BIOSTATS 3.5 - CH. 21: THE CHI-SQUARE TEST FOR GOODNESS OF FIT

• The chi-square test for goodness of fit is covered in Chapter 21

Comparing Two Proportions

• Two-sample problems involve proportions
• The sampling distribution assesses the difference between two proportions.
• Large sample confidence intervals are used for comparing proportions.
• More accurate confidence intervals assist in comparing proportions.
• Hypothesis tests facilitate the comparison of proportions.
• Relative risk and odds ratio relate to comparing two proportions

Large Sample CI for Two Proportions

• Approximate level C confidence interval (CI) for p1 - p2 with two independent simple random samples (SRS) of sizes n1 and n2, and sample proportions of successes p̂1 and p̂2: (p̂1 - p̂2) ± m
• m is the margin of error
• m = zSEdiff = z** √(p̂1(1 - p̂1)/n1 + p̂2(1 - p̂2)/n2)
• C signifies the area under the standard Normal curve between -z* and z*.
• Use the method when the number of successes and failures are at least 10 in each sample.

Hypothesis Tests for Two Proportions

• The null hypothesis (H0) is that p1 = p2 = p
• The information from both samples can be pooled to estimate p if H0 is true, assuming sampling is done twice from the same population.
• The pooled sample proportion is p̂ = total successes / total observations = (count1 + count2) / (n1 + n2)
• Z = p̂1 - p̂2 / √(p̂(1 - p̂)(1/n1 + 1/n2))
• This is appropriate when all counts (successes and failures in each sample) are 5 or more.

Chi-Square Test for Goodness of Fit

• The chi-square test for goodness of fit is applied to a categorical variable (one SRS) with any number of levels.
• The null hypothesis posits that all population proportions are equal (uniform hypothesis).
• A goodness-of-fit hypothesis example question: Are hospital births uniformly distributed in the week? H0: p1 = p2 = p3 = p4 = p5 = p6 = p7 = 1/7
• Population proportions may be equal to specific values, provided they sum to 1 in H0. Co-dominant phenotype H0 example: When crossing homozygote parents expressing two co-dominant phenotypes A and B, expect in F2 H0: pA = 1/4, pAB = 1/2, pB = 1/4, where AB is an intermediate phenotype.
• The chi-square test is used when data are categorical. It checks how different the observed data are from expectation under H0.

Chi-Square Statistic

• The chi-square statistic (X2) compares observed and expected counts.
• Observed counts: The actual number of observations of each type.
• Expected counts: The number of observations expected of each type under the null hypothesis.
• X2 = Σ ((observed count - expected count)² / expected count), calculated separately for each type and then summed.
• Statistically significant large X2 values represent strong deviations from the expected distribution under H0.

Chi-Square Distributions

• Chi-square distributions are a family of right-skewed distributions that take only positive values, characterized by specific degrees of freedom.
• Published tables and software provide upper-tail area for critical values across many chi-square distributions.

Chi-Square Test for Goodness of Fit

• With k proportions, the chi-square statistic for goodness of fit measures the divergence between observed and expected counts.
• It follows the chi-square distribution with k - 1 degrees of freedom, as described by: X2 = Σ ((count of outcome i – npio)² / npio)
• The P-value is the tail area under the chi-square distribution, where df = k − 1.

Goodness of Fit Example

• Aphids avoid ladybugs by dropping off leaves. An experiment examines this mechanism.
• In the experiment, live aphids landed on their ventral side in 95% of the trials (19 out of 20) when dropped upside-down from tweezers; dead aphids landed on their ventral side in 52.2% of the trials (12 out of 23).
• Question: At a 5% significance level, is there evidence live aphids’ landing right side up (on their ventral side) is not a chance event?
• H0: pventral = 0.5 and pdorsal = 0.5
• Ha: H0 is not true.
• Expected counts are large enough (both are greater than 5.0).
• X2 contributions calculation: Σ (obs-exp)² / exp = 8.1 + 8.1 = 16.2. Since X2 > 12.12, P < 0.0005 (software presents P-value = 0.000057).
• This is a highly significant result, there is very strong evidence the righting behavior of live aphids is not chance (P < 0.0005). The H0 is rejected.

Interpreting Significant Chi-Square Results

• The individual values summed in the X2 statistic are the chi-square components (or contributions).
• When the test is statistically significant, the largest components indicate conditions differing most from the expected H0.
• Compare observed and expected counts to interpret the findings.
• The actual proportions can be compared qualitatively in a graph.

Lack of Significance

• A non-significant P-value is not conclusive.
• H0 could be true (or not).
• This is relevant in the chi-square goodness of fit test where we are often interested in the hypothesis that the data fit a particular model.
• A significant P-value indicates the data do not follow that model.
• A non-significant P-value is NOT validation of the null hypothesis.
• It does NOT indicate data follow the hypothesized model, shows the data are not inconsistent with the model.

Lack of Significance Example

• The Frizzle fowl chicken variety has curled feathers, a genetic cross between Frizzle and Leghorn (straight-feathered).
• F1 chickens produced all had slightly frizzled feathers.
• Phenotype counts include 23 Frizzled, 50 Slightly frizzled, and 20 Straight.
• The most likely genetic model is that of a single gene locus with two co-dominant alleles producing a 1:2:1 ratio in F2.
• This raises the question as to whether the data is consistent with such a model.
• H0: pF = 1/4; psF = 2/4; ps = 1/4
• Ha: H0 is not true
• X2 = ((23-23.25)²/23.25)+((50-46.5)²/46.5)+((20-23.25)²/23.25) ≈ 0.72
• Test results are 0.002688 for Frizzled, 0.263441 for Slightly frizzled, and 0.454301 for Straight

Conditions for the Goodness of Fit Test

• The chi-square test for goodness of fit is used when there is a single SRS from a population.
• The variable is categorical with k mutually exclusive levels.
• The chi-square test is safely used when all expected counts have values ≥ 1.0, and no more than 20% of the k expected counts have values < 5.0.

Chi-Square Statistics Example

• Consider a plant geneticist analyzing 100 progeny from a cross with a 3:1 phenotypic ratio hypothesis of yellow-flowered to green-flowered plants.
• The observed ratio of 84 yellow: 16 green is compared to the 75 yellow: 25 green ratio predicted by the geneticist’s hypothesis.
• The core question is whether the observed frequencies significantly deviate from the expected frequencies if the hypothesis were accurate.
• The null hypothesis in this case is that the population has a 3:1 ratio of yellow-flowered to green-flowered plants.
• The alternative hypothesis would be that the population has a flower-color ratio which is not 3 yellow: 1 green.
• 100 observations are made: X2 = ((84 - 75)2 / 75) + ((16 - 25)2 / 25) = 4.320.
• Larger disagreements between observed and expected frequencies result in larger X2 values. This calculation measures goodness of fit.
• The test statistic generalizes for more than just 2 categories.
• A X2 value of 0 indicates a perfect fit between the observed and the expected.