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Bacteria Colonies: Poisson Distribution A pathologist has been studying the frequency of bacterial colonies within the field of a microscope using samples of throat cultures from healthy adults. Long-term history indicates that there is an average of 2.80 bacteria colonies per field. Let \(r\) be a random variable that represents the number of bacteria colonies per ficld. Let \(O\) represent the number of observed bacteria colonies per field for throat cultures from healthy adults. A random sample of 100 healthy adults gave the following information. $$\begin{array}{c|cccccc} \hline \boldsymbol{r} & 0 & 1 & 2 & 3 & 4 & 5 \text { or more } \\ \hline \boldsymbol{o} & 12 & 15 & 29 & 18 & 19 & 7 \\ \hline \end{array}$$ (a) The pathologist wants to use a Poisson distribution to represent the probability of \(r\), the number of bacteria colonies per field. The Poisson distribution is \(P(r)=\frac{e^{-\lambda} \lambda^{r}}{r !}\) where \(\lambda=2.80\) is the average number of bacteria colonies per field. Compute \(P(r)\) for \(r=0,1,2,3,4,\) and 5 or more. (b) Compute the expected number of colonies \(E=100 P(r)\) for \(r=0,1,2,3,4,\) and 5 or more. (c) Compute the sample statistic \(\chi^{2}=\Sigma \frac{(O-E)^{2}}{E}\) and the degrees of freedom. (d) Test the statement that the Poisson distribution fits the sample data. Use a \(5 \%\) level of significance.

Step by step solution

01

Calculate Poisson Probabilities

We use the formula for the Poisson distribution: \(P(r) = \frac{e^{-\lambda} \lambda^r}{r!}\), where \(\lambda = 2.80\). Calculate \(P(r)\) for \(r=0, 1, 2, 3, 4, \) and \(r \geq 5\):- \(P(0) = \frac{e^{-2.80} \cdot 2.80^0}{0!} \approx 0.0608\)- \(P(1) = \frac{e^{-2.80} \cdot 2.80^1}{1!} \approx 0.1701\)- \(P(2) = \frac{e^{-2.80} \cdot 2.80^2}{2!} \approx 0.2382\)- \(P(3) = \frac{e^{-2.80} \cdot 2.80^3}{3!} \approx 0.2223\)- \(P(4) = \frac{e^{-2.80} \cdot 2.80^4}{4!} \approx 0.1556\)- \(P(\geq 5) = 1 - (P(0) + P(1) + P(2) + P(3) + P(4)) \approx 0.1530\)

02

Compute Expected Frequencies

Calculate the expected number of colonies \(E = 100 \cdot P(r)\) for each \(r\).- \(E(0) = 100 \cdot 0.0608 = 6.08\)- \(E(1) = 100 \cdot 0.1701 = 17.01\)- \(E(2) = 100 \cdot 0.2382 = 23.82\)- \(E(3) = 100 \cdot 0.2223 = 22.23\)- \(E(4) = 100 \cdot 0.1556 = 15.56\)- \(E(\geq 5) = 100 \cdot 0.1530 = 15.30\)

03

Calculate Chi-Square Statistic

Use the formula \(\chi^2 = \Sigma \frac{(O - E)^2}{E}\) to calculate the chi-square statistic.- For \(r=0:\) \(\frac{(12 - 6.08)^2}{6.08} \approx 5.71\)- For \(r=1:\) \(\frac{(15 - 17.01)^2}{17.01} \approx 0.24\)- For \(r=2:\) \(\frac{(29 - 23.82)^2}{23.82} \approx 1.09\)- For \(r=3:\) \(\frac{(18 - 22.23)^2}{22.23} \approx 0.81\)- For \(r=4:\) \(\frac{(19 - 15.56)^2}{15.56} \approx 0.75\)- For \(r \geq 5:\) \(\frac{(7 - 15.30)^2}{15.30} \approx 4.50\)Summing these gives \(\chi^2 \approx 13.10\).

04

Determine Degrees of Freedom

The degrees of freedom (df) for the chi-square test is calculated as \(\text{df} = k - 1\), where \(k\) is the number of categories, minus the number of parameters estimated (which is 1 for \(\lambda\)). For this problem, \(k = 6\) (0, 1, 2, 3, 4, ≥5), thus \(\text{df} = 5\).

05

Conduct Chi-Square Test

Using a chi-square distribution table, find the critical value for \(\text{df} = 5\) at \(5\%\) significance level (\(\alpha = 0.05\)). The critical value is approximately 11.07. Since the calculated \(\chi^2 = 13.10\) is greater than 11.07, we reject the null hypothesis. This suggests that the Poisson distribution does not fit the sample data well.

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Key Concepts

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

Chi-Square Test

The Chi-Square test is a statistical method used to determine if there is a significant difference between expected and observed frequencies in one or more categories. In this context, it helps us evaluate whether the Poisson distribution adequately represents our bacterial colony data.

The Chi-Square statistic formula is:

• \( \chi^2 = \sum \frac{(O - E)^2}{E} \)

Here,

• \( O \) is the observed frequency of bacteria colonies,
• \( E \) is the expected frequency calculated from the Poisson distribution.

Each category (0, 1, 2, 3, 4, 5 or more) contributes to the total chi-square value. We sum these contributions to determine the final \( \chi^2 \) value. If this value is greater than the critical value from the Chi-Square distribution table, we reject the Poisson distribution as a model for our data.

In this example, the calculated \( \chi^2 \) was 13.10, which exceeded the critical value of 11.07, leading us to conclude that the model doesn't fit our data well.

Degrees of Freedom

Degrees of Freedom, often abbreviated as df, is a key concept in statistical tests like the Chi-Square test. It determines the distribution of the test statistic and influences the critical value used for hypothesis testing.

In a Chi-Square test, the degrees of freedom is generally calculated as:

• \( \text{df} = k - 1 - p \)

where:

• \( k \) is the number of categories or distinct groups in the data,
• \( p \) is the number of parameters estimated to fit the distribution, which is 1 in this Poisson distribution case (\( \lambda \)).

For our bacteria colony exercise with categories: 0, 1, 2, 3, 4, and 5 or more, we have 6 distinct groups. Thus, with one parameter, our degrees of freedom work out as \( 6 - 1 - 1 = 4 \).

A greater number of degrees of freedom will increase the Chi-Square distribution span, affecting the cut-off for significance. For such tests, understanding degrees of freedom helps us interpret statistical analysis results accurately.

Probability

Probability is a foundational concept in statistics, quantifying the chances of an event. In the context of the Poisson distribution, it gives us the likelihood of observing a particular number of bacteria colonies under predefined conditions.

For the Poisson distribution:

• \( P(r) = \frac{e^{-\lambda} \cdot \lambda^r}{r!} \)

This formula estimates the probability, \( P(r) \), of \( r \) bacteria colonies in a given field.

• \( \lambda \) represents the average rate of occurrence (2.80 in our case).
• \( e \) is the base of the natural logarithm, approximately 2.718.
• \( r! \) or 'r factorial' accounts for all possible permutations of \( r \).

We compute probability for each \( r \), helping forecast expected outcomes. The calculated probabilities are then used to derive expected frequencies, essentially setting a benchmark to compare against what was actually observed. Understanding probability in this context allows us to rigorously test fitting models, such as if our sample data conforms to a Poisson distribution.