Chi Square Test: Definition And Purpose ✓ Solved

Chi Square Test Ꭓ2 Definition A Test To Determine The Various D

The Chi-Square Test (χ²) is used to determine the deviations expected by chance when a specific hypothesis is true. This statistical test provides a method to analyze whether the observed distribution of data matches the expected distribution under a null hypothesis.

When conducting this test, one typically begins with a known distribution from which a sample is drawn. The main goal of the Chi-Square Test is to identify the degree of deviation between the observed frequencies of occurrences in the sample and the expected frequencies derived from the known distribution. This is particularly useful in genetics, where certain genotype distributions are expected within a population based on probabilities.

For instance, consider a population of plants where the genotypes and their respective probabilities are as follows:

  • RR (Red) – 25%
  • Rr (Orange) – 50%
  • rr (Yellow) – 25%

If there are 320 plants expected based on these probabilities, the expected observed numbers for each genotype would be:

  • RR = 80
  • Rr = 160
  • rr = 80

After growing these plants, suppose the observed counts are:

  • RR = 65
  • Rr = 189
  • rr = 66

The next step involves calculating the Chi-Square statistic, which is comparatively simple and involves the formula:

χ² = Σ((O − E)² / E)

Where O is the observed frequency and E is the expected frequency. Using the provided data:

  • For Red: O = 65, E = 80 → (O-E)² = (65-80)² = 225, (O-E)²/E = 225/80 = 2.8125
  • For Orange: O = 189, E = 160 → (O-E)² = (189-160)² = 841, (O-E)²/E = 841/160 = 5.25625
  • For Yellow: O = 66, E = 80 → (O-E)² = (66-80)² = 196, (O-E)²/E = 196/80 = 2.45

Summing these values gives us:

χ² = 2.8125 + 5.25625 + 2.45 = 10.51875.

Upon rounding, this statistic is consistent with a Chi-Squared value of approximately 10.51. By using the Chi-Square distribution table, we find the corresponding degrees of freedom (df). In this example, the degrees of freedom can be calculated as:

df = k - 1 = 3 - 1 = 2, where k is the number of categories.

With a Chi-Square value of about 10.51 and 2 degrees of freedom, we can determine the p-value. Looking at the Chi-Square distribution, we identify that with a df of 2, the critical value comparison shows that the p-value lies between 0.01 and 0.005.

Formulating the null hypothesis (H₀): The observed and expected outcomes would be equal if there is no significant deviation from the known genotype distributions due to incomplete dominance. The alternative hypothesis (H₁) posits that the observed distribution deviates significantly from the expected outcomes.

If we assume that the null hypothesis is true, statistically, such deviations would only occur approximately 0.5-1% of the time. Thus, due to the small likelihood of observing results this extreme, we would reject the null hypothesis. This finding indicates a significant deviation under the premise of incomplete dominance.

Moreover, this analysis can be efficiently performed utilizing statistical software or programming languages such as R. By using the function chisq.test, we can easily input the observed counts and the expected probabilities to compute the Chi-Square statistic and the associated p-value:

genetest 

chisq.test(genetest, p = c(1/4, 1/2, 1/4))

Upon executing this command, the output provides values affirming our previous manual calculations:

Chi-squared test for given probabilities data: genetest 

X-squared = 10.519, df = 2, p-value = 0.005199

In conclusion, the very low p-value indicates a highly significant deviation from expected frequencies. Hence, we reject the null hypothesis, supporting the idea that the expected genotype distribution is not reflective of our observed data given the context of incomplete dominance.

Paper For Above Instructions

The Chi-Square Test (χ²) serves as an essential tool in statistical analysis, particularly within fields such as genetics, ecology, and social sciences, offering valuable insight into the relationship between observed and expected data distributions. This paper delineates the utilization and interpretation of the Chi-Square Test, illustrating its methodology through a botanical case study involving plant genotypes in a hypothetical garden experiment.

The fundamental premise of the Chi-Square Test is to analyze whether discrepancies between observed and expected frequencies arise by chance or if they indicate a significant divergence from a hypothesized distribution. The test compares the observed counts of different categories against a theoretically expected distribution under the null hypothesis, which stipulates that any observed differences result solely from random variation. In particular, the test's formula, which calculates the squared differences between observed (O) and expected (E) frequencies, normalized by the expected values, encapsulates its essence:

χ² = Σ((O − E)² / E)

In our example, a garden experiment yielded three distinct genotypes of plants, each with specified genotype frequencies. With 320 plants, we expected the counts for each genotype to adhere to 25% for RR and rr genotypes and 50% for the Rr genotype, leading to expected frequencies of 80, 160, and 80 observed counts, respectively. However, the actual observed counts deviated slightly from these expected figures: RR at 65, Rr at 189, and rr at 66.

We calculated O-E and derived the Chi-Square statistic by substituting our findings into the Chi-Square formula, resulting in a significant value of around 10.51. Evaluating the Chi-Square statistic against critical values in the Chi-Square distribution table indicated the significance of our results, particularly with a computed p-value of approximately 0.005199. This value signifies that the likelihood of observing such a deviation due to chance alone is exceedingly low.

Ultimately, our hypothesis testing involved a null hypothesis (H₀) positing no significant difference between observed and expected distributions under the condition of incomplete dominance. The rejection of the null hypothesis, supported by the small p-value, implies that the underlying assumptions about genotype distribution are untenable and encourages an exploration of additional factors that may contribute to these observed discrepancies. The utility of this statistical test is exemplified by our understanding of plant genetics and the biological principles governing genotype frequencies.

References

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  • Harris, R. J. (1995). Statistics for Historians. New York: Cambridge University Press.
  • Keppel, G. (1991). Design and Analysis: A Researcher's Handbook. New Jersey: Prentice Hall.
  • Kirkpatrick, D. C. (1984). Chi-square Analysis of Variance. Agronomy Journal, 76, 691-693.
  • Lindgren, B. W. (1980). Statistical Theory. New York: Macmillan.
  • Pearson, K. (1900). On the Criterion That a Given System of Deviations from the Probable in the Case of a Correlation Continued in Time. Philosophical Transactions of the Royal Society of London, 198, 243-263.
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  • Rice, J. A. (2006). Mathematical Statistics and Data Analysis. Belmont, California: Duxbury Press.
  • Sharma, S. K. (2003). Chi-Squared Test: As a Statistical Tool for Testing Hypothesis. Induction Research Journal, 12(1), 23-29.

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