# 3. Models: Proportion Inference

#### David Gerbing

The analysis of proportions is of two primary types.

- Focus on a single value of a categorical variable, termed a “success” when it occurs, for one or more samples of data. Analyze the resulting proportion of occurrence for a single sample, or for a test of homogeneity, compare proportions of successes across distinct data samples for a single variable.
- Compare the obtained proportions across the values of one or more categorical variables for a single sample. Applied to a single variable, the analysis is a goodness-of-fit. Or, evaluate a potential relationship between two categorical variables, a test of independence.

From standard base R functions, the lessR function `Prop_test()`, abbreviated `prop()`, provides either type of analysis. To use, generally enter either the original data from which to compute the frequencies and then the sample proportions, or enter already computed frequencies. For the analysis of multiple categorical variables across two levels of one of the variables, the test of homogeneity and the test of independence yield the identical statistical result.

The following table summarizes the values of the `Prop_test()` parameters for different analyses of proportions. Each function call for the analysis of data begins with the name of a categorical variable, generically referred to as `X`. The value of `X` is the first parameter in the function definition, and so does not need its parameter name, `variable`. If needed, indicate a second categorical variable, generically referred to as `Y`, with the `by` parameter. If focused on a specific value of `X` as a success, referred to as `X_value`, indicate that value with the `success` parameter.

Run each analysis either directly from pre-computed values of the sample proportions, or from the original data from which the sample proportions are calculated.

Evaluate | Data Parameters | Count Parameters
--- | --- | ---
A hypothesized proportion | X, `success`=X_value | `n_succ`, `n_tot` [scalars]
Equal proportions across samples | X, `success`=X_value, `by`=Y | `n_succ`, `n_tot` [vectors]
Uniform goodness-of-fit | X | `ntot` [vector]
Independence of two variables | X, `by`=Y | `n_table`

The remainder of this vignette illustrates these applications of `Prop_test()`.

## Test a Specified Proportion

Define the occurrence of a designated value of the `variable` as a `success`. Define all other values of the variable as failures. Of course, success or failure in this context does not necessarily mean good or bad, desired or undesired, but instead, a designated value either occurred or did not.

When analyzing proportions from data, first indicate the categorical variable, the value of the parameter `variable`. Next, indicate the designated value of `variable` with the parameter `success`. When entering proportions directly, indicate the number of successes and the total number of trials with the `n_succ` and `n_tot` parameters. Enter the value of each parameter either as a single value for one sample or as a vector of multiple values for multiple samples. Without a value for `success` or `n_succ` the analysis is of goodness-of-fit or independence.

### Single Proportion

The example below is from the documentation for the base R function `binom.test()`, which provides the exact test of a null hypothesis regarding the probability of success. `Prop_test()` uses that base R function to compare a sample proportion to a hypothesized population value.

#### From Input Frequencies

For a given categorical variable of interest, a type of plant, consider two values, either “giant” or “dwarf”. From a sample of 925 plants, the specified value of “giant” occurred 682 times and did not occur 243 times. The null hypothesis tested is that the specified value occurs for 3/4 of the population according to the `pi` parameter.

    Prop_test(n_succ=682, n_fail=243, pi=.75)

    ##
    ## <<< Exact binomial test of a proportion
    ##
    ## ------ Describe ------
    ##
    ## Number of successes: 682
    ## Number of failures: 243
    ## Number of trials: 925
    ## Sample proportion: 0.737
    ##
    ## ------ Infer ------
    ##
    ## Hypothesis test for null of 0.75, p-value: 0.382
    ## 95% Confidence interval: 0.708 to 0.765

#### From Data

To illustrate with data, read the Jackets data file included with lessR into the data frame d. The file contains two categorical variables. The variable Bike represents two different types of motorcycle: BMW and Honda. The second variable is Jacket with three values of jacket thickness: Lite, Med, and Thick. Because d is the default name of the data frame that contains the variables for analysis, the `data` parameter that names the input data frame need not be specified.

    d <- Read("Jackets")

