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The Z-test is a statistical method used to determine if there is a significant difference between the sample mean and a known population mean. Because it involves standard normal distribution, it's employed when the population variance is known, or the sample size is large enough (usually \( n \geq 30 \)).
To apply the Z-test, you need:

• The sample mean (\( \bar{x} \)).
• The population mean (\( \mu \)).
• The population standard deviation (\( \sigma \)).
• The sample size (\( n \)).

The formula for the Z-test statistic is:\[ Z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}} \]In this formula:

• \( \bar{x} \) is the average of your sample, which represents an estimate of the population mean.
• \( \mu \) is the given population mean you are testing against.
• \( \sigma \) is the known standard deviation of the population.
• \( n \) is the number of sample observations.

Once you have your Z-score, you can proceed to find the p-value by looking up this score in a Z-table or using a calculator.

In hypothesis testing, the null hypothesis \((H_0)\) and the alternative hypothesis \((H_a)\) are essential. They set the framework for the statistical test.
The null hypothesis is a statement asserting that there is no effect or no difference, and it is presumed true until evidence suggests otherwise. For the exercise involving Jones & Ryan, the null hypothesis is:\[ H_0: \mu = 125,500 \]Here, \( \mu \) represents the mean year-end bonus for employees at Jones & Ryan. This hypothesis indicates that there is no difference between the mean bonus at this firm and the known population mean.
On the other hand, the alternative hypothesis suggests the presence of an effect or a difference. It is what you aim to support. For the same scenario:\[ H_a: \mu eq 125,500 \]This shows a two-tailed test, meaning that the possibility could be higher or lower. If evidence strongly disputes \( H_0 \), \( H_a \) is accepted as more plausible. However, if no substantial evidence against \( H_0 \) is found, it is not necessarily proven true—it simply remains unrefuted.

Once the Z-score is determined, the next step is to compute the p-value. This value helps to make decisions regarding the null hypothesis. A p-value, in simple terms, is the probability that the observed results, or something more extreme, would occur if the null hypothesis is true.
For a two-tailed test like the example of bonus payments, you calculate it using the Z-score and a standard normal distribution table. Here’s how:

• Find the probability corresponding to the Z-score (\( Z = -1.58 \)) in a standard normal distribution table.
• Since it's a two-tailed test, double this probability to account for both ends of the distribution.
• The computed p-value is \( 2 \times P(Z < -1.58) \), resulting in approximately 0.1142.

The p-value is then compared to the level of significance (\( \alpha \)). If the p-value is less than \( \alpha \), typically 0.05, you reject \( H_0 \). Conversely, a p-value greater than \( \alpha \) leads to a failure to reject the null hypothesis. In the exercise context, since 0.1142 is larger than 0.05, the conclusion is to not reject the null hypothesis.