91Ó°ÊÓ

A Test Goes Wrong. Software can generate samples from (almost) exactly Normal distributions. Here is a random sample of size 5 from the Normal distribution with mean 20 and standard deviation 2.5: Inll random \(\begin{array}{lllll}22.94 & 17.04 & 17.58 & 20.96 & 19.29\end{array}\) These data match the conditions for a \(z\) test better than real data will: the population is extremely close to Normal and has known standard deviation \(\alpha=2.5\), and the population mean is \(\mu=20\). Although we know the true value of \(\mu\), suppose we pretend that we do not and we test the hypotheses $$ \begin{aligned} &H_{0}: \mu=17.5 \\ &H_{a}: \mu \neq 17.5 \end{aligned} $$ a. What are the \(z\) statistic and its \(P\)-value? Is the test significant at the \(5 \%\) level? b. We know that the null hypothesis does not hold, but the test failed to give strong evidence against \(H_{00}\). Explain why this is not surprising.

Step by step solution

01

Compute the Sample Mean

First, calculate the sample mean \( \bar{x} \) from the provided data. The formula for the sample mean is \( \bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i \), where \( n = 5 \) in this case. Substituting the values: \( \bar{x} = \frac{22.94 + 17.04 + 17.58 + 20.96 + 19.29}{5} \approx 19.562 \).

02

Compute the Standard Error of the Mean

The standard error of the mean (SE) is given by the formula \( SE = \frac{\sigma}{\sqrt{n}} \), where \( \sigma = 2.5 \) is the population standard deviation and \( n = 5 \). Hence, \( SE = \frac{2.5}{\sqrt{5}} \approx 1.118 \).

03

Compute the z Statistic

The \( z \) statistic is calculated using \( z = \frac{\bar{x} - \mu_0}{SE} \), where \( \mu_0 = 17.5 \) is the null hypothesis mean. Substituting the values, \( z = \frac{19.562 - 17.5}{1.118} \approx 1.845 \).

04

Determine the P-value

The \( P \)-value in a two-tailed test is the probability that the \( z \)-statistic is as extreme as the calculated \( z \)-statistic under the null hypothesis. Using the standard normal distribution table, find the \( P \)-value for \( z = 1.845 \). It is approximately 0.065, which is greater than 0.05.

05

Interpretation at 5% Significance Level

Since the \( P \)-value (0.065) is larger than the significance level of 0.05, we fail to reject the null hypothesis \( H_0: \mu = 17.5 \). Thus, the test is not significant at the 5% level.

06

Explanation of Test Result

The test did not provide strong evidence against \( H_0 \) because the sample size is small (\( n = 5 \)), resulting in a higher standard error. Smaller samples have less power to detect differences from the null hypothesis, especially when the population mean is close to the null hypothesis value.

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Key Concepts

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

Normal Distribution

A normal distribution is a key concept in statistics, known for its bell-shaped curve. It's symmetrically centered around the mean, where most of the data points are concentrated. In any normal distribution:

• The mean, median, and mode are all equal.
• The curve has a characteristic bell shape due to its properties.
• The total area under the curve is equal to 1, representing the whole probability.

In our exercise, we start with a sample from a normally distributed population with a mean (\( \mu \)) of 20 and a standard deviation (\( \sigma \)) of 2.5. This makes our assumptions for further tests valid as the normal distribution allows statisticians to make inferences about populations through sampling.

Sample Mean

The sample mean (\( \bar{x} \)) is essentially the average value of the data points in a sample, providing an estimate of the population mean. Here's why it's important:

• A sample mean is the central value of a data set collected from a specific sample size.
• It serves as an unbiased estimate of the true population mean when the sample is random.
• Calculation involves summing all the observed values and dividing by the number of observations (\( n \)).

In the context of our problem, the sample mean calculated from the sample of 5 scores is approximately 19.562. This average helps us compare against the hypothesized population mean of 17.5 in the hypothesis test.

Standard Error

The standard error (SE) is crucial for understanding how much variability exists in the sample mean compared to the actual population mean. It acts as a measure of the spread of sample means if multiple samples were taken. Key points include:

• The formula for standard error is \( SE = \frac{\sigma}{\sqrt{n}} \).
• A smaller SE indicates a more precise estimate of the population mean.
• The larger the sample size, the smaller the SE, which reduces deviations from the population mean.

In our situation, with a population standard deviation of 2.5 and a sample size of 5, the standard error computes to approximately 1.118. This SE is part of calculating the z-score, helping us to understand the sample mean's position within the expected population distribution.

Hypothesis Testing

Hypothesis testing is a statistical method used to make decisions about a population parameter based on sample data, involving null and alternative hypotheses. Key elements of this process include:

• The null hypothesis (\( H_0 \)) posits there is no effect or difference. In our problem, \( H_0 = \mu = 17.5 \).
• The alternative hypothesis (\( H_a \)) suggests that there is a significant effect or difference, here \( H_a: \mu eq 17.5 \).
• A significance level (commonly 5%) determines how conservative or liberal we are willing to be about incorrectly rejecting the null hypothesis.

In the exercise, after computing the z-statistic and determining the \( P \)-value, we concluded that the \( P \)-value was greater than 0.05. Thus, we failed to reject the null hypothesis, meaning there wasn't significant evidence to suggest \( H_0 \) was false. This reflects the challenge of small samples which can make statistical tests less powerful.