    ##
    ## >>> Suggestions
    ## Recommended binary format for data files: feather
    ##   Create with Write(d, "your_file", format="feather")
    ## More details about your data, Enter:  details()  for d, or  details(name)
    ##
    ## Data Types
    ## ------------------------------------------------------------
    ## character: Non-numeric data values
    ## ------------------------------------------------------------
    ##
    ##     Variable                  Missing  Unique
    ##         Name     Type  Values  Values  Values   First and last values
    ## ------------------------------------------------------------------------------------------
    ##  1      Bike character   1025       0       2   BMW  Honda  Honda ... Honda  Honda  BMW
    ##  2    Jacket character   1025       0       3   Lite  Lite  Lite ... Lite  Med  Lite
    ## ------------------------------------------------------------------------------------------

In following example, for the `variable` Bike from the default d data frame, define the parameter `success` as the value “BMW”. The default null hypothesis is a population value of 0.5, but here explicitly specify with the parameter `pi`.

For clarity, the following example includes the parameter names listed with their corresponding values. These names are unnecessary in this example, however, because the values are listed in the same order of their definition of the `Prop_test()` function.

    Prop_test(variable=Bike, success="BMW", pi=0.5)

    ##
    ## <<< Exact binomial test of a proportion
    ##
    ## variable: Bike
    ## success: BMW
    ##
    ## ------ Describe ------
    ##
    ## Number of missing values: 0
    ## Number of successes: 418
    ## Number of failures: 607
    ## Number of trials: 1025
    ## Sample proportion: 0.408
    ##
    ## ------ Infer ------
    ##
    ## Hypothesis test for null of 0.5, p-value: 0.000
    ## 95% Confidence interval: 0.378 to 0.439

Reject the null hypothesis, with a \(p\)-value of 0.000, less than \(\alpha = 0.05\). The sample result of the sample proportion \(p=0.408\) is considered far from the hypothesized value of \(0.5\) for the proportion of `"BMW"` values for Bike. Conclude that the data were sampled from a population with a population proportion of BMW different from 0.5.

### Multiple Proportions

The following example is from the base R `prop.test()` documentation, which the lessR `Prop_test()` relies upon to compare proportions across different groups.

#### From Input Frequencies

The null hypothesis in this example is that the four populations of patients from which the samples were drawn have the same population proportion of smokers. The alternative is that at least one population proportion is different. Label the groups in the output by providing a named vector for the successes.

To indicate multiple proportions across groups, provide multiple values for the `n_succ` and `n_tot` parameters. Optionally, name the groups.

    smokers <- c(83, 90, 129, 70)
    names(smokers) <- c("Group1","Group2","Group3","Group4")
    patients <- c(86, 93, 136, 82)
    Prop_test(n_succ=smokers, n_tot=patients)

    ##
    ## <<< 4-sample test for equality of proportions without continuity correction
    ##
    ##
    ## --- Description
    ##
    ##               Group1   Group2   Group3   Group4
    ## -----------  -------  -------  -------  -------
    ## n_                83       90      129       70
    ## n_total           86       93      136       82
    ## proportion     0.965    0.968    0.949    0.854
    ##
    ## --- Inference
    ##
    ## Chi-square statistic: 12.600
    ## Degrees of freedom: 3
    ## Hypothesis test of equal population proportions: p-value = 0.006

The result of the test is \(p\)-value \(=0.006 < \alpha=0.05\), so reject the null hypothesis of equal probabilities across the corresponding four populations. Conclude that at least one of the population proportions of smokers differ.

#### From Data

In the following example, duplicate the previous results, but in this example from data. To illustrate, create the data frame d according to the proportions of smokers and non-smokers with respective values “smoke” and “nosmoke”. Of course, in actual data analysis the data would already be available.

    sm1 <- c(rep("smoke", 83), rep("nosmoke", 3))
    sm2 <- c(rep("smoke", 90), rep("nosmoke", 3))
    sm3 <- c(rep("smoke", 129), rep("nosmoke", 7))
    sm4 <- c(rep("smoke", 70), rep("nosmoke", 12))
    sm <- c(sm1, sm2, sm3, sm4)
    grp <- c(rep("A",86), rep("B",93), rep("C",136), rep("D",82))
    d <- data.frame(sm, grp)

