Jan. 19 Statistic for the day:Number of Wisconsin’s 33Senators who voted in favor of a 1988 bill that allows the blind to hunt: 27

Assignment: Read Chapter 8 and do Assignment: Read Chapter 8 and do exercises 1, 2, 3, 8, 10exercises 1, 2, 3, 8, 10

Source: Legislative Hotline, Madison, Wisc.

These slides were created by Tom Hettmansperger and in some cases modified by David Hunter

Age at Death of English Rulers

Revisited

60, 50, 47, 53, 48, 33, 71, 43, 65, 34, 60, 50, 47, 53, 48, 33, 71, 43, 65, 34, 56, 59, 49, 81, 67, 68, 49, 16, 86, 6756, 59, 49, 81, 67, 68, 49, 16, 86, 67

Turning this data into information.

Shape: Histogram

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Age at death of a sample of 20 rulers of England

Histogram but different shape(The number of intervals has been changed from 8 to 10.)

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Sample of Rulers of England n = 20 The numberof intervals has been changed to 10.

Alternatives to median and IQR:

Mean or average of the data. Mean or average of the data. Standard deviation of the data.Standard deviation of the data.

The mean is easy. Just add up the numbers and divide by the sample size.

The standard deviation is a pain. Generally you will use a calculator or computer.

Rough way to approximate the standard deviation:

Look at the histogram and estimate the Look at the histogram and estimate the range of the middle 95%range of the middle 95% of the data. of the data.

The standard deviation is about The standard deviation is about

¼¼ of this range of this range

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Age at death of a sample of 20 rulers of England

Take 95% range = 85 – 15 = 70Estimate of standard deviation = .25x70 = 17.5

Using the calculation formula std dev = 16.7

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Range of middle 95% is roughly 24 – 16 = 8Standard deviation is roughly .25x8 = 2Standard deviation from formula 1.927

In Summary:

The standard deviation is roughly .25 times the range ofthe middle 95% of the data. Look at the histogram or stem and leaf. The mean is the arithmetic average.

If you want to visualize a histogram and you only know the mean and the standard deviation:1. Put the center at the mean or the median.2. Go out 2 standard deviations on either side of the center.3. Draw a histogram humped up at the mean and droppingoff on either side within part 2.

Smoothing the histogram: The Normal Curve (Chapt 8)

The histogram is rough and noisy.

Replace it with a bell shaped curve.

Center the bell at the mean.

The middle 95% of the bell should be 4 standard deviations.

Makes more accurate predictions provided the bell shape is appropriate for the underlying population.

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Histogram of HandSpan, with Normal Curve

Mean = 20.86Standard deviation = 1.927

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Histogram of HandSpan, with Normal Curve

Mean = 20.86Standard deviation = 1.927

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Histogram of Height, with Normal Curve

Mean = 68 inches or 5 feet 8 inchesStandard deviation = 4 inches

Research Question 1: How high should I build my doorways so that 99% of the people will not

have to duck?

Secondary Question 2: If I built my Secondary Question 2: If I built my doors 75 inches (6 feet 3 inches) doors 75 inches (6 feet 3 inches) high, what percent of the people high, what percent of the people

would have to duck?would have to duck?

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Histogram of Height, with Normal Curve

Question 1Question 2

Find the value at Question 1 so that 99% of the distributionis below it.The value at Question 2 is 75; find the amount of distributionabove it.

Z-Scores: Measurement in Standard Deviations

Given the mean (68), the standard deviation Given the mean (68), the standard deviation (4), and a value (height say 75) compute (4), and a value (height say 75) compute

75.14

687575 SDmean

Z

This says that 75 is 1.75 standard deviations above the mean.

1. How many standard deviations are you above or below the mean.

Use:Mean = 68 inchesStandard deviation = 4 inches

2. Now use the table from the book to determinewhat percentile you are.

Compute your Z-score.

Answer to Question 2: What percent of people would have to duck if I built my doors 75 inches high?

Recall: 75 has a Z-score of 1.75

From the standard normal table in the book: .96 or 96% of the distribution is below 1.75. Hence, .04 or 4% is above 1.75.

So 4% of the distribution is above 75 inches.

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Histogram of Height, with Normal Curve

Question 2

4% in here

75

The value at Question 2 is 75; find the amount of distributionabove it. Convert 75 to Z = 1.75 and use Table 8.1 in book.

Question 1: What is the value so that 99% of thedistribution is below it? Called the 99th percentile.

1. Look up the Z-score that corresponds to the 99th percentile. From the table: Z = 2.33.

2. Now convert it over to inches:

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3.77433.26899

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Since 77 inches is 6 feet 5 inches, 99% of the distributionis shorter than 77 inches and they will not have to duck.

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Histogram of Height, with Normal Curve

Question 177.3 inches is the 99th percentile

99% in here

Find the value at Question 1 so that 99% of the distributionis below it. Look up Z-score for 99th percentile and convert it back to inches.

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Stat 100 students Sp01

n=78MalesStat 100

n=30Steelers

n=12PSU BB

n=14Lakers

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Heights in Inches ( red circle is my doorway 77 inches)

78