91Ó°ÊÓ

Daily Wages The daily wages in a particular industry are normally distributed with a mean of \(\$ 94\) and a standard deviation of \(\$ 11.88\). Suppose a company in this industry employs 40 workers and pays them \(\$ 91.50\) per week on the average. Can these workers be viewed as a random sample from among all workers in the industry? a. What are the null and alternative hypotheses to be tested? b. Use the Large-Sample Test of a Population Mean applet to find the observed value of the test statistic. c. Use the Large-Sample Test of a Population Mean applet to find the \(p\) -value for this test. d. If you planned to conduct your test using \(\alpha=.01\), what would be your test conclusions? e. Was it necessary to know that the daily wages are normally distributed? Explain your answer.

Step by step solution

01

Define null and alternative hypotheses

The null hypothesis (\(H_{0}\)) assumes that the workers in the company can be considered a random sample from the entire industry, and thus have a mean daily wage equal to the population mean. The alternative hypothesis (\(H_{1}\)) states that the workers are not a random sample, and their mean daily wage is not equal to the population mean. $$ H_{0}: \mu = \$94 \\ H_{1}: \mu \neq \$94 $$

02

Calculate the test statistic

The test statistic (z) for large-sample test is calculated as follows: $$ z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} $$ Here, - \(\bar{x}\) = sample mean = \(\$91.50\) - \(\mu\) = population mean = \(\$94\) - \(\sigma\) = population standard deviation = \(\$11.88\) - \(n\) = sample size = \(40\) Plugging in the values: $$ z = \frac{91.50 - 94}{\frac{11.88}{\sqrt{40}}} = -1.672 $$

03

Find the p-value

Since this is a two-tailed test (\(\mu \neq \$94\)), we need to find the p-value for both tails. Using a Z-table or an appropriate applet, we can find the p-value corresponding to the test statistic: $$ p = 2 * P(Z \leq -1.672) \approx 0.095 $$

04

Test conclusions at \(\alpha = 0.01\)

We are given a significance level (\(\alpha\)) of 0.01. To determine our test conclusion, we compare the calculated p-value to the given significance level: - If \(p \leq \alpha\), we reject the null hypothesis. - If \(p > \alpha\), we fail to reject the null hypothesis. Since \(0.095 > 0.01\), we fail to reject the null hypothesis. We can conclude that the workers can be considered a random sample from the entire industry at the \(\alpha = 0.01\) significance level.

05

Discuss the normality of the daily wage distribution

It was necessary to know that the daily wages are normally distributed because the hypothesis testing method used in this exercise relies on the underlying distribution of the population. If the data were not normally distributed, we might have needed to use a different approach, such as non-parametric hypothesis tests or rely on the Central Limit Theorem if the sample size was large enough. In this case, knowing that the daily wages are normally distributed allowed us to perform the hypothesis test appropriately.

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Key Concepts

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

Normal Distribution

The concept of a normal distribution is fundamental in statistics, especially when dealing with large datasets. This distribution is represented as a symmetrical bell-shaped curve that describes how values of a variable are spread. Some key characteristics include:

• Most values cluster around a central point known as the mean, \ \( \mu \ \).
• The spread of the values is described by the standard deviation, \ \( \sigma \ \).
• It is symmetric. This means the left side of the curve is a mirror image of the right side.

In our exercise, the daily wages follow this normal distribution with a mean of \(94 and a standard deviation of \)11.88. Knowing this shape allows us to employ hypothesis tests under the assumption of normally distributed data. This makes calculations and predictions about data behavior much more accurate and reliable.

Population Mean

The population mean is a statistical measure that represents the average value of all possible observations in the population. In this example, the population mean (\( \mu \) ) is $94, representing the average daily wage across the entire industry.

Understanding the population mean is crucial because it serves as a benchmark for hypothesis testing. When we compare the sample mean to the population mean, we can determine if the sample truly represents the population or if there are significant differences. A discrepancy suggests that the sample might not be taken from the intended population or reflects some variation due to sampling error.

Test Statistic

The test statistic is a crucial component in hypothesis testing, helping us make decisions about our hypotheses. For normally distributed data, the Z-score is commonly used to determine how far and in what direction, a sample statistic (e.g., sample mean) deviates from the population mean, in units of the population standard deviation. In this context, the test statistic is calculated using the formula:

\[ z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \]

Where:

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

In our exercise, this calculation results in a Z-score of -1.672, representing how distinctly the sample mean (\(91.50) deviates from the population mean (\)94). The further from 0 the Z-score is, the stronger the evidence against the null hypothesis.

P-Value

The p-value is pivotal in determining the strength of the results in hypothesis testing. It provides insight into the probability of observing a test statistic at least as extreme as the one calculated, assuming the null hypothesis is true.

A p-value helps us decide whether to accept or reject the null hypothesis. In our context, the p-value associated with the test statistic of -1.672 is approximately 0.095, which is relatively high compared to common significance levels, such as \( \alpha = 0.01 \).

• If the p-value is less than \( \alpha \), it suggests rejecting the null hypothesis, indicating significant differences.
• If the p-value is greater than \( \alpha \), it leads us to fail to reject the null hypothesis, suggesting no strong evidence against it.

Here, since 0.095 > 0.01, we do not reject the null hypothesis, indicating these workers can indeed be seen as a random sample from the broader industry population.