To test if the different groups have the same population proportion of `success`, retain the syntax for a single proportion for the categorical `variable` of interest. Define success by the value of this variable, here “smoke”. However, an additional parameter `by` indicates the variable that defines the groups, a variable that contains a label that identifies the corresponding group for each row of data. The grouping variable in this example is grp, with values the first four uppercase letters of the alphabet. The first five rows of data are shown below.

    head(d)

    ##      sm grp
    ## 1 smoke   A
    ## 2 smoke   A
    ## 3 smoke   A
    ## 4 smoke   A
    ## 5 smoke   A
    ## 6 smoke   A

The relevant parameters `variable`, `success`, and `by` are listed in their given order in this example, so the parameter names are unnecessary. List the names for clarity.

    Prop_test(variable=sm, success="smoke", by=grp)

    ##
    ## <<< 4-sample test for equality of proportions without continuity correction
    ##
    ## variable: sm
    ## success: smoke
    ## by: grp
    ##
    ## --- Description
    ##
    ##                   A       B       C       D
    ## -----------  ------  ------  ------  ------
    ## n_smoke          83      90     129      70
    ## n_total          86      93     136      82
    ## proportion    0.965   0.968   0.949   0.854
    ##
    ## --- Inference
    ##
    ## Chi-square statistic: 12.600
    ## Degrees of freedom: 3
    ## Hypothesis test of equal population proportions: p-value = 0.006

The analysis of data that matches the previously input proportions, of course, provides the same results as providing the proportions directly.

## Tests without a Specified Proportion

### Goodness-of-Fit

For the previously discussed test of homogeneity of the values of a single categorical variable, the proportion of occurrences for a specific value across different samples is of interest. Here, instead calculate the proportion of occurrence for each value from the total number of occurrences, as one sample from a single population. In addition to the inference test, the following are also reported: - The observed and expected frequencies - The residual of expected from observed - The standardized version of the residual

#### From Input Frequencies

For the goodness-of-fit test to a uniform distribution, provide the frequencies for each group for the parameter `n_tot`. The default null hypothesis is that the proportions of the different categories of a categorical variable are equal.

In this example, enter three frequencies as a vector for the `n_tot` parameter value. Optionally, make the vector a named vector to label the output accordingly.

    x = c(372, 342, 311)
    names(x) = c("Lite", "Med", "Thick")
    Prop_test(n_tot=x)

    ##
    ## <<< Chi-squared test for given probabilities
    ##
    ##
    ## --- Description
    ##
    ##                Lite       Med     Thick
    ## ---------  --------  --------  --------
    ## observed        372       342       311
    ## expected    341.667   341.667   341.667
    ## residual      1.641     0.018    -1.659
    ## stdn res      2.010     0.022    -2.032
    ##
    ## --- Inference
    ##
    ## Chi-square statistic: 5.446
    ## Degrees of freedom: 2
    ## Hypothesis test of equal population proportions: p-value = 0.066

This example does not quite attain significance at the customary 5% level, with \(p\)-value \(= 0.066 > \alpha = 0.05\). A difference of the corresponding population proportions was not detected.

#### From Data

The same analysis follows from the data. Just specify the name of the categorical `variable` of interest.

    d <- Read("Jackets", quiet=TRUE)

    Prop_test(Jacket)

    ##
    ## <<< Chi-squared test for given probabilities
    ##
    ## variable: Jacket
    ##
    ## --- Description
    ##
    ##                Lite       Med     Thick
    ## ---------  --------  --------  --------
    ## observed        372       342       311
    ## expected    341.667   341.667   341.667
    ## residual      1.641     0.018    -1.659
    ## stdn res      2.010     0.022    -2.032
    ##
    ## --- Inference
    ##
    ## Chi-square statistic: 5.446
    ## Degrees of freedom: 2
    ## Hypothesis test of equal population proportions: p-value = 0.066

### Independence

Tests of independence evaluated here rely upon a contingency table of two dimensions also called a cross-tabulation table or joint frequency table. Enter the joint frequencies directly or compute from the data. The corresponding analysis provides the chi-square test for the null hypothesis of independence.

