval is (64533, 65467), which means we can be 95% confident the true mean cost of the competitor’s vehicles are between those two values.
• Find Confidence Intervals of Proportions
Example
An student wants to estimate what proportion of the student body eats on campus. The student randomly samples 200 students and finds 120 eat on campus. Using a 95% confidence interval, estimate the true proportion of students who eat on campus. Write a statement about the confidence level and interval.
Week 7 Lecture
• Example Solution
p hat = 120/200 = 0.60
q hat = 1- 0.60 = 0.40
n p hat = 200 * 0.60 = 120
n q hat = 200 * 0.40 = 80
Using E = Zc* Square Root ((p hat * q hat)/n)
= 1.96 * Square Root ((0.60*0.40)/200)
=0.0679
Now we subtract this from the mean for the left side of the interval and add it to the mean for the right side. (0.60 – 0.0679, 0.60 + 0.0679) = (0.5321, 0.6679)
So with 95% confidence, we can say the population proportion of students who eat lunch on campus is (0.5321, 0.6679) or between 53.21% and 66.79%.
Week 7 Lecture
• Link to charts will be posted at