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The Banner Mattress and Furniture Company wishes to study the number of credit applications received per day for the last 300 days. The information is reported in the following table.To interpret, there were 50 days on which no credit applications were received, 77 days on which only one application was received, and so on. Would it be reasonable to conclude that the population distribution is Poisson with a mean of \(2.0 ?\) Use the .05 significance level. Hint: To find the expected frequencies use the Poisson distribution with a mean of \(2.0 .\) Find the probability of exactly one success given a Poisson distribution with a mean of 2.0 . Multiply this probability by 300 to find the expected frequency for the number of days in which there was exactly one application. Determine the expected frequency for the other days in a similar manner.

Step by step solution

01

Understand the Problem

We are given the number of credit applications received per day for 300 days. The data needs to be compared against a Poisson distribution with a mean of \(\lambda = 2.0\). The goal is to determine if it’s reasonable to conclude that this is the actual distribution of the data.

02

Organize Given Data

The provided data is a frequency distribution of credit applications received per day. - 0 applications: 50 days - 1 application: 77 days - 2 applications: 81 days - 3 applications: 48 days - 4 applications: 31 days - 5 applications: 12 days - 6 applications: 1 day (This sums to 300 days total).

03

Calculate Poisson Probabilities

For each number of applications (0-6), calculate the probability using the Poisson probability mass function: \[ P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!} \]Substitute \(\lambda = 2.0\) for each \(k\) from 0 to 6.

04

Calculate Expected Frequencies

Multiply each probability by the total number of days (300) to find the expected frequency.- For 0 applications: \(P(X=0)\times300\)- For 1 application: \(P(X=1)\times300\)- Continue this calculation for each observed frequency.

05

Formulate Hypotheses for Chi-Square Test

The null hypothesis \(H_0\) assumes that the observed frequencies come from a Poisson distribution with mean 2.0. The alternative hypothesis \(H_a\) states that the observed data does not follow this distribution.

06

Compute Chi-Square Test Statistic

The formula for the chi-square statistic is: \[ \chi^2 = \sum \frac{(O_i - E_i)^2}{E_i} \]Where \(O_i\) is the observed frequency and \(E_i\) is the expected frequency calculated previously. Compute this for all categories.

07

Determine Degrees of Freedom

The degrees of freedom is given by the number of categories minus 1 minus the number of parameters estimated from the data (which is 1 in this case since we only estimate the mean). Hence, \(df = 6 - 1 = 5\).

08

Decision Rule and Conclusion

Using a chi-square distribution table at a significance level of 0.05 and 5 degrees of freedom, find the critical value. Compare it with the computed chi-square statistic to decide whether to reject or not 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 check whether observed data fits a certain distribution. In simple terms, it helps us decide if the distribution we see in our data aligns with what we'd expect from a statistical model. For Banner Mattress and Furniture Company, the Chi-Square Test helps compare the actual number of daily credit applications to what would be expected under a Poisson distribution.

Here's how it works:

• First, you list each category's observed frequency. In the exercise, these are the daily counts of credit applications.
• Next, you calculate the expected frequencies using the Poisson distribution. This tells you how often you would expect each outcome, given the assumptions of the model.
• Then you compute the chi-square statistic. This involves summing up a series of values—each one quantifies how much the observed frequency differs from the expected frequency for a category.

If this statistic is too high, it suggests the data does not match the expected distribution.

The technique ultimately lets us test the null hypothesis: that our data follows a Poisson distribution with a mean of 2. A large chi-square statistic compared to a critical value leads us to reject this hypothesis, indicating the data does not fit the model well.

Hypothesis Testing

Hypothesis Testing is the process of making inferences about a population using sample data. It involves setting up a hypothesis that we test with data. For Banner Mattress and Furniture Company, hypothesis testing involves checking if the credit applications follow a Poisson distribution.

The steps in hypothesis testing generally include:

• Formulating Hypotheses: Set up both a null hypothesis (\(H_0\)) and an alternative hypothesis (\(H_a\)). For this exercise, \(H_0\) states the data fits a Poisson distribution with mean 2, while \(H_a\) indicates it does not.
• Decision Rule: Determine a significance level (\(\alpha\)), here is 0.05. This level helps us decide when to reject \(H_0\).
• Compute Test Statistic: Calculate a relevant statistic—in this case, the chi-square test statistic—that tells us how far our observed data is from what we expect.
• Make a Conclusion: Compare the test statistic to a critical value from the chi-square distribution. If the test statistic exceeds this critical value, we reject the null hypothesis.

Overall, hypothesis testing helps us evaluate assumptions about how data should behave.

Probability Mass Function

The Probability Mass Function (PMF) is a function that gives the probability of each possible outcome for a discrete random variable. It’s what makes the Poisson distribution tick, giving each outcome a likelihood of occurrence. This function plays a key role in understanding Poisson distribution, as was needed for the Banner Mattress and Furniture's credit application exercise.

In the Poisson distribution, this function is represented as:\[ P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!} \]where:

• \( P(X=k) \) is the PMF, providing the probability of observing \( k \) outcomes.
• \( \lambda \) refers to the average rate (or mean number of occurrences) which is 2 in this exercise.
• \( e \) is a mathematical constant approximately equal to 2.71828.
• \( k! \) is the factorial of \( k \), which is the number of occurrences.

This function allows us to compute how likely it is to observe each number of applications in a day. Simply put, it frames our expectations for the data and is essential for calculating expected frequencies in a statistical test. Understanding PMF is crucial for interpreting statistical distributions in hypothesis testing and frequency analysis.