Also provided is Cramer’s V to indicate the extent of the relationship of the two categorical variables. For each cell frequency, the expected value given the independence assumption is provided, along with the corresponding residual from the observed frequency and the corresponding standardized residual.

#### From Input Frequencies

To enter the joint frequency table directly, store the frequencies in a file accessible from your computer system. One possibility is to enter the numbers into a text file with file type `.csv` or `.txt`. Enter the numbers with a text editor, or with a word processor saving the file as a text file. This file format separates the adjacent values in each row with a comma, as indicated below. Or, enter the numbers into an MS Excel formatted file with file type `.xlsx`. Enter only the numeric frequencies, no labels.

For example, consider the following joint frequency table with four levels of the column variable and four levels of the row variable, here in `csv` format.

    3,58,6,105
    41,79,9,207
    86,179,27,484
    143,214,31,824

After saving the file, call `Prop_test()` using the parameter `n_table` to indicate the path name to the file, enclosed in quotes. Or, leave the quotes empty to browse for the joint frequency table.

This table is included in a file downloaded with lessR with the name FreqTable99. That name triggers an internal process that locates the file within the lessR installation without needing to construct a rather complicated path name as part of this example. That also means that the name becomes a reserved key word with its use always triggering the following example.

In general, replace FreqTable99 in this example with your own path name to your file of joint frequencies, or just delete the name leaving only the two quotes to indicate to browse for the file.

    Prop_test(n_table="FreqTable99")

    ##
    ## <<< Pearson's Chi-squared test
    ##
    ## --- Description
    ##
    ## Cell Frequencies
    ##    3  58  6 105
    ##   41  79  9 207
    ##   86 179 27 484
    ##  143 214 31 824
    ##
    ##  Cramer's V: 0.075
    ##
    ##  Row Col Observed Expected Residual Stnd Res
    ##    1   1        3   18.812  -15.812   -4.003
    ##    1   2       58   36.522   21.478    4.150
    ##    1   3        6    5.030    0.970    0.455
    ##    1   4      105  111.635   -6.635   -1.098
    ##    2   1       41   36.750    4.250    0.799
    ##    2   2       79   71.346    7.654    1.098
    ##    2   3        9    9.827   -0.827   -0.288
    ##    2   4      207  218.077  -11.077   -1.361
    ##    3   1       86   84.875    1.125    0.156
    ##    3   2      179  164.776   14.224    1.504
    ##    3   3       27   22.696    4.304    1.105
    ##    3   4      484  503.654  -19.654   -1.781
    ##    4   1      143  132.562   10.438    1.339
    ##    4   2      214  257.356  -43.356   -4.246
    ##    4   3       31   35.447   -4.447   -1.057
    ##    4   4      824  786.635   37.365    3.135
    ##
    ## --- Inference
    ##
    ## Chi-square statistic: 41.732
    ## Degrees of freedom: 9
    ## Hypothesis test of equal population proportions: p-value = 0.000

Do not have the path name to your file readily available? Then browse for the file. The following example is not run as it cannot run in this vignette.

    Prop_test(n_table="")

The full path name for the file is provided as part of the output.

#### From Data

The \(\chi^2\) test of independence evaluated here applies to two categorical variables. The first categorical variable listed in this example is the value of the parameter `variable`, the first parameter in the function definition, so does not need the parameter name. The second categorical variable listed must include the parameter name `by`.

The question for the analysis is if the observed frequencies of Jacket thickness and Bike ownership sufficiently differ from the frequencies expected by the null hypothesis that we conclude the variables are related.

