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Chapter 10: Problem 21

According to the local union president, the mean gross 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 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
Reject the null hypothesis. The mean income is not \$45,000 (\(p = 0.0078\)).

Step by step solution

01

State the Hypotheses

We need to define our null and alternative hypotheses. \( H_0: \mu = \\(45,000 \), this states that the mean income is \\)45,000. \( H_a: \mu eq \\(45,000 \), this states that the mean income is not \\)45,000.
02

Determine the Test Statistic

The test statistic for the mean is calculated using the formula: \[ z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \] Here, \( \bar{x} = 45,500 \), \( \mu = 45,000 \), \( \sigma = 3,000 \), and \( n = 120 \). Substitute these values into the formula: \[ z = \frac{45,500 - 45,000}{\frac{3,000}{\sqrt{120}}} \approx 2.89 \]
03

Determine the Critical Value

Given the significance level \( \alpha = 0.10 \) for a two-tailed test, we find the critical \( z \)-values. For \( \alpha/2 = 0.05 \) on each tail, the critical \( z \)-value is approximately \( \pm 1.645 \).
04

Compare Test Statistic with Critical Values

With a calculated \( z \)-value of 2.89 and critical values of \( \pm 1.645 \), the \( z \)-value falls outside the critical region. Therefore, we reject the null hypothesis.
05

Calculate the p-Value

For \( z = 2.89 \), the \( p \)-value corresponds to the probability of observing a value as extreme or more extreme in the \( z \)-table. This value is about 0.0039 for one tail. For two-tailed tests, \( p = 2(0.0039) = 0.0078 \).
06

Conclusion

Since the \( p \)-value (0.0078) is less than the significance level (0.10), we conclude that there is sufficient evidence to reject the null hypothesis and infer that the mean income is not \$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 plays a significant role in statistics and hypothesis testing. It refers to a probability distribution that is symmetric about the mean, depicting that data near the mean are more frequent in occurrence than data far from the mean. Think of it as the classic bell-shaped curve.
  • Mean: This is the central point of the curve. In our case, the assumed mean income for plumbers is \( \\( 45,000 \).
  • Standard Deviation: This measures the spread of the data around the mean. A smaller value of standard deviation means the data is closely clustered, while a larger value spreads out the data. Here, it is \( \\) 3,000 \).
  • Symmetry: A normal distribution is perfectly symmetric around the mean.
Understanding the normal distribution helps determine probabilities and make inferences about a population from a sample.
p-Value Calculation
In hypothesis testing, the p-value plays a crucial role in deciding whether to reject the null hypothesis. The p-value represents "the probability of observing your data, or something more extreme, if the null hypothesis is true."
  • A small p-value (typically 鈮� 0.05) suggests that it is unlikely that the observed data would be obtained if the null hypothesis were true, leading to rejecting the null.
  • A larger p-value (> 0.10) indicates weak evidence against the null hypothesis, so it wouldn't be rejected.
In this exercise, we calculated a p-value of 0.0078. Since this is relatively small compared to our significance level of 0.10, it suggests that the mean income significantly differs from \( \$ 45,000 \).
To compute the p-value, the test statistic was compared against the standard normal distribution, determining how extreme the observed value is.
Significance Level
The significance level, denoted by \( \alpha \), is a topic of great importance in hypothesis testing. It dictates the criteria for rejecting the null hypothesis and is set as a threshold probability.
  • Common significance levels are 0.05, 0.01, and 0.10.
  • It represents the probability of rejecting the null hypothesis when it is actually true (Type I error).
  • In our exercise, the significance level is set to 0.10, meaning there is a 10% risk we are willing to take to reject a true null hypothesis.
Choosing a significance level depends on how much risk you are willing to accept. In scientific experiments, stricter levels like 0.01 are often used to maintain high confidence in results.
z-Test
The z-test is a statistical procedure used to determine if there is a significant difference between sample and population means. It is particularly used when the sample size is large (usually \( n > 30 \)) and you know the population standard deviation.
  • The z-test uses the standard normal distribution to evaluate the test statistic.
  • Calculated using: \[z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}\]where \( \bar{x} \) is the sample mean, \( \mu \) is the population mean, \( \sigma \) is the population standard deviation, and \( n \) is the sample size.
  • In this context, the calculated z-value was approximately 2.89.
The z-test allowed us to compare our sample mean of \( \\( 45,500 \) against the population mean of \( \\) 45,000 \) and test the validity of our hypotheses.

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Most popular questions from this chapter

Given the following hypotheses: $$\begin{array}{l}H_{0}: \mu=400 \\\H_{1}: \mu \neq 400\end{array}$$ A random sample of 12 observations is selected from a normal population. The sample mean was 407 and the sample standard deviation was 6 . Using the .01 significance level: a. State the decision rule. b. Compute the value of the test statistic. c. What is your decision regarding the null hypothesis?

The management of White Industries is considering a new method of assembling its golf cart. The present method requires a mean time of 42.3 minutes to assemble a cart. The mean assembly time for a random sample of 24 carts, using the new method, was 40.6 minutes, and the standard deviation of the sample was 2.7 minutes. Using the .10 level of significance, can we conclude that the assembly time using the new method is faster?

Given the following hypotheses: $$\begin{array}{l}H_{0}: \mu=100 \\\H_{1}: \mu \neq 100\end{array}$$ A random sample of six resulted in the following values: \(118,105,112,119,105,\) and 111 . Assume a normal population. Using the .05 significance level, can we conclude the mean is different from \(100 ?\) a. State the decision rule. b. Compute the value of the test statistic. c. What is your decision regarding the null hypothesis? d. Estimate the \(p\) -value.

A Washington, D.C., "think tank" announces the typical teenager sent 67 text messages per day in 2017 . To update that estimate, you phone a sample of 12 teenagers and ask them how many text messages they sent the previous day. Their responses were: \(\begin{array}{llllllllll}51 & 175 & 47 & 49 & 44 & 54 & 145 & 203 & 21 & 59 & 42 & 100\end{array}\) At the .05 level, can you conclude that the mean number is greater than \(67 ?\) Estimate the \(p\) -value and describe what it tells you.

The American Water Works Association reports that the per capita water use in a single-family home is 69 gallons per day. Legacy Ranch is a relatively new housing development. The builders installed more efficient water fixtures, such as low-flush toilets, and subsequently conducted a survey of the residences. Thirty-six owners responded, and the sample mean water use per day was 64 gallons with a standard deviation of 8.8 gallons per day. At the .10 level of significance, is that enough evidence to conclude that residents of Leqacy Ranch use less water on average?
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