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9.34 Plant Genetics A peony plant with red petals was crossed with another plant having streaky petals. A geneticist states that \(75 \%\) of the offspring resulting 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. a. What hypothesis should you use to test the geneticist's claim? b. Calculate the test statistic and its \(p\) -value. Use the \(p\) -value to evaluate the statistical significance of the results at the \(1 \%\) level.

Short Answer

Expert verified
Answer: Yes, the hypothesis test provides sufficient evidence to reject the geneticist's claim at the 1% significance level.

Step by step solution

01

Formulate the null and alternative hypotheses

We want to test the geneticist's claim that \(75\%\) of the offspring will have red flowers. The null hypothesis (\(H_0\)) would be that the proportion of plants with red flowers in the population is equal to \(0.75\), and the alternative hypothesis (\(H_1\)) would be that the proportion is not equal to \(0.75\). \(H_0: p = 0.75\) \(H_1: p \neq 0.75\)
02

Calculate the test statistic

The test statistic for testing a proportion can be calculated using the following formula: \(Z=\dfrac{(\hat{p}-p_0)}{\sqrt{\dfrac{p_0(1-p_0)}{n}}}\) Here, \(\hat{p}\) is the observed proportion in the sample, \(p_0\) is the hypothesized proportion under the null hypothesis, and \(n\) is the sample size. In this case, \(\hat{p} = \dfrac{58}{100} = 0.58\), \(p_0 = 0.75\), and \(n = 100\). Plug these values into the formula: \(Z = \dfrac{(0.58 - 0.75)}{\sqrt{\dfrac{0.75(1-0.75)}{100}}}= -4.62\)
03

Calculate the p-value

The p-value is the probability of obtaining a test statistic as extreme or more extreme as the observed value under the null hypothesis. Since this is a two-tailed test (alternative hypothesis is \(p \neq 0.75\)), the p-value is equal to the combined probability of obtaining a Z-score less than or equal to \(-4.62\) or a Z-score greater than or equal to \(4.62\). Using a standard normal distribution table or a calculator: \(p-value = 2 \times P(Z \leq -4.62) = 2 \times (1-0.99981) = 0.00038\)
04

Compare the p-value to the significance level

We are given a significance level of \(1\%\), which means that we would reject the null hypothesis if the p-value is less than or equal to \(0.01\). Since the p-value we calculated (\(0.00038\)) is less than the significance level (\(0.01\)), we reject the null hypothesis.
05

Conclusion

Based on our hypothesis test, we reject the null hypothesis and conclude that there is sufficient evidence at the \(1\%\) significance level to doubt the geneticist's claim that \(75\%\) of the offspring resulting from this cross will have red flowers.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Null Hypothesis
The null hypothesis is a foundational concept in hypothesis testing. It represents a general statement or default position that there is no relationship between two measured phenomena or no association among groups. In statistical terms, it is a statement we test, often expressed as an equation. In the context of the peony plant genetic experiment, the null hypothesis (\(H_0\)) states that \(75\%\) of the offspring will have red flowers. This hypothesis presumes that the observed effect is due to chance.Some key points about the null hypothesis:
  • It is usually denoted as \(H_0\).
  • It always contains an equality statement (e.g., \(p = 0.75\)).
  • The goal of hypothesis testing is to determine whether there is enough evidence to reject the null hypothesis.
By assuming the null hypothesis is true, we can then assess whether the real-world data we collect fits within this assumption. If not, we might consider rejecting it in favor of something else.
Alternative Hypothesis
The alternative hypothesis is the statement we want to test against the null hypothesis. It represents a new theory or claim that the experiment or investigation is seeking to prove. In the peony plant scenario, the alternative hypothesis (\(H_1\)) suggests that the proportion of plants with red flowers is different from \(75\%\). Essentially, it challenges the geneticist's initial claim.Some important points about the alternative hypothesis:
  • It is symbolized as \(H_1\) or \(H_a\).
  • It does not contain an equality statement; instead, it shows inequality, such as \(p eq 0.75\), \(p > 0.75\), or \(p < 0.75\).
  • Acceptance of this hypothesis suggests evidence against the null hypothesis.
The decision of accepting the alternative hypothesis over the null hypothesis is based on the observed data. If the data is statistically significant, then the alternative hypothesis is a more probable explanation than the null hypothesis.
P-Value
The p-value is a critical component in hypothesis testing, providing a means to evaluate the strength of the evidence against the null hypothesis. It indicates the probability of obtaining the observed data, or something more extreme, if the null hypothesis were true. In this exercise, a p-value is calculated to see how likely the observed difference (58% vs 75% red flowers) could have occurred by random chance.Key aspects of p-value:
  • It helps to determine the statistical significance of the test results.
  • A smaller p-value (usually \(\leq 0.05\)) suggests stronger evidence against the null hypothesis.
  • It provides a "threshold" for determining statistical significance, which is usually set by the researcher.
In the peony plant example, the p-value was found to be \(0.00038\), which is very low, indicating that the difference from \(75\%\) is unlikely to have occurred by chance. As a result, this low p-value leads us to reject the null hypothesis, suggesting an alternative explanation for the observed data.
Statistical Significance
Statistical significance helps determine whether the observed effect is real or happened by chance. It provides a cut-off point that researchers use to decide whether the results may suggest evidence against the null hypothesis. This cut-off value is known as the significance level and is often set at \(\alpha = 0.05\) or \(\alpha = 0.01\), representing \(5\%\) or \(1\%\) risk of error.Significance in the peony plant case:
  • We use a threshold of \(1\%\) significance level (\(\alpha = 0.01\)).
  • The p-value of \(0.00038\) is much lower than \(0.01\), suggesting the results are statistically significant.
  • As a result, we have enough evidence to reject the null hypothesis.
Statistical significance does not measure the size of an effect or the importance of a result, just the likelihood that the result could have occurred by chance. In this case, due to the small p-value, it suggests strong evidence against the null hypothesis that \(75\%\) of the offspring will have red flowers, favoring an alternative explanation.

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