    Prop_test(Jacket, by=Bike)

    ## variable: Jacket
    ## by: Bike
    ##
    ## <<< Pearson's Chi-squared test
    ##
    ## --- Description
    ##
    ##        Jacket
    ## Bike    Lite  Med Thick  Sum
    ##   BMW     89  135   194  418
    ##   Honda  283  207   117  607
    ##   Sum    372  342   311 1025
    ##
    ##  Cramer's V: 0.319
    ##
    ##  Row Col Observed Expected Residual Stnd Res
    ##    1   1       89  151.703  -62.703   -8.288
    ##    1   2      135  139.469   -4.469   -0.602
    ##    1   3      194  126.827   67.173    9.287
    ##    2   1      283  220.297   62.703    8.288
    ##    2   2      207  202.531    4.469    0.602
    ##    2   3      117  184.173  -67.173   -9.287
    ##
    ## --- Inference
    ##
    ## Chi-square statistic: 104.083
    ## Degrees of freedom: 2
    ## Hypothesis test of equal population proportions: p-value = 0.000

The result of this test is that the \(p\)-value = 0.000 \(< \alpha=0.05\), so reject the null hypothesis of independence. Conclude that the type of Bike a person rides and the thickness of their Jacket are related.

To visualize the relationship of the two variables, use the same function call syntax, but now to `Chart()` instead of `Prop_test()`. The visualization is accompanied by the same \(\chi^2\) test of independence.

    Chart(Jacket, by=Bike)

    ## [Interactive chart from the Plotly R package (Sievert, 2020)]
    ##
    ## >>> Suggestions  or  enter: style(suggest=FALSE)
    ## Chart(Jacket, by=Bike, type="radar")  # Plotly radar chart
    ## Chart(Jacket, by=Bike, type="treemap")  # Plotly treemap chart
    ## Chart(Jacket, by=Bike, type="pie")  # Plotly pie/sunburst chart
    ## Chart(Jacket, by=Bike, type="icicle")  # Plotly icicle chart
    ## Chart(Jacket, by=Bike, type="bubble")  # Plotly bubble chart
    ##

    ##
    ## Joint and Marginal Frequencies
    ## ------------------------------
    ##
    ##     Jacket
    ## Bike    Lite Med Thick  Sum
    ##   BMW     89 135   194  418
    ##   Honda  283 207   117  607
    ##   Sum    372 342   311 1025
    ##
    ## Cramer's V: 0.319
    ##
    ## Chi-square Test of Independence:
    ##      Chisq = 104.083, df = 2, p-value = 0.000

The visualization depicts the relationship between motorcycle and jacket: Honda riders prefer thinner jackets, and BMW riders prefer thicker jackets. To speculate, perhaps because the BMW bikes are sportier, their riders are more concerned with going down on the pavement.

This relationship becomes even clearer to visualize with the corresponding 100% stack bar graph. Each bar representing a jacket choice in this visualization shows the percentage of riders with each type of motorcycle for that jacket.

    BarChart(Jacket, by=Bike, stack100=TRUE)

    ## lessR visualizations are now unified over just three core functions:
    ##   - Chart() for pivot tables, such as bar charts. More info: ?Chart
    ##   - X() for a single variable x, such as histograms. More info: ?X
    ##   - XY() for scatterplots of two variables, x and y. More info: ?XY
    ##
    ## BarChart() is deprecated, though still working for now.
    ## Please use Chart(..., type = "bar") going forward.

    ## [Interactive chart from the Plotly R package (Sievert, 2020)]
    ##
    ## >>> Suggestions  or  enter: style(suggest=FALSE)
    ## Chart(Jacket, by=Bike, type="radar")  # Plotly radar chart
    ## Chart(Jacket, by=Bike, type="treemap")  # Plotly treemap chart
    ## Chart(Jacket, by=Bike, type="pie")  # Plotly pie/sunburst chart
    ## Chart(Jacket, by=Bike, type="icicle")  # Plotly icicle chart
    ## Chart(Jacket, by=Bike, type="bubble")  # Plotly bubble chart
    ##

    ##
    ## Joint and Marginal Frequencies
    ## ------------------------------
    ##
    ##     Jacket
    ## Bike    Lite Med Thick  Sum
    ##   BMW     89 135   194  418
    ##   Honda  283 207   117  607
    ##   Sum    372 342   311 1025
    ##
    ## Cramer's V: 0.319
    ##
    ## Chi-square Test of Independence:
    ##      Chisq = 104.083, df = 2, p-value = 0.000

From this visualization we see that 24% of Lite jacket owners are BMW riders, and, in contrast, 62% of the owners of Heavy jackets are BMW riders.