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    John Henry O. Valencia, RN, RM, MANc

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    We often have occasions to make comparisons between two

    characteristics of something to see if they are linked or related to each

    other.

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    One way to do this is to work out what we would expect to find if there

    was no relationship between them (the usual null hypothesis) and what

    we actually observe.

    The test we use to measure the differences between what is observed

    and what is expected according to an assumed hypothesis is called the

    chi-square test.

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    First and foremost, make sure you pronounce it

    correctly, chi-square as in kite NOT chee as

    in cheetah (Chee-Square?) or chaye as in ChaiTea (Chai-Square?)

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    Test for goodness of fit

    Test for independence of attributes

    Testing homogeneity

    Testing given population variance

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    Identifies significant differences among

    the observed frequencies and the expected

    frequencies of a particular group.

    Do the number of individuals or objects that fall in each category differ

    from the number you would expect?

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    Attempts to identify whether any differences

    between the expected and observed

    frequencies are due to chance, or some

    other factor that is affecting it.

    Is this difference between the expected and observed are due to sampling

    error, or is it a REALdifference?

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    Data- 2 types

    Numerical data- in form of numbers. (ex.

    1,2,3,4) Categorical data- comes in form of divisions.

    (ex. Yes or no)

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    Observed Frequencies the observed frequencies are the frequencies actually

    obtained in each cell of the table, from our random sample.

    When conducting a chi-squared test, the term observed

    frequencies is used to describe the actual data in the

    contingency table.

    Observed frequencies are compared with the expectedfrequencies and differences between them suggest that the

    model expressed by the expected frequencies does not

    describe the data well.

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    Expected Frequencies

    values for parameters that are hypothesized to

    occur

    are the frequencies that you would predict

    ('expect') in each cell of the table, if you knew

    only the row and column totals, and if you

    assumed that the variables under comparison

    were independent.

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    can be determined through:

    a) Hypothesizing that the frequencies are equal for

    each category.

    b) Hypothesizing the values on the basis of some

    prior knowledge.

    c) A mathematical method

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    Contingency Tables Frequency table in which a sample from a population is classified

    according to two attributes, which are divided in to two or more

    classes .

    A contingency table is used to summarise categorical data. It may beenhanced by including the percentages that fall into each category.

    What you find in the rows of a contingency table is contingent upon

    (dependent upon) what you find in the columns.

    Can be:

    2 x 2

    h x k

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    Chi-square test for Independence

    -This tests whether the categoryfrom which the data comes from

    affects the data.

    -May also be thought of as testing whether the categories in the

    experiment prefercertain kinds of data.

    Example: Is there a difference in the car choices of male andfemales?

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    Chi-square test for goodness-of-fit

    -This tests whether the observeddata fitthe expecteddata.

    Example: Do the car sales this year match the car sales last year?

    (ie. Did we still sell around 50 blue cars? 25 red cars?)

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    The values of the parameters to be compared are

    quantitative and nominal.

    There should be one or more categories in the setup.

    The observations should be independent of each other.

    An adequate sample size. (At least 10)

    Most of the time, it is the frequency of the observations

    that are used.

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    Simple random sample

    All observations must be used

    The expected frequency in any one cell of thetable must be greater than 5.

    The total number of observations must be

    greater than 20.

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    2= (O E)2

    E

    2

    = The value of chi squareO= The observed valueE= The expected value (O E)2= all the values of (O E) squared then addedtogether

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    1. State the hypotheses

    2. State the level of significance

    3. Set up a contingency table

    4. Compute for the expected frequencies

    5. Rearrange the table to show the observed and expected frequencies

    on the columns

    6. Determine the degree of freedom

    7. Check the tabular Chi-squared value with your df and level of

    significance

    8. State your conclusion

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    A Nutrition and Dietetics Student wants to see whether the food

    preferences of males and females differed. He tried to see whether males

    or females had a general difference in the preference for cooked and raw

    foods. A survey was conducted with the following results:

    Twelve males preferred Cooked foods.

    Eight males preferred Raw foods.

    Five females preferred Cooked foods.

    Five females preferred Caw foods.

