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# Lesson 13 - PowerPoint PPT Presentation

Practical Statistics. Chi-Square Statistics. There are six statistics that will answer 90% of all questions! Descriptive Chi-square Z-tests Comparison of Means Correlation Regression. Chi-square: Chi-square is a simple test for counts….. . Chi-square:

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### Practical Statistics

Chi-Square Statistics

• There are six statistics that will

• Descriptive

• Chi-square

• Z-tests

• Comparison of Means

• Correlation

• Regression

Chi-square is a simple test for counts…..

Chi-square is a simple test for counts…..

Which means: nominal data

and… if some cases…

Ordinal data

• Chi-square:

• There are three types:

• Test for population variance

• Test of “goodness-of-fit”

• Contingency table analysis

• Chi-square:

• There are three types:

• Test for population variance

• Chi-square:

• There are three types:

• Test for population variance

• Test of “goodness-of-fit”

Where o = frequency of actual observation, and

e = frequency you expected to find

of a Monkey Shine Restaurant is made up of 30%

stop in while shopping, 30% Chinese business men,

and 10% tourists. A random sample of 600 customers

at the Kowloon Monkey Shine found 150 Western

and 65 women who were shopping.

Is the clientele at this establishment different

than the norm of the this company?

With (4-1) degrees of freedom

The chi-square distribution is highly skewed

and dependent upon how many degrees of

freedom (df) a problems has.

Chi-square = 34.45, df = 3

By looking in a table, the critical value of

Chi-square with df = 3 is 7.82. The probability

that the researched frequency equals the

frequency found in the MR project was p < .001.

http://www.fourmilab.ch/rpkp/experiments/analysis/chiCalc.html

the largest contribution to chi-square came from

the tourists.

= 5.00 + 0.56 + 2.22 + 26.67 = 34.45 df = 3

Hence, the Kowloon property is attracting more

tourist than what would be expected at the Monkey

Shine.

• Chi-square:

• There are three types:

• Test for population variance

• Test of “goodness-of-fit”

• Contingency table analysis

Where o = frequency of actual observation, and

e = frequency you expected to find

A contingency table

is a table with numbers grouped by frequency.

A contingency table

is a table with numbers grouped by frequency.

There are three groups: brand loyal customers,

Each is asked if they like the taste of new

product over the old. They answer with a “yes”

or a “no.”

A contingency table would look like this:

A contingency table

is a table with numbers grouped by frequency.

All the numbers in the table are “observed”

frequencies (o).

So, what are the expected values?

The expected values (e) would be a random

distribution of frequencies.

The expected values(e) would be a random

distribution of frequencies. These can be calculated

by multiplying the row frequency by the column

frequency and dividing by the total number of

observations.

For example, the expected values (e) of “loyal”

And “yes” would be (150 X 90)/270 = 50

For example, the expected values (e) of “regular”

And “no” would be (120 X 100)/270 = 44.4

The expected values (e) for the entire table

would be:

The chi-square value is calculated for every cell,

and then summed over all the cells.

The chi-square value is calculated for every cell:

For Cell A: (50-50)^2/50 = 0

For Cell D: (40-44.4)^/44.4 = 0.44

The chi-square value is calculated for every cell:

The chi-square value is calculated for every cell:

Chi-square = 0 + 0 + .35 + .44 + .44 + .54 = 1.77

The df = (r-1)(c-1) = 1 X 2 = 2

A chi-square with a df = 2 has a critical value

of 5.99, this chi-square = 1.77, so the results

are nonsignificant.

http://www.fourmilab.ch/rpkp/experiments/analysis/chiCalc.html

The probability = 0.4127.

This means that the distribution is random, and

there is no association between customer type

And taste preference.

A chi-square with a df = 2 has a critical value

of 5.99, this chi-square = 1.77, so the results

are nonsignificant.

This means that the distribution is random, and

there is no association between customer type

And taste preference.

Note: This type of chi-square is a test of

association using nothing but

counts (frequency);