91Ó°ÊÓ

Wall Street securities firms paid out record year-end bonuses of \(\$ 125,500\) per employee for 2005 (Fortune, February 6,2006 ). Suppose we would like to take a sample of employees at the Jones \& Ryan securities firm to see whether the mean year-end bonus is different from the reported mean of \(\$ 125,500\) for the population. a. State the null and alternative hypotheses you would use to test whether the year-end bonuses paid by Jones \& Ryan were different from the population mean. b. Suppose a sample of 40 Jones \& Ryan employees showed a sample mean year- end bonus of \(\$ 118,000 .\) Assume a population standard deviation of \(\sigma=\$ 30,000\) and \(\mathrm{com}-\) pute the \(p\) -value. c. With \(\alpha=.05\) as the level of significance, what is your conclusion? d. Repeat the preceding hypothesis test using the critical value approach.

Step by step solution

01

State the Hypotheses

The null hypothesis (H0) states that the mean year-end bonus for Jones & Ryan employees is equal to the population mean of \(125,500. Thus, \(H_0: \mu = 125,500\). The alternative hypothesis (H1) states that the mean year-end bonus for Jones & Ryan employees is different from \)125,500. Thus, \(H_1: \mu eq 125,500\).

02

Calculate the Test Statistic

Given the sample mean \(\bar{x} = 118,000\), population mean \(\mu = 125,500\), population standard deviation \(\sigma = 30,000\), and sample size \(n = 40\), use the formula for the z-test statistic: \[ z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}} \]Substituting the values: \[ z = \frac{118,000 - 125,500}{30,000/\sqrt{40}} \] \[ z \approx -1.5811 \]

03

Compute the p-value

Since the test is two-tailed, we find the p-value by referring to the standard normal distribution table for \( z = -1.5811 \). This gives us a p-value of approximately 0.114. Since this is a two-tailed test, we double it to get the total p-value, resulting in \(p \approx 0.228\).

04

Conclusion Using p-value Approach

Given \(\alpha = 0.05\), compare \(p = 0.228\) with \(\alpha\). Since \(p > \alpha\), we fail to reject the null hypothesis. This means we do not have enough evidence to state that the mean year-end bonuses at Jones & Ryan are different from $125,500.

05

Critical Value Approach

Using a standard normal distribution, identify the critical z-values for a two-tailed test with \(\alpha = 0.05\). These critical values are approximately \(z = \pm 1.96\). Since the calculated \(z = -1.5811\) is within the range of \(-1.96\) and \(1.96\), we again fail to reject the null hypothesis.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Null and Alternative Hypotheses

In hypothesis testing, we start by stating two opposing ideas about a population parameter. These are known as the null and alternative hypotheses.

• The null hypothesis ( \(H_0\)) generally states that there is no effect or difference in the population, or that a certain parameter equals a specific value. For our example, it claims that the mean year-end bonus at Jones & Ryan equals the population mean of \(125,500: \(H_0: \mu = 125,500\).
• The alternative hypothesis ( \(H_1\)), on the other hand, suggests the contrary—there is an effect or difference. For this exercise, it posits that the average bonus differs from \)125,500: \(H_1: \mu eq 125,500\).

Consider these hypotheses as two possible "stories" about the data. Testing involves checking which story is supported by your sample.

P-Value Interpretation

The p-value is an essential part of hypothesis testing. It helps us decide whether there is enough evidence to reject the null hypothesis. The p-value represents the probability of obtaining a sample result at least as extreme as the one observed if the null hypothesis were true. In this example, with a calculated p-value of approximately 0.228:

• A small p-value (typically ≤ 0.05) indicates strong evidence against the null hypothesis, leading us to reject it.
• A large p-value (greater than 0.05) suggests weak evidence against the null hypothesis, so we do not reject it.

In our case, since 0.228 is greater than the common significance level of 0.05, it implies insufficient evidence to reject the null hypothesis. Thus, we conclude that the mean bonuses at Jones & Ryan are not significantly different from the population mean.

Z-Test Statistic

The z-test statistic is a pivotal element in hypothesis testing, especially when dealing with population means. It's a measure of how many standard deviations our sample mean is from the null hypothesis population mean.For this exercise, we calculated the z-test statistic as follows:\[ z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}} \]Here:

• \(\bar{x} = 118,000\) (sample mean)
• \(\mu = 125,500\) (population mean)
• \(\sigma = 30,000\) (population standard deviation)
• \(n = 40\) (sample size)

Substituting these values gives:\[ z \approx \frac{118,000 - 125,500}{30,000/\sqrt{40}} \approx -1.5811\]A negative sign simply indicates that the observed sample mean is less than the assumed population mean under the null hypothesis.

Statistical Significance

Statistical significance helps us determine the strength of the results obtained from the hypothesis test. When we say a result is statistically significant, it implies that it's unlikely to have occurred by random chance alone, according to a pre-determined threshold called the significance level (\(\alpha\)). In most scenarios, this is set at 0.05.Steps to determine statistical significance:

• If the p-value ≤ 0.05, we have enough evidence to reject the null hypothesis, indicating statistical significance.
• If the p-value > 0.05, we fail to reject the null hypothesis, suggesting a lack of statistical significance.

In the current example, with a p-value of 0.228, which is higher than 0.05, we consider the result not statistically significant, leading to the conclusion that there's no strong evidence that the Jones & Ryan bonuses significantly differ from the reported mean.