91Ó°ÊÓ

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 field. 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. $$ \begin{array}{l|rrrrrr} \hline \boldsymbol{r} & 0 & 1 & 2 & 3 & 4 & 5 \text { or more } \\ \hline 0 & 12 & 15 & 29 & 18 & 19 & 7 \\ \hline \end{array} $$ (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 Probability for r = 0, 1, 2, 3, 4, 5

The Poisson probability formula is given by: \[ P(r) = \frac{e^{-\lambda} \lambda^{r}}{r!} \] where \( \lambda = 2.8 \). We calculate for each value of \( r \):- \( r = 0 \), \( P(0) = \frac{e^{-2.8} 2.8^{0}}{0!} = e^{-2.8} \approx 0.0608 \)- \( r = 1 \), \( P(1) = \frac{e^{-2.8} 2.8^{1}}{1!} = 2.8 \times e^{-2.8} \approx 0.1703 \)- \( r = 2 \), \( P(2) = \frac{e^{-2.8} 2.8^{2}}{2!} = \frac{2.8^2}{2} \times e^{-2.8} \approx 0.2384 \)- \( r = 3 \), \( P(3) = \frac{e^{-2.8} 2.8^{3}}{3!} = \frac{2.8^3}{6} \times e^{-2.8} \approx 0.2223 \)- \( r = 4 \), \( P(4) = \frac{e^{-2.8} 2.8^{4}}{4!} = \frac{2.8^4}{24} \times e^{-2.8} \approx 0.1556 \)- \( r \geq 5 \), calculate the cumulative probability \( P(5) = 1 - (P(0) + P(1) + P(2) + P(3) + P(4)) \approx 0.1526 \) after summation and subtraction.

02

Compute Expected Number of Colonies (E)

For each \( r \), multiply the probability \( P(r) \) by the sample size (100) to find the expected number of observations:- \( E(0) = 100 \times 0.0608 \approx 6.08 \)- \( E(1) = 100 \times 0.1703 \approx 17.03 \)- \( E(2) = 100 \times 0.2384 \approx 23.84 \)- \( E(3) = 100 \times 0.2223 \approx 22.23 \)- \( E(4) = 100 \times 0.1556 \approx 15.56 \)- \( E(5) = 100 \times 0.1526 \approx 15.26 \) (for 5 or more colonies)

03

Calculate Chi-Square Statistic

Use the chi-square formula \[ \chi^{2} = \sum \frac{(O - E)^{2}}{E} \] to calculate the statistic:- For \( r = 0 \), \( \frac{(12 - 6.08)^{2}}{6.08} \approx 5.712 \)- For \( r = 1 \), \( \frac{(15 - 17.03)^{2}}{17.03} \approx 0.242 \)- For \( r = 2 \), \( \frac{(29 - 23.84)^{2}}{23.84} \approx 1.071 \)- For \( r = 3 \), \( \frac{(18 - 22.23)^{2}}{22.23} \approx 0.805 \)- For \( r = 4 \), \( \frac{(19 - 15.56)^{2}}{15.56} \approx 0.757 \)- For \( r = 5 \), \( \frac{(7 - 15.26)^{2}}{15.26} \approx 4.478 \)Summing these values, \( \chi^{2} \approx 13.065 \).

04

Determine Degrees of Freedom and Test Statistic

The degrees of freedom (df) is calculated as \( k - 1 - m \), where \( k \) is the number of categories (6 in this case, including combined 5 or more), and \( m \) is the number of parameters estimated (1 in this case, \( \lambda \)). Thus, \( df = 6 - 1 - 1 = 4 \).Using a chi-square distribution table, find the critical value for \( df = 4 \) at a 5% significance level, which is approximately \( 9.488 \). Since \( \chi^{2} = 13.065 \) is greater than \( 9.488 \), we reject the null hypothesis.

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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 the expected frequencies and the observed frequencies in one or more categories. This is particularly useful for categorical data and helps verify if a specific distribution, such as the Poisson distribution, fits the observed data properly. To perform a Chi-Square Test, follow these general steps:

• Calculate Expected Frequencies: These are derived from the hypothesized distribution. In our case, use the Poisson distribution probabilities multiplied by the sample size.
• Compute the Chi-Square Statistic: Use the formula \( \chi^{2} = \sum \frac{(O - E)^{2}}{E} \), where \( O \) is the observed frequency, and \( E \) is the expected frequency for each category. The sum is over all categories.
• Compare to Critical Value: Use a Chi-Square distribution table to find the critical value at the desired significance level and degrees of freedom to assess significance.

By comparing the calculated Chi-Square statistic to this critical value, you can conclude whether to reject or accept your null hypothesis.

Probability

Probability is a fundamental concept in statistics that quantifies the likelihood of an event occurring. It ranges from 0 to 1, where 0 means the event will not happen and 1 indicates certainty that the event will happen. In the context of the Poisson distribution used in our exercise:

• Poisson Probability describes the probability of a given number of events happening in a fixed interval of time or space. It is particularly useful for events that occur independently at a constant rate.
• The Poisson probability formula is stated as \( P(r) = \frac{e^{-\lambda} \lambda^{r}}{r!} \), where \( \lambda \) is the average number of occurrences in the interval, and \( r \) denotes the number of occurrences you want to find the probability for. \( e \) is the base of the natural logarithm, approximately equal to 2.718.

What's special about a Poisson distribution is that its mean is also equal to its variance, providing a unique characteristic that fits specific distributions of data quite efficiently.

Degrees of Freedom

Degrees of freedom (df) are the number of independent values in a calculation, which can vary within a statistical analysis. It is an essential concept in statistical testing, including the Chi-Square Test, as it influences the shape of the distribution curve used for hypothesis testing.

• Formula for Degrees of Freedom in Chi-Square: Calculate \( df \) as \( k - 1 - m \), where \( k \) is the number of categories and \( m \) is the number of parameters estimated. In the example exercise, this results in \( df = 6 - 1 - 1 = 4 \).
• Role in Critical Value: Degrees of freedom are used to locate the critical value on the Chi-Square distribution table, essential for determining the threshold at which we can reject the null hypothesis.
• Conceptual Role: A higher degrees of freedom usually indicates greater flexibility in the statistical model, which can potentially handle larger data variability.

Understanding degrees of freedom helps in making informed decisions about the validity of statistical analyses.