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The mean salary of federal government employees on the General Schedule is \(\$ 59,593 .\) The average salary of 30 randomly selected state employees who do similar work is \(\$ 58,800\) with \(\sigma=\$ 1500\). At the 0.01 level of significance, can it be concluded that state employees earn on average less than federal employees?

Short Answer

Expert verified
State employees earn less on average than federal employees at the 0.01 significance level.

Step by step solution

01

Define the Hypotheses

We need to formulate the null and alternative hypotheses. The null hypothesis is that there is no difference in average salaries, i.e., \( H_0: \mu = 59,593 \). The alternative hypothesis states that the state employees earn less, i.e., \( H_a: \mu < 59,593 \).
02

Determine the Significance Level

The significance level, \( \alpha \), is 0.01, as stated in the problem. This means we are 99% confident in rejecting the null hypothesis.
03

Calculate the Test Statistic

We will use a z-test for single mean due to the known population standard deviation (\( \sigma = 1500 \)). The formula for the test statistic is: \[ z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \] where \( \bar{x} = 58,800 \), \( \mu = 59,593 \), \( \sigma = 1500 \), and \( n = 30 \). Calculating gives: \[ z = \frac{58,800 - 59,593}{\frac{1500}{\sqrt{30}}} \approx -2.89 \]
04

Determine the Critical Value

We look up the z-table for \( \alpha = 0.01 \) in a one-tailed test. The critical z-value for \( 0.01 \) significance in the left tail is approximately \( -2.33 \).
05

Make the Decision

We compare the calculated z-value to the critical value. Since \( z = -2.89 \) is less than \( -2.33 \), we reject the null hypothesis.
06

Conclusion

Based on the test, we conclude that at the 0.01 level of significance, there is sufficient evidence to suggest state employees earn less on average than federal employees.

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Key Concepts

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

Understanding the Z-Test
In hypothesis testing, the z-test is a crucial statistical tool used for determining whether two population means are different when the variances are known and the sample size is sufficient. It's perfect for testing differences in means with a known standard deviation, or standard deviation of a sample when the population standard deviation is unknown.

The z-test works by transforming the sample data into a z-score, which tells us how many standard deviations the sample mean is from the population mean. A high absolute value of the z-score shows that the sample data point lies far from the population mean, indicating a significant difference. In our exercise, we're dealing with a known population standard deviation (00"). We then use this to compute the z-value by comparing the sample mean (800") to the population mean (93").
  • If the z-value is beyond a certain critical threshold, it implies a significant difference.
  • A positive z-score indicates a value above the mean, while a negative z-score points to a value below the mean.
Significance Level in Hypothesis Testing
The significance level, often denoted by \( \alpha \), serves as a threshold for deciding whether to reject the null hypothesis. It's a pre-determined probability that stipulates how extreme data must be before we reject the null hypothesis.

A smaller significance level, like 0.01, indicates a stringent criterion for rejecting the null hypothesis. In our case, \( \alpha = 0.01 \) suggests that we require a very high confidence (99%) to reject the null hypothesis.
  • Common significance levels are 0.05, 0.01, and 0.001.
  • The significance level directly affects the probability of making a Type I error, which is mistakenly rejecting a true null hypothesis.

Thus, a 1% significance level implies that there is only a 1% risk of concluding that there is a difference when there is none.
Identifying the Critical Value
The critical value is a point on the scale of the test statistic beyond which we reject the null hypothesis, and it's determined based on the significance level. For a z-test, you'd refer to the z-distribution table to find this critical value for a given \( \alpha \).

In the exercise, our significance level of 0.01 and one-tailed test led us to a critical z-value of -2.33. This means:
  • If our computed z-score is less than -2.33, we can reject the null hypothesis.
  • The critical value essentially demarcates the extreme values that would lead us to reject the null hypothesis.

Remember, the position of the critical value depends on whether the test is one-tailed or two-tailed.
- For a one-tailed test, you look at one end of the distribution.
- In a two-tailed test, values on both extremes are considered for rejection.
The Concept of Null Hypothesis
The null hypothesis, symbolized as \( H_0 \), is a statement of no effect or no difference. It represents a default position that there is nothing unusual happening; it's a starting assumption for any hypothesis test.

