A biologist uses the chi-square test to determine whether the observed distribution of 5
species matches an expected theoretical distribution. The chi-square test yields a test
statistic of 7.779 .
What is the null hypothesis?
The observed distribution and the expected distribution do not differ v
What is the alternative hypothesis?
The mean of the random variable is zero
What is the number of degrees of freedom? Ex: 5 .
What is the -value? Ex: 1.23 .
Does the data differ from expected distribution at the significance level?

Ask by Wade Jimenez. in the United States

Dec 17,2024

Question

A biologist uses the chi-square test to determine whether the observed distribution of 5
species matches an expected theoretical distribution. The chi-square test yields a test
statistic of 7.779 .
What is the null hypothesis?
The observed distribution and the expected distribution do not differ v
What is the alternative hypothesis?
The mean of the random variable is zero
What is the number of degrees of freedom? Ex: 5 .
What is the -value? Ex: 1.23 .
Does the data differ from expected distribution at the significance level?

Ask by Wade Jimenez. in the United States

Dec 17,2024

UpStudy AI · Accepted Answer

To determine whether the observed distribution of 5 species matches an expected theoretical distribution using the chi-square test, we need to consider the following:

1. Null Hypothesis ($$ H_{0} $$): The observed distribution and the expected distribution do not differ significantly.
2. Alternative Hypothesis ($$ H_{1} $$): The observed distribution differs significantly from the expected distribution.
3. Number of Degrees of Freedom: The number of degrees of freedom is calculated as $$ k-1 $$, where $$ k $$ is the number of categories or groups.
4. $$ p $$-value: The $$ p $$-value is the probability of obtaining a test statistic as extreme as the one observed, assuming the null hypothesis is true.
5. Significance Level ($$ \alpha $$): The significance level is the threshold used to determine whether to reject the null hypothesis.

Given:
- $$ \chi^{2} $$ test statistic = 7.779
- Number of species = 5
- Significance level $$ \alpha = 0.05 $$

Let's calculate the number of degrees of freedom and the $$ p $$-value to determine if the data differs from the expected distribution at the $$ \alpha = 0.05 $$ significance level.
Calculate the value by following steps:
- step0: Calculate:
  $$5-1$$
- step1: Subtract the numbers:
  $$4$$
The number of degrees of freedom is 4.

Next, we need to calculate the $$ p $$-value for the given $$ \chi^{2} $$ test statistic of 7.779. Since the $$ p $$-value is not directly provided, we will calculate it using the chi-square distribution table or a statistical software.
Calculate the value by following steps:
- step0: Calculate:
  $$1-0.9$$
- step1: Subtract the numbers:
  $$0.1$$
The $$ p $$-value is 0.1.

Now, we compare the $$ p $$-value (0.1) with the significance level $$ \alpha = 0.05 $$ to determine if the data differs from the expected distribution at the $$ \alpha = 0.05 $$ significance level.

Since the $$ p $$-value (0.1) is greater than the significance level (0.05), we fail to reject the null hypothesis. This means that there is not enough evidence to conclude that the observed distribution differs significantly from the expected distribution at the $$ \alpha = 0.05 $$ significance level.
1. Null Hypothesis ($$ H_{0} $$): The observed distribution and the expected distribution do not differ significantly. 2. Alternative Hypothesis ($$ H_{1} $$): The observed distribution differs significantly from the expected distribution. 3. Number of Degrees of Freedom: 4. 4. $$ p $$-value: 0.1. 5. Conclusion: At $$ \alpha = 0.05 $$, the null hypothesis is not rejected, indicating no significant difference between the observed and expected distributions.

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