91Ó°ÊÓ

In hypothesis testing, the null hypothesis, denoted as \(H_0\), is a statement that assumes no effect or no difference. It serves as a starting point for statistical testing. In this exercise, the null hypothesis \(H_0\) is \(\mu = 250\). This means we assume the population mean is 250. The objective is to test this assumption against an alternative hypothesis.
The alternative hypothesis, \(H_a\), is what you want to prove. Here, it is \(\mu < 250\), indicating we suspect the population mean is less than 250.
The purpose of testing these hypotheses is to infer whether to reject the null hypothesis based on the sample data. If evidence strongly suggests that the assumption under \(H_0\) is unlikely, \(H_0\) may be rejected in favor of \(H_a\).

• The null hypothesis is a key concept as it provides a foundation for statistical testing.
• It often represents a status quo or null effect condition.
• Testing \(H_0\) involves calculating a test statistic and comparing it to a critical value or p-value.

The sample mean, denoted as \(\bar{x}\), is the average value of a sample taken from a population. It acts as an estimate of the population mean, \(\mu\). To calculate \(\bar{x}\), one sums all the sample values and then divides by the number of values in the sample.
In this specific problem, the sample mean is found through manipulation of the test statistic formula. Given:

• \(z^\star = -1.18\)
• population standard deviation \(\sigma = 22.6\)
• sample size \(n = 85\)
• the hypothesized population mean \(\mu = 250\)

You can rearrange the formula: \[\bar{x} = z^\star \cdot \left( \frac{\sigma}{\sqrt{n}} \right) + \mu\]Substituting the known values into the formula provides the sample mean, \(\bar{x}\). By using the sample mean, we can then determine sums and differences relevant to testing hypotheses.

A test statistic offers a standardized way to decide whether the observed data deviates significantly under the null hypothesis. It translates a sample outcome into a single value on a statistical distribution, aiding in hypothesis testing.
In this scenario, the test statistic used is the z-statistic, computed using: \[ Z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \]Where:

• \(\bar{x}\) is the sample mean
• \(\mu\) is the population mean stated in the null hypothesis
• \(\sigma\) is the population standard deviation
• \(n\) is the sample size

This test statistic value is compared against critical values or used to find the p-value, helping decide if \(H_0\) should be rejected. In the existing exercise, \(z^\star = -1.18\), informing how much our sample mean differs from the hypothesized population mean.

Standard deviation, denoted as \(\sigma\), measures how spread out numbers are in a data set. It provides insight into the variability of the data values around the mean.
In hypothesis testing, knowing the population standard deviation allows for more accurate test statistic computation, particularly when utilizing the z-test as seen in this exercise.
The formula for standard deviation measurement is: \[\sigma = \sqrt{\frac{\sum (X_i - \mu)^2}{N}}\]Where:

• \(X_i\) represents each data point
• \(\mu\) is the mean of the data
• \(N\) is the number of data points

In this task, the standard deviation \(\sigma = 22.6\) is used to find the test statistic. Accurately understanding standard deviation is crucial as it affects the width of confidence intervals and helps determine the reliability of the sample mean as an indicator of the population mean.