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According to the local union president, the mean income of plumbers in the Salt Lake City area follows the normal probability distribution with 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 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
The mean income is likely different; \( p \approx 0.068 \), reject \( H_0 \).

Step by step solution

01

Identify the Hypotheses

The null hypothesis \( H_0 \) is that the mean income of plumbers in Salt Lake City is \( \mu = \\(45,000 \). The alternative hypothesis \( H_a \) is that the mean income is not \( \mu = \\)45,000 \). This is a two-tailed test.
02

Determine the Test Statistic

We'll use the formula for the Z-test statistic for the sample mean: \[ Z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \] Where:- \( \bar{x} = 45,500 \) is the sample mean,- \( \mu = 45,000 \) is the population mean,- \( \sigma = 3,000 \) is the population standard deviation,- \( n = 120 \) is the sample size.
03

Calculate the Test Statistic

Substitute the given values into the Z formula: \[ Z = \frac{45,500 - 45,000}{\frac{3,000}{\sqrt{120}}} \] Calculate this step by step:- First, calculate the denominator: \( \frac{3,000}{\sqrt{120}} \approx 273.86 \).- Next, calculate the numerator: \( 45,500 - 45,000 = 500 \).- Therefore, \( Z = \frac{500}{273.86} \approx 1.826 \).
04

Determine the Critical Z-Value

For a two-tailed test with a significance level of 0.10, the critical Z-values are approximately \( \pm 1.645 \).
05

Compare the Test Statistic to the Critical Value

The calculated Z-value is \( 1.826 \), which is greater than the critical value of \( 1.645 \). Therefore, we have enough evidence to reject the null hypothesis at the 0.10 significance level.
06

Calculate the p-value

Using a standard normal distribution table or calculator, the p-value corresponding to a Z-value of 1.826 is about 0.034. Since this is a two-tailed test, we multiply by 2 to get \( p \approx 0.068 \).
07

Conclusion

Since the p-value \( 0.068 \) is less than the significance level \(0.10\), we reject the null hypothesis. It is reasonable to conclude that the mean income of plumbers is different from \( \$45,000 \).

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Key Concepts

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

Normal Distribution
The concept of normal distribution is fundamental in statistics. It describes how the values of a variable are likely to be distributed in a population. Picture a bell-shaped curve—this is the graphical representation of a normal distribution. It is symmetric, with most data points clustered around the central peak and fewer points at the extremes.

In our exercise, the mean income of plumbers was stated to follow this type of distribution. Key features of a normal distribution are its mean and standard deviation:
  • Mean (\( \mu \)): The average of all values. Here, it's \( \\(45,000 \).
  • Standard Deviation (\( \sigma \)): This measures the spread of the values around the mean. Smaller values mean data points are closer to the mean, while larger values indicate a wider spread. In our case, it's \( \\)3,000 \).
Understanding these helps in determining the shape and spread of the income data among plumbers.
Two-tailed Test
In hypothesis testing, determining whether to use a one-tailed or two-tailed test is crucial. A two-tailed test checks for the possibility of an effect in both directions, either above or below the mean.

For our scenario, the alternative hypothesis is that the mean income of plumbers is "not equal to" \( \$45,000 \). This is why a two-tailed test is appropriate. We're not just interested in whether the mean is greater or lesser, but simply whether it's different.
  • If we thought the mean could only increase, a one-tailed test might be used instead.
  • The significance level determines how often we're willing to be wrong if the null hypothesis is actually true. Here, it is set at 0.10.
Using a two-tailed test accounts for uncertainty in both directions, confirming robust statistical testing through this approach.
P-value
A key outcome of hypothesis testing is the p-value, which tells us how likely we are to observe the data we have, assuming the null hypothesis is true.

In simpler terms, it's the probability of obtaining results at least as extreme as the ones observed during the test. You calculate and interpret this value to decide whether or not to reject the null hypothesis.
  • A smaller p-value means stronger evidence against the null hypothesis. Typical thresholds are 0.05 or 0.10.
  • In our example, the p-value is approximately 0.068, which is less than the significance level of 0.10, indicating enough evidence to reject the null hypothesis.
This process ultimately helps confirm that changes or differences in data are statistically significant, not just due to random chance.
Z-test
The Z-test is used to determine if there's a significant difference between sample and population means. It is ideal when the population variance is known and the sample size is large.

The formula for the Z-test statistic plays a major role:\[Z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}\]Here, each component has a specific purpose:
  • \(\bar{x}\): Sample mean, which in this case is \(\\(45,500\).
  • \(\mu\): Population mean, set at \(\\)45,000\).
  • \(\sigma\): Standard deviation, which is \(\$3,000\).
  • \(n\): Sample size, given as 120.
After substituting the values, we calculate the Z value to be 1.826.

This calculated Z-value is then compared to the critical Z-values to determine whether the sample mean significantly deviates from the population mean. For our two-tailed test at 0.10 significance, the critical Z-value is \( \pm 1.645 \), showing that our calculated Z is indeed significant.

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Most popular questions from this chapter

Watch Corporation of Switzerland claims that its watches on average will neither gain nor lose time during a week. A sample of 18 watches provided the following gains (+)or losses(-)in seconds per week. $$ \begin{array}{|rrrrrrrrr|}\hline-0.38 & -0.20 & -0.38 & -0.32 & +0.32 & -0.23 & +0.30 & +0.25 & -0.10 \\\\-0.37 & -0.61 & -0.48 & -0.47 & -0.64 & -0.04 & -0.20 & -0.68 & +0.05 \\\\\hline\end{array}$$ Is it reasonable to conclude that the mean gain or loss in time for the watches is \(0 ?\) Use the .05 significance level. Estimate the \(p\) -value.

Chicken Delight claims that 90 percent of its orders are delivered within 10 minutes of the time the order is placed. A sample of 100 orders revealed that 82 were delivered within the promised time. At the .10 significance level, can we conclude that less than 90 percent of the orders are delivered in less than 10 minutes?

An urban planner claims that, nationally, 20 percent of all families renting condominiums move during a given year. A random sample of 200 families renting condominiums in the Dallas Metroplex revealed that 56 had moved during the past year. At the .01 significance level, does this evidence suggest that a larger proportion of condominium owners moved in the Dallas area? Determine the \(p\) -value.

eGolf.com receives an average of 6.5 returns per day from online shoppers. For a sample of 12 days, it received the following number of returns $$\begin{array}{|llllllllllll|}\hline 0 & 4 & 3 & 4 & 9 & 4 & 5 & 9 & 1 & 6 & 7 & 10 \\\\\hline\end{array}$$ At the .01 significance level, can we conclude the mean number of returns is less than \(6.5 ?\)

(a) Is this a one- or two-tailed test? (b) What is the decision rule? (c) What is the value of the test statistic? (d) What is your decision regarding \(H_{0} ?\) (e) What is the \(p\) -value? Interpret it. A sample of 36 observations is selected from a normal population. The sample mean is \(21,\) and the population standard deviation is \(5 .\) Conduct the following test of hypothesis using the .05 significance level. $$\begin{array}{l}H_{0}: \mu \leq 20 \\\H_{1}: \mu>20\end{array}$$

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