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On Friday, Wall Street traders were anxiously awaiting the federal government's release of numbers on the January increase in nonfarm payrolls. The early consensus estimate among economists was for a growth of 250,000 new jobs (CNBC, February 3,2006 ). However, a sample of 20 economists taken Thursday afternoon provided a sample mean of 266,000 with a sample standard deviation of 24,000 . Financial analysts often call such a sample mean, based on late-breaking news, the whisper number. Treat the "consensus estimate" as the population mean. Conduct a hypothesis test to determine whether the whisper number justifies a conclusion of a statistically significant increase in the consensus estimate of economists. Use \(\alpha=.01\) as the level of significance.

Step by step solution

01

State Hypotheses

We need to set up our null and alternative hypotheses. The null hypothesis is that the true mean increase in payrolls is equal to the consensus estimate, i.e., \( H_0: \mu = 250,000 \). The alternative hypothesis is that the true mean increase is greater than the consensus estimate, i.e., \( H_1: \mu > 250,000 \).

02

Calculate the Test Statistic

We will use a one-sample t-test since we are comparing the sample mean to the known population mean. The formula for the t statistic is: \( t = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}} \), where \( \bar{x} = 266,000 \), \( \mu_0 = 250,000 \), \( s = 24,000 \), and \( n = 20 \). Substituting these values results in \( t = \frac{266,000 - 250,000}{\frac{24,000}{\sqrt{20}}} \).

03

Compute the t Value

Perform the calculation: \( t = \frac{16,000}{5,366.56} \approx 2.981 \). This is the calculated t value.

04

Determine the Critical Value

For a one-tailed t-test at \( \alpha = 0.01 \) and \( df = 19 \) (since \( df = n - 1 \)), we refer to the t distribution table to find the critical t value. The critical t value is approximately 2.539.

05

Compare t Value to Critical Value

Comparing the calculated t value (2.981) to the critical t value (2.539), we see that 2.981 is greater than 2.539.

06

Make a Decision

Since the calculated t value exceeds the critical value, we reject the null hypothesis \( H_0 \). This means there is sufficient evidence to conclude that there is a statistically significant increase in the payrolls.

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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 compare the mean of a sample to a known population mean. It helps determine if there is a significant difference between the two means. In this exercise, a one-sample t-test is employed because we are comparing the whisper number to the consensus estimate. Here's how it works:

• Identify the sample mean (\( \bar{x} \)). In the given problem, it is 266,000.
• Identify the population mean (\( \mu_0 \)). This is the consensus estimate, 250,000.
• Use the sample standard deviation (s = 24,000) and the sample size (n = 20) in the formula for the test statistic.

The objective is to see if the sample mean differs significantly from the population mean, leading us to decide between retaining or rejecting the null hypothesis.

level of significance

The level of significance (\( \alpha \)) in hypothesis testing is the threshold chosen by the researcher to decide whether or not to reject the null hypothesis. It quantifies the probability of making a Type I error, which occurs when the null hypothesis is wrongly rejected.

• In this problem, \( \alpha = 0.01 \), meaning there is a 1% risk of concluding that there is a significant difference when there isn't one.
• A smaller \( \alpha \) (like 0.01) indicates a stricter criterion, reducing the risk of a Type I error, but making it harder to detect a true effect.

The level of significance is essential as it influences the critical value against which the test statistic is compared.

null and alternative hypotheses

Hypothesis testing involves creating two opposing hypotheses. The null hypothesis (\( H_0 \)) typically represents a statement of no effect or no difference, and the alternative hypothesis (\( H_1 \)) suggests there is an effect or a difference.

• In this scenario, \( H_0: \mu = 250,000 \), implying the true mean increase does not differ from the consensus estimate.
• The alternative hypothesis, \( H_1: \mu > 250,000 \), suggests the actual mean increase in payrolls is greater than the consensus estimate.

These hypotheses guide the testing process, as the goal is to determine which is more reasonably supported by the data.

critical value

The critical value is a point on the test distribution that divides the region where we reject the null hypothesis from the region where we do not reject it. This value depends on the level of significance and the degrees of freedom, which are calculated as the sample size minus one (\( df = n - 1 \)).

• For this exercise, with \( \alpha = 0.01 \) and \( df = 19 \), we find the critical value by using a t-distribution table. The critical value found is approximately 2.539.
• Once the critical value is obtained, compare it with the calculated t value (2.981). Since 2.981 exceeds 2.539, the null hypothesis is rejected.

Thus, the critical value acts as a benchmark for making the decision about the null hypothesis based on the data.