In our example, the null hypothesis is \( \mu = 59,593 \), meaning there is no difference in the average salaries of state and federal employees. This is the claim we start with and aim to test through our calculated z-score and the critical value.
  • We always assume the null hypothesis is true until we have sufficient evidence against it.
  • Failing to reject the null hypothesis does not prove it; it only suggests that we do not have enough evidence to support a difference.
Exploring the Alternative Hypothesis
The alternative hypothesis, denoted as \( H_a \), is what researchers aim to support. It proposes that there is a new effect, or a difference exists contrary to the null hypothesis.

In the given exercise, the alternative hypothesis is \( \mu < 59,593 \), suggesting that state employees indeed earn less than federal employees. This hypothesis becomes the center of the test as we gather evidence to see if there's enough to shift from the null to the alternative hypothesis.
  • The alternative hypothesis can be tested as one-tailed or two-tailed, depending on the research question.
  • Accepting the alternative means the evidence is strong enough against the null hypothesis, justifying a significant conclusion.

Understanding both the null and alternative hypotheses is vital for making informed conclusions from hypothesis testing.

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Most popular questions from this chapter

Perform each of the following steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Find the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Assume that the population is approximately normally distributed. Teens are reported to watch the fewest total hours of television per week of all the demographic groups. The average television viewing for teens on Sunday from 1: 00 to 7: 00 P.M. is 58 minutes. A random sample of local teens disclosed the following times for Sunday afternoon television viewing. At \(\alpha=0.01\), can it be concluded that the average is greater than the national viewing time? (Note: Change all times to minutes.) $$ \begin{aligned} &\begin{array}{llll} 2: 30 & 2: 00 & 1: 30 & 3: 20 \end{array}\\\ &\begin{array}{llll} 1: 00 & 2: 15 & 1: 50 & 2: 10 \end{array}\\\ &\begin{array}{ll} 1: 30 & 2: 30 \end{array} \end{aligned} $$

A researcher knew that before cell phones, a person made on average 2.8 calls per day. He believes that the number of calls made per day today is higher. He selects a random sample of 30 individuals who use a cell phone and asks them to keep track of the number of calls that they made on a certain day. The mean was 3.1. At \(\alpha=0.01\) is there enough evidence to support the researcher's claim? The standard deviation for the population found by a previous study is 0.8 . Would the null hypothesis be rejected at \(\alpha=0.05 ?\)

Perform each of the following steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Find the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Assume that the population is approximately normally distributed. The National Novel Writing Association states that the average novel is at least 50,000 words. A particularly ambitious writing club at a college- preparatory high school had randomly selected members with works of the following lengths. At \(\alpha=0.10\), is there sufficient evidence to conclude that the mean length is greater than 50,000 words? $$ \begin{array}{lll} 48,972 & 50,100 & 51,560 \\ 49,800 & 50,020 & 49,900 \\ 52,193 & & \end{array} $$

Find the critical value (or values) for the \(t\) test for each. a. \(n=12, \alpha=0.01,\) left-tailed b. \(n=16, \alpha=0.05,\) right-tailed c. \(n=7, \alpha=0.10,\) two-tailed d. \(n=11, \alpha=0.025,\) right-tailed e. \(n=10, \alpha=0.05,\) two-tailed

Perform each of the following steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Find the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Assume that the population is approximately normally distributed. The U.S. Bureau of Labor and Statistics reported that a person between the ages of 18 and 34 has had an average of 9.2 jobs. To see if this average is correct, a researcher selected a random sample of 8 workers between the ages of 18 and 34 and asked how many different places they had worked. The results were as follows: $$ \begin{array}{llllllll} 8 & 12 & 15 & 6 & 1 & 9 & 13 & 2 \end{array} $$ At \(\alpha=0.05,\) can it be concluded that the mean is \(9.2 ?\) Use the \(P\) -value method. Give one reason why the respondents might not have given the exact number of jobs that they have worked.

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