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Consider the following frequency table of observations on the random variable \(X\). $$\begin{array}{lrrrrr}\text { Values } & 0 & 1 & 2 & 3 & 4 \\\\\text { Observed frequency } & 24 & 30 & 31 & 11 & 4\end{array}$$ (a) Based on these 100 observations, is a Poisson distribution with a mean of 1.2 an appropriate model? Perform a goodness-of-fit procedure with \(\alpha=0.05 .\) (b) Calculate the \(P\) -value for this test.

Short Answer

Expert verified
A Poisson distribution with mean 1.2 is an appropriate model, as the chi-square test shows no significant difference (P-value ≈ 0.071).

Step by step solution

01

Define the Hypotheses

For the goodness-of-fit test, we define the null hypothesis, \( H_0 \), that the data follows a Poisson distribution with \( \lambda = 1.2 \). The alternative hypothesis, \( H_a \), is that the data does not follow a Poisson distribution with this mean.
02

Calculate Expected Frequencies

Using the Poisson probability mass function \( P(X=k) = \frac{\lambda^k e^{-\lambda}}{k!} \), we calculate the expected frequencies for each value of \(X\). Multiply each probability by 100 (the total number of observations) to find the expected frequencies:- For \(X = 0\): \( P(X=0) = \frac{1.2^0 e^{-1.2}}{0!} \approx 0.3012 \), so expected frequency is \(100 \times 0.3012 = 30.12\).- For \(X = 1\): \( P(X=1) = \frac{1.2^1 e^{-1.2}}{1!} \approx 0.3614 \), so expected frequency is \(100 \times 0.3614 = 36.14\).- For \(X = 2\): \( P(X=2) = \frac{1.2^2 e^{-1.2}}{2!} \approx 0.2168 \), so expected frequency is \(100 \times 0.2168 = 21.68\).- For \(X = 3\): \( P(X=3) = \frac{1.2^3 e^{-1.2}}{3!} \approx 0.0867 \), so expected frequency is \(100 \times 0.0867 = 8.67\).- For \(X = 4\) and beyond, combine with \( X = 4+ \): \( P(X \geq 4) \approx 1 - (P(X=0) + P(X=1) + P(X=2) + P(X=3)) \approx 0.034 \), so expected frequency is \(100 \times 0.034 = 3.39\).
03

Perform Chi-Square Test

Calculate the chi-square statistic with formula:\[ \chi^2 = \sum \frac{(O_i - E_i)^2}{E_i} \]where \( O_i \) are the observed frequencies and \( E_i \) are the expected frequencies:- For \( X=0 \): \( (24-30.12)^2 / 30.12 \approx 1.248 \)- For \( X=1 \): \( (30-36.14)^2 / 36.14 \approx 1.044 \)- For \( X=2 \): \( (31-21.68)^2 / 21.68 \approx 3.978 \)- For \( X=3 \): \( (11-8.67)^2 / 8.67 \approx 0.651 \)- For \( X=4+ \): \( (4-3.39)^2 / 3.39 \approx 0.108 \)Sum these to obtain \( \chi^2 \approx 7.029 \).
04

Determine Critical Value and Decision

With degrees of freedom \( df = k - 2 = 5 - 2 = 3 \) (since parameter \( \lambda \) was estimated), and \( \alpha = 0.05 \), find the critical value from the chi-square table \(\chi^2_{0.05,3} \approx 7.815\). Since \(7.029 < 7.815\), we do not reject \( H_0 \).
05

Calculate P-value

The P-value is the probability of observing a chi-square statistic as extreme as 7.029 under the null hypothesis. From a chi-square distribution with \(3\) degrees of freedom, the P-value is approximately 0.071.
06

Interpret Results

Since the P-value \(0.071 > 0.05\), we do not reject the null hypothesis. It suggests that a Poisson distribution with mean \(1.2\) is an appropriate model for this data.

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Key Concepts

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

Poisson distribution
A Poisson distribution is a probability distribution used to model the occurrence of events over a specified interval of time or space. It is particularly useful when events occur randomly and independently, and when they are rare overall within large samples. Poisson distributions are characterized by their mean \( \lambda \), which is also equal to the variance of the distribution. This distribution is appropriate for modeling count data, such as the number of phone calls at a call center in an hour or the number of decay events per second from a radioactive source.
To use a Poisson distribution, it is assumed that:
  • Events are independent of each other, meaning the occurrence of one event does not affect the probability of another.
  • The average rate (mean number of events) is constant over time.
  • Two events can’t occur at the exact same instant.
When using the Poisson distribution in hypothesis testing, the formula \( P(X=k) = \frac{\lambda^k e^{-\lambda}}{k!} \) is applied to calculate the probabilities for different numbers of occurrences \( k \). For instance, if \( \lambda = 1.2 \), the random variable \( X \) represents the number of occurrences, and different probabilities are computed for each \( k \) to derive expected frequencies.
Chi-square test
The Chi-square test is a statistical test used to examine the differences between categorical variables from a random sample. In the context of goodness-of-fit tests, it helps us determine whether our observed data fits a particular distribution, like the Poisson distribution in the exercise.
In performing a Chi-square goodness-of-fit test, you first calculate expected frequencies based on the assumed distribution (e.g., Poisson). Then, using the formula \[ \chi^2 = \sum \frac{(O_i - E_i)^2}{E_i} \] where \( O_i \) stands for the observed frequencies and \( E_i \) are the expected frequencies, you compute the Chi-square test statistic. This can be interpreted as a measure of how much the observed data deviates from what would be expected under the assumed distribution.
The steps are as follows:
  • Calculate the expected frequencies for each observed category or event based on the theoretical distribution.
  • Apply the formula to find the Chi-square value.
  • Find the degrees of freedom, typically the number of categories minus the number of estimated parameters (plus one).
  • Compare the calculated \( \chi^2 \) value to a critical value from a Chi-square distribution table using the chosen significance level \( \alpha \).
A Chi-square test with a result below the critical value indicates that the observed distribution and the expected distribution do not have a significant difference. If it is above, the difference is considered significant.
Hypothesis testing
Hypothesis testing is a fundamental statistical method used to make decisions about data. It involves two hypotheses: the null hypothesis \( H_0 \), which reflects what is presumed to be true about the dataset, and the alternative hypothesis \( H_a \), representing what you want to test for.
In the context of this exercise, we defined our hypotheses as follows:
  • \( H_0 \): The data follows a Poisson distribution with a mean of 1.2.
  • \( H_a \): The data does not follow a Poisson distribution with this mean.
Performing hypothesis testing involves:
  • Calculating a test statistic from the sample data, here it's the Chi-square statistic.
  • Determining the corresponding \( p \)-value, which indicates the probability of observing the test statistic as extreme as the one calculated, assuming \( H_0 \) is true.
  • Comparing the \( p \)-value to the significance level \( \alpha \) (commonly 0.05). If the \( p \)-value is less than \( \alpha \), the null hypothesis is rejected, suggesting evidence for the alternative hypothesis.
In this exercise, the calculated \( p \)-value was 0.071. Since it was greater than the significance level of 0.05, we did not reject \( H_0 \), suggesting that the Poisson distribution with a mean of 1.2 might indeed be a suitable fit for the observed data.

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