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The hourly wages in a particular industry are normally distributed with mean \(\$ 13.20\) and standard deviation \(\$ 2.50 .\) A company in this industry employs 40 workers, paying them an average of \(\$ 12.20\) per hour. Can this company be accused of paying substandard wages? Use an \(\alpha=.01\) level test.

Short Answer

Expert verified
Yes, the company pays substandard wages as the average is significantly lower than the industry mean.

Step by step solution

01

Define the Hypotheses

In this problem, we need to determine whether the company pays substandard wages. We set up the null and alternative hypotheses as follows: - Null Hypothesis (H_0): The mean hourly wage of the company's employees is equal to the industry's mean, \( \mu = 13.20 \).- Alternative Hypothesis (H_a): The mean hourly wage of the company's employees is less than the industry's mean, \( \mu < 13.20 \).
02

Identify the Test Statistic

To assess the hypothesis, we will use a t-test for a sample mean since we know the sample size is 40, and the population standard deviation is given. The formula for the t-test statistic is:\[ t = \frac{\bar{x} - \mu}{s / \sqrt{n}} \]where \( \bar{x} = 12.20 \) is the sample mean, \( \mu = 13.20 \) is the population mean, \( s = 2.50 \) is the population standard deviation, and \( n = 40 \) is the sample size.
03

Calculate the Test Statistic

Now plug the values into the t-test formula:\[ t = \frac{12.20 - 13.20}{2.50 / \sqrt{40}} \]Calculate the denominator first:\( 2.50 / \sqrt{40} = 0.395 \).Now compute the t value:\( t = \frac{-1.00}{0.395} \approx -2.53 \).
04

Determine the Critical Value

At \(\alpha = 0.01 \), we use a t-table to find the critical value for a one-tailed test with \( n-1 = 39 \) degrees of freedom. Approximating to 38 degrees of freedom, the critical value \( t_{critical} \approx -2.426 \).
05

Compare Test Statistic with Critical Value

Compare the calculated t value \(-2.53\) with the critical value \(-2.426\). Since \(-2.53\) is less than \(-2.426\), the test statistic falls into the rejection region of the null hypothesis.
06

Conclusion

Since the calculated t-value is in the rejection region, we reject the null hypothesis at the \(\alpha = 0.01\) significance level. This implies that there is sufficient evidence to claim that the company's average hourly wage is significantly lower than the industry's average, indicating substandard wages.

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Key Concepts

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

t-test
A t-test is a statistical test used to determine whether there is a significant difference between the means of two groups, which may be related to certain features in the data. In the context of the exercise, we are using a t-test to assess if the mean hourly wage of the company's employees is significantly different from the industry mean.

The t-test formula is given by \[ t = \frac{\bar{x} - \mu}{s / \sqrt{n}} \] where: - \( \bar{x} \) is the sample mean, - \( \mu \) is the population mean, - \( s \) is the population standard deviation, - \( n \) is the sample size.

The t-test is robust for small sample sizes, specifically useful when the sample size is below 30 but still applicable with larger samples. It's especially useful when the population standard deviation is unknown, but in this specific exercise, it is known.
significance level
The significance level, denoted by \( \alpha \), is the probability of rejecting the null hypothesis when it is true. It essentially measures the confidence we have in our statistical test by determining the threshold for statistical significance.

In this exercise, the significance level is \( \alpha = 0.01 \), which means we are willing to accept a 1% chance of mistakenly concluding that the company pays substandard wages when it actually pays average wages.
  • A lower \( \alpha \) level (like 0.01) means a more rigorous criterion for evidence and strengthens our confidence in the conclusion.
Choosing an appropriate significance level is crucial as it affects the likelihood of Type I errors, where the null hypothesis is incorrectly rejected.
normal distribution
The normal distribution, or Gaussian distribution, is a continuous probability distribution that is symmetrical around its mean, showing that data near the mean are more frequent in occurrence than data far from the mean. In the context of the exercise, hourly wages in the industry are said to follow a normal distribution, simplifying the application of the t-test.
  • Mean: Average value of the data, central point of the distribution.
  • Standard Deviation: Measures the dispersion of data points from the mean.

This exercise assumes that hourly wages follow a normal distribution because this underlying assumption allows for the use of parametric testing methods. The normal distribution also supports the application of the Central Limit Theorem, which justifies the normality of the sample mean for a large sample size like 40.
null and alternative hypotheses
In hypothesis testing, null and alternative hypotheses are essentially educated guesses about a statistical relationship between data sets. The null hypothesis (\(H_0\)) is generally a statement of no effect or no difference, and the alternative hypothesis (\(H_a\)) represents what we aim to prove or the effect we suspect exists.

For this problem:- **Null Hypothesis** (\(H_0\)): The mean hourly wage of the company's employees equals the industry mean (\( \mu = 13.20 \)).
- **Alternative Hypothesis** (\(H_a\)): The mean hourly wage of the company's employees is less than the industry mean (\( \mu < 13.20 \)).

Hypothesis testing steps involve:
  • Formulating these hypotheses.
  • Using a statistical test (like the t-test) to make decisions about which hypothesis to reject.
  • A critical part of hypothesis testing is the P-value approach, where we compare calculated values to threshold levels to determine the hypotheses' validity.
Understanding these two types of hypotheses is fundamental as they form the basis upon which statistical decisions and interpretations are made.

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Most popular questions from this chapter

A merchant figures her weekly profit to be a function of three variables: retail sales (denoted by X), wholesale sales (denoted by \(Y\) ), and overhead costs (denoted by \(W\) ). The variables \(X, Y\), and \(W\) are regarded as independent, normally distributed random variables with means \(\mu_{1}, \mu_{2},\) and \(\mu_{3}\) and variances \(\sigma^{2}, a \sigma^{2},\) and \(b \sigma^{2},\) respectively, for known constants \(a\) and \(b\) but unknown \(\sigma^{2}\). The merchant's expected profit per week is \(\mu_{1}+\mu_{2}-\mu_{3}\). If the merchant has made independent observations of \(X, Y,\) and \(W\) for the past \(n\) weeks, construct a test of \(H_{0}: \mu_{1}+\mu_{2}-\mu_{3}=k\) against the alternative \(H_{a}: \mu_{1}+\mu_{2}-\mu_{3} \neq k,\) for a given constant \(k\). You may specify \(\alpha=.05\)

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