91Ó°ÊÓ

When \(X_{1}, X_{2}, \ldots, X_{n}\) are independent Poisson variables, each with parameter \(\mu\), and \(n\) is large, the sample mean \(\bar{X}\) has approximately a normal distribution with \(\mu=E(\bar{X})\) and \(V(\bar{X})=\mu / n\). This implies that $$ Z=\frac{\bar{X}-\mu}{\sqrt{\mu / n}} $$ has approximately a standard normal distribution. For testing \(H_{0}: \mu=\mu_{0}\), we can replace \(\mu\) by \(\mu_{0}\) in the equation for \(Z\) to obtain a test statistic. This statistic is actually preferred to the large- sample statistic with denominator \(S / \sqrt{n}\) (when the \(X_{i}\) 's are Poisson) because it is tailored explicitly to the Poisson assumption. If the number of requests for consulting received by a certain statistician during a 5-day work week has a Poisson distribution and the total number of consulting requests during a 36-week period is 160 , does this suggest that the true average number of weekly requests exceeds 4.0? Test using \(\alpha=.02\).

Step by step solution

01

Define the Hypotheses

First, we define the null and alternative hypotheses. Our null hypothesis \(H_0\) is that the average number of weekly requests \(\mu = 4.0\). The alternative hypothesis \(H_a\) is that \(\mu > 4.0\), which implies a one-tailed test since we are testing if the true average exceeds 4.0.

02

Calculate the Sample Mean

We know the total number of requests over 36 weeks is 160. Hence, the sample mean \(\bar{X}\) can be calculated as \(\bar{X} = \frac{160}{36} \approx 4.44\).

03

Calculate the Test Statistic

Substitute \(\bar{X} = 4.44\), \(\mu_0 = 4.0\), and \(n = 36\) into the test statistic formula: \[ Z = \frac{\bar{X} - \mu_0}{\sqrt{\frac{\mu_0}{n}}} = \frac{4.44 - 4.0}{\sqrt{\frac{4.0}{36}}} \approx \frac{0.44}{0.333} \approx 1.32. \]

04

Determine the Critical Value

For a significance level of \(\alpha = 0.02\), we determine the critical value from the standard normal distribution table. A one-tailed test at \(\alpha = 0.02\) gives a critical value of approximately 2.05.

05

Make a Decision

Since the calculated test statistic \(Z \approx 1.32\) is less than the critical value 2.05, we do 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.

Normal Approximation

When dealing with Poisson-distributed variables, like in our textbook exercise, the normal approximation is a handy tool. If you have a data set from a Poisson distribution with a known parameter \( \mu \), and your sample size \( n \) is large enough, you can approximate your data using a normal distribution.
This conversion is possible because the Central Limit Theorem tells us that the sampling distribution of the sample means approximates a normal distribution as the sample size increases. In our exercise, we're looking at consulting requests, which initially follow a Poisson distribution.

• The mean of this distribution, \(\mu \), stays the same when approximated normally.
• The variance becomes \( \mu / n \), where \( n \) is our sample size.

Knowing this, we can transform our data into a format we're more comfortable with: the normal distribution. This transformation allows us to employ hypothesis tests using standard normal distribution techniques.

Hypothesis Testing

Hypothesis testing is a method used to decide whether your data strongly supports one assumption over another. In the context of Poisson distributions and normal approximations, you first identify your null hypothesis (\(H_0\)) and your alternative hypothesis (\(H_a\)).
In our original exercise, the null hypothesis is \( H_0: \mu = 4.0 \), meaning we initially assume the average number of weekly requests is equal to 4. For the alternative hypothesis \( H_a: \mu > 4.0 \), we suspect the real number could be higher.
Here's what you generally do during hypothesis testing:

• Define the hypotheses based on your observation.
• Calculate the test statistic, which involves using your sample mean and expected value.
• Compare the test statistic to a critical value to make your decision.

The objective is to see if the evidence (sample data) is strong enough to reject the null hypothesis, typically using a threshold known as the significance level.

Test Statistic

The test statistic is a standardized value that helps you determine where your sample data sits in terms of probability. In the context of our exercise involving a Poisson distribution, once we compute our sample mean, we can proceed to calculate the test statistic \( Z \) using this formula:
\[ Z = \frac{\bar{X} - \mu_0}{\sqrt{\frac{\mu_0}{n}}} \]
It compares the observed sample mean \( \bar{X} \) to the null hypothesis mean \( \mu_0 \).
The process is straightforward. You subtract the hypothesized mean (in this case, 4.0) from your sample mean (in our example, 4.44), and then divide by the standard deviation, \( \sqrt{\frac{\mu_0}{n}} \).
This calculation outputs the test statistic, in our exercise, approximately 1.32.

• Higher values can indicate strong evidence against the null hypothesis.
• Lower values might mean the data is consistent with the null hypothesis.

Using the Z-score, you can decide the statistical significance of your findings.

Significance Level

The significance level, denoted as \(\alpha\), sets the threshold for accepting or rejecting the null hypothesis. It is essentially the probability of making a Type I error, which is falsely rejecting the null hypothesis when it is actually true.
In hypothesis testing, especially with a normal distribution approximation like in our example, you compare your computed test statistic with this threshold. For our exercise, we used \(\alpha = 0.02\). This means:

• There is a 2% risk of concluding that the average number of weekly requests exceeds 4 when it actually does not.
• This low \(\alpha\) level signifies the investigator's preference for a low probability of error.

You then find the corresponding critical value for this significance level from the standard normal distribution. If your test statistic is greater than the critical value, you reject \(H_0\).
In our instance, since 1.32, the test statistic, is less than the critical value at \(\alpha = 0.02\), which was about 2.05, the null hypothesis was not rejected. Thus, the evidence was insufficient to claim the average number of requests exceeds 4.