Question

Peonies A peony plant with red petals was crossed with another plant having streaky petals. A geneticist states that \(75 \%\) of the offspring from this cross will have red flowers. To test this claim, 100 seeds from this cross were collected and germinated, and 58 plants had red petals. Use the chi- square goodness-of-fit test to determine whether the sample data confirm the geneticist's prediction.

Short answer

Answer: No, there is not enough evidence to confirm the geneticist's prediction. Since the p-value (less than 0.001) is less than the standard level of significance (0.05 or 5%), we reject the null hypothesis, concluding that the sample data does not confirm the geneticist's prediction that 75% of the offspring from the cross will have red flowers.

Step-by-step solution

Calculate Expected Frequencies

The geneticist claims that 75% of offspring will have red petals. With a sample size of 100, we can calculate the expected number of offspring with red and streaky petals as follows: Red offspring: \(100\times 0.75=75\) (expected number of red-petaled offspring). Streaky offspring: \(100\times(1-0.75)=100\times 0.25=25\) (expected number of streaky-petaled offspring).

Compute Chi-Square Test Statistic

Using the formula \(\chi^2 = \sum{\frac{(O - E)^2}{E}}\), where O is the observed frequency and E is the expected frequency, we can calculate the chi-square test statistic: For red-petaled offspring: \(\frac{ (58-75)^2 }{75} = \frac{ (-17)^2 }{ 75} = \frac{ 289 }{ 75} =3.8533\) (rounded to four decimal places). For streaky-petaled offspring: \(\frac{(42-25)^2}{25} = \frac{(17)^2}{25} = \frac{ 289 }{ 25} = 11.56\) (rounded to two decimal places). Now, we sum these values to find the total chi-square test statistic: \(\chi^2 = 3.8533 + 11.56 = 15.4133\) (rounded to four decimal places).

Determine Degrees of Freedom and Critical Value

Since there are two categories of offspring (red and streaky), the degrees of freedom for this test is \(df = 2-1 = 1\).

Find the p-value

Using the chi-square table or a calculator, we can find the p-value associated with our test statistic and degrees of freedom. With a chi-square value of 15.4133 and 1 degree of freedom, we find that the p-value is less than 0.001.

Conclusion

Since the p-value (less than 0.001) is less than the standard level of significance (0.05 or 5%), we reject the null hypothesis. This means that there is enough evidence to conclude that the sample data does not confirm the geneticist's prediction that 75% of the offspring from the cross will have red flowers.

Key concepts

Genetics

Genetics is the fascinating study of heredity and variation in organisms. In simple terms, it is all about how traits are passed from parents to their offspring. Each parent contributes genetic material to the offspring, and these combinations can lead to different outcomes, just like the red and streaky petals of peony plants in the exercise.

In this case, the geneticist made a prediction based on known genetic patterns, assuming that 75% of the offspring would have red petals due to a particular genetic cross. These predictions often rely on dominant and recessive traits observed over generations. Understanding genetics helps us deduce such probabilities, making it clear why a 75% prediction was expected.

While genetics can appear complex with terms like dominant, recessive, and heterozygous, at its core, it simply seeks to explain how living organisms inherit features and characteristics. Today's experiment uses the principles of genetics, alongside statistical analysis, to either confirm or challenge these predictions.

Expected Frequency

Expected frequency refers to the predicted number of occurrences for each outcome in a specific test or experiment. It plays a central role in statistical tests like the chi-square goodness-of-fit test.

In the exercise with peony plants, we calculated the expected number of offspring with red petals based on the geneticist's claim. This is done by multiplying the total number of seeds by the probability (75%) of the predicted outcome. So, for red-petaled plants, expected frequency is 75, while for streaky-petaled plants, it's 25.

These expectations help us compare actual outcomes against predictions, using mathematical formulas. Observing deviations between expected and actual frequencies can tell us whether our genetic predictions hold true or if there's more to explore.

Degrees of Freedom

Degrees of freedom is a statistical concept that determines the number of values in a calculation that have the freedom to vary. It plays a crucial role in a chi-square test.

In the chi-square goodness-of-fit test, degrees of freedom help us understand how many independent pieces of information are available to estimate another piece. For the peony's test, we look at two categories—red and streaky. The formula for degrees of freedom is the number of categories minus one. Here, it is 2-1, giving us 1 degree of freedom.

Degrees of freedom are important because they influence the shape of the chi-square distribution, which is crucial for finding the p-value and making decisions about hypothesis testing.

P-value

The p-value in a statistical test is a number that helps you understand whether your results are significant. It is a probability that measures the strength of the evidence against the null hypothesis.

In our scenario, a p-value less than 0.001 is found, indicating a very low probability that the observed differences or more extreme differences occur just by chance under the null hypothesis.

If the p-value is less than a predetermined significance level (commonly 0.05), we reject the null hypothesis. This means the sample provides enough evidence to challenge the geneticist's initial claim. Since our p-value is well below 0.05, it's clear there's a significant difference between the predicted and observed outcomes, suggesting that the geneticist's prediction may not be accurate.