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    H0: There is no significant difference between the food

    preferences of males and females. / Food preference is

    independent of gender.

    H1: There is a significant difference between the food

    preferences of males and females. / Food preference is

    affected by gender.

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    = 0.05

    0.05 is the level of significance for mostscientific experiments.

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    The contingency table summarizes the data.

    The categories on the columns are the preferencesthat you are checking.

    The categories on the rows are the populationswhose preferences are being

    checked. A row total and column total is always included as well.

    Preference Male Female Total (Row)

    Cooked 12 5 17

    Raw 8 5 13

    Total (Column) 20 10 30

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    The chi-square test for independence usually uses the third method of getting

    expected frequencies.

    Expected Frequency = (Row Total)(Column Total)

    Grand total

    This expected frequency is computed for EACH cell.

    Preference Male Female Total (Row)

    Cooked (20)(17)/30 =11.33 (10)(17)/30 =5.67 17

    Raw (13)(20)/30 =8.67

    (13)(10)/30 =4.33

    13

    Total (Column) 20 10 30

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    Preference O E (O-E) (O-E)2 (O-E)2

    E

    Cooked Male 12 11.33 0.67 0.4489 0.0396

    Cooked Female 5 5.67 -0.67 0.4489 0.0792

    Raw Male 8 8.67 -0.67 0.4489 0.518

    Raw Female 5 4.33 0.67 0.4489 0.1037

    Total 0.2743

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    The degrees of freedom is:

    df = (Rows 1)(Columns 1)

    df = (21)(2

    1) = 1

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    Checking the table, we see that the tabular

    chi-squared value for df = 1, and = 0.05 is

    3.841.

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    Since our calculated chi-square is less than this, the

    conclusion is to ACCEPT THE NULL HYPOTHESIS.

    Hence, food preference is independent of gender.

    If it were greater, we would reject the null hypothesis.

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    Coding in SPSS:

    Preference Male

    (1)

    Female

    (2)

    Total (Row)

    Cooked

    (1)

    12 5 17

    Raw

    (2)

    8 5 13

    Total (Column) 20 10 30

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    A hospital employs a number of visiting surgeons to undertake

    particular operations. If complications occur during or after the operation

    the patient has to be transferred to a larger hospital nearby where the

    required back up facilities are available.

    A hospital administrator, worried by the effects of this on costs,

    examines the records of the three surgeons. Surgeon A had 6 out of her

    last 47 patients were transferred, Surgeon B, 4 out of his last 72 patients

    and surgeon C, 14 out of his last 41.

    Form the data into a 2 x 3 contingency table and test at the 5%

    significance level, whether the proportion transferred is independent of

    the surgeon.

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    H0: Patient transfer is independent of surgeons choice

    or there is no association on transfer rate and the

    surgeon

    H1: patient transfer is affected by surgeons choice or

    there is an association on transfer rate and the surgeon

    Alpha Level:

    = 0.05

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    SURGEON A B C TOTAL(ROW)

    TRANSFERRED 6 4 14 24

    NOT-TRANSFERRED 41 68 27 136

    Total

    (Column)

    47 72 41 160

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    Since our calculated Chi Square is greater than the

    critical value, the conclusion is to reject the null

    hypotheses, hence, transfer rate is affected by

    surgeonschoice / preference.

    The difference between what we expected and our

    observation were too great / large to be explained by

    Chance alone.

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    A marketing firm producing detergents is interested

    in studying the consumer behavior in the context ofpurchase decision of detergents in a specific market. It

    would like to know in particular whether the income level

    of the consumers influence their choice of the brand.

    Currently there are two brands in the market. Brand 1 is

    the premium brand while Brand 2 is the economy brand.

    Income level was classified as Lower, Middle, Upper

    Middle and High and random sampling procedure wasadopted covering the entire market. A sample of 300

    consumers participated in this study. The following data

    emerged from the study.

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    Analyze the data using chi-square testand draw your conclusions.

    Income Level Brand 1 Brand 2

    Lower 25 65Middle 30 30

    Upper Middle 50 22

    High 60 18

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