WEEK 7 LECTURESTATISTICS FOR

DECISION MAKING

B. Heard

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WEEK 7 LECTURE 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

Since this is the Cumulative Distribution Function, it “fills” from left to right. Therefore, you could say his catch was in the “Top 1 %”

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 1.35 pounds. What could you say about the catch?

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 %”

WEEK 7 LECTURE Other types of questions If you have a normal distribution with a mu

= 100 and sigma = 15, what number corresponds to a z = -2

-2 = (x – 100)/15Multiply both sides by 15 to get-30 = x – 100Add 100 to each side to get70 = x

So “70” is my answer, I just did a little Algebra.

WEEK 7 LECTURE 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.

(Answer next chart)

WEEK 7 LECTURE Answer 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.

Variance = 25/81 or 0.309Standard Deviation = Square Root(25/81) = 5/9 = 0.556

WEEK 7 LECTURE 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

Answer is 0.322 rounded to three decimals. Remember the distribution fills from left to right.

WEEK 7 LECTURE 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

Get results for Both 32 and then 44.

WEEK 7 LECTURE Answer

Subtract 0.655422 – 0.211855To get0.443567Or 0.444 rounded tothree decimals

P(32 < x < 44) = 0.444 based onthe given mean and std deviation.

WEEK 7 LECTURE 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 ZMake sure to click Options and set to 95%

WEEK 7 LECTURE

WEEK 7 LECTURE Click your OK buttons…

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.

WEEK 7 LECTURE Find Confidence Intervals of Proportions Example

A 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)=0.0679Now 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%.