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According to the local union president, the mean income of plumbers in the Salt Lake City follows the normal distribution. This normal distribution has a mean of \(\$ 45,000\) and a standard deviation of \(\$ 3,000\). A recent investigative reporter for KYAK TV found, for a sample of 120 plumbers, the mean gross income was \(\$ 45,500\). At the .10 significance level, is it reasonable to conclude that the mean income is not equal to \(\$ 45,000\) ? Determine the \(p\) -value.

Short Answer

Expert verified
It is reasonable to conclude that the mean income is not equal to $45,000.

Step by step solution

01

State the Hypotheses

We begin by defining our null and alternative hypotheses. The null hypothesis \[ H_0: \mu = 45000 \]and the alternative hypothesis\[ H_a: \mu eq 45000 \]where \( \mu \) is the true mean income of plumbers.
02

Identify the Relevant Information

We are given that the population mean \( \mu = 45000 \), the standard deviation \( \sigma = 3000 \), the sample mean \( \bar{x} = 45500 \), and the sample size \( n = 120 \). Our significance level is \( \alpha = 0.10 \).
03

Calculate the Test Statistic

Use the formula for the Z-test statistic:\[ Z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}} \]Substitute the values:\[ Z = \frac{45500 - 45000}{3000 / \sqrt{120}} \]Calculate the standard error:\[ \text{Standard Error} = \frac{3000}{\sqrt{120}} \approx 273.86 \]Then the Z-value:\[ Z \approx \frac{500}{273.86} \approx 1.826 \]
04

Determine the Critical Value and Decision Rule

Since we use a two-tailed test at the \(0.10\) significance level, we need to find the critical Z-values that correspond to \( \alpha/2 = 0.05 \) in each tail. So, from the Z-table, the critical values are approximately \(-1.645\) and \(+1.645\).
05

Make a Decision

Compare the calculated Z-value with the critical values. Since the calculated Z-value \(1.826\) is greater than the critical value \(1.645\), it falls in the rejection region. We reject the null hypothesis.
06

Find the p-value

The Z-score of \(1.826\) corresponds to a p-value of about \(0.034\) for one-tailed. Since it is a two-tailed test, multiply this by 2:\[ p\text{-value} \approx 2 \times 0.034 = 0.068 \]
07

Conclusion

Since the p-value \(0.068\) is less than the significance level of \(0.10\), we reject the null hypothesis. There is sufficient evidence to conclude that the mean income is not equal to \(\$ 45,000\).

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Key Concepts

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

Normal Distribution
The normal distribution is a fundamental concept in statistics that describes how data points are distributed or spread out. Imagine it as a smooth, bell-shaped curve where most values cluster around a central peak, and the probabilities for values taper off symmetrically on either side of this peak. In the context of our exercise, the normal distribution serves to model the income distribution of plumbers in Salt Lake City.
Two parameters define a normal distribution:
  • Mean ( \( \mu \) ): This is the center of the distribution. In our example, the mean income is \( \\( 45,000 \) , which suggests that most plumbers have incomes close to this amount.
  • Standard Deviation ( \( \sigma \) ): This measures how spread out the values are around the mean. A smaller standard deviation means the values are clustered closely around the mean, while a larger one indicates a wider spread. In this case, \( \\) 3,000 \) is the standard deviation.
The importance of the normal distribution in hypothesis testing is that it helps determine probabilities and critical values necessary for making statistical decisions.
Z-test
A Z-test is a statistical method used to determine if there is a significant difference between the observed data and what is expected under the null hypothesis. In simple terms, it helps you decide if the sample mean is different from the population mean.
The Z-test is particularly useful when you know the standard deviation of the population and are working with a large sample size. Here is what you do in a Z-test:
  • Calculate the Z-value using the formula:
    \[ Z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}} \] where:
    \( \bar{x} \) is the sample mean, \( \mu \) is the population mean, \( \sigma \) is the standard deviation, and \( n \) is the sample size.
  • Compare the Z-value to critical values based on your significance level to determine if you should reject or fail to reject the null hypothesis.
In our example, the calculated Z-value is approximately 1.826, which we use to make a decision about the plumbers' income.
p-value
The p-value is a measure that helps you decide whether your results are statistically significant. Essentially, it tells you the probability of obtaining your observed data, or more extreme, if the null hypothesis is true. A smaller p-value indicates stronger evidence against the null hypothesis.
Here's how to interpret the p-value:
  • If the p-value is less than the significance level ( \( \alpha \) ), you reject the null hypothesis. It suggests that the observed difference is unlikely to have occurred by random chance.
  • If the p-value is greater than \( \alpha \) , you fail to reject the null hypothesis, meaning there isn't enough evidence against it.
In our case, we found a p-value of around 0.068, which is less than the significance level of 0.10; thus, we have sufficient evidence to reject the null hypothesis.
Significance Level
The significance level, denoted by \( \alpha \) , is the threshold used in hypothesis testing to determine when you should reject the null hypothesis. It represents the probability of making a Type I error, which is rejecting a true null hypothesis.
Common significance levels include 0.05, 0.01, and 0.10. Your choice of \( \alpha \) should reflect the context and requirements of your analysis. Each level indicates the risk you are willing to take in making an incorrect decision.
  • \( \alpha = 0.10 \) : Here, you allow a 10% chance of incorrectly rejecting the null hypothesis. It is often chosen for exploratory or preliminary studies where being overly cautious is not essential.
  • \( \alpha = 0.05 \) : This is the most common level, often used in scientific studies for a balance between risk and evidence.
In our exercise, a significance level of 0.10 was used, which leans toward being slightly more lenient compared to other common thresholds, allowing for a wider margin in accepting evidence against the null hypothesis.

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Most popular questions from this chapter

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