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Consider the following hypothesis test: \\[ \begin{array}{l} H_{0}: \mu \geq 45 \\ H_{\mathrm{a}}: \mu<45 \end{array} \\] A sample of 36 is used. Identify the \(p\) -value and state your conclusion for each of the following sample results. Use \(\alpha=.01\) a. \(\bar{x}=44\) and \(s=5.2\) b. \(\bar{x}=43\) and \(s=4.6\) c. \(\bar{x}=46\) and \(s=5.0\)

Short Answer

Expert verified
Part a: Fail to reject \( H_0 \). Part b: Reject \( H_0 \). Part c: Fail to reject \( H_0 \).

Step by step solution

01

Define the hypothesis

We have the null hypothesis \( H_0: \mu \geq 45 \) and the alternative hypothesis \( H_a: \mu < 45 \). We are using a left-tailed test with a significance level of \( \alpha = 0.01 \).
02

Understand the test statistic

The test statistic is calculated using the formula: \[ z = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}} \] where \( \bar{x} \) is the sample mean, \( \mu_0 = 45 \) is the hypothesized population mean, \( s \) is the sample standard deviation, and \( n = 36 \) is the sample size.
03

Calculate the test statistic for part a

For part a, where \( \bar{x} = 44 \) and \( s = 5.2 \):\[ z = \frac{44 - 45}{\frac{5.2}{\sqrt{36}}} = \frac{-1}{\frac{5.2}{6}} = \frac{-1}{0.867} \approx -1.153 \]
04

Find the p-value for part a

We find the p-value using the z-score from Step 3. The p-value associated with \( z = -1.153 \) (using a z-table or calculator) is approximately 0.1251.
05

Conclusion for part a

With a p-value of 0.1251, which is greater than \( \alpha = 0.01 \), we fail to reject the null hypothesis. There is not enough evidence to support \( H_a \).
06

Calculate the test statistic for part b

For part b, where \( \bar{x} = 43 \) and \( s = 4.6 \):\[ z = \frac{43 - 45}{\frac{4.6}{\sqrt{36}}} = \frac{-2}{\frac{4.6}{6}} = \frac{-2}{0.767} \approx -2.608 \]
07

Find the p-value for part b

We find the p-value using the z-score from Step 6. The p-value associated with \( z = -2.608 \) is approximately 0.0046.
08

Conclusion for part b

With a p-value of 0.0046, which is less than \( \alpha = 0.01 \), we reject the null hypothesis. There is enough evidence to support \( H_a \).
09

Calculate the test statistic for part c

For part c, where \( \bar{x} = 46 \) and \( s = 5.0 \):\[ z = \frac{46 - 45}{\frac{5.0}{\sqrt{36}}} = \frac{1}{\frac{5.0}{6}} = \frac{1}{0.833} \approx 1.200 \]
10

Find the p-value for part c

We find the p-value using the z-score from Step 9. The p-value associated with \( z = 1.200 \) is approximately 0.8849 (since it's a left-tailed test and we need to find the area on the left-hand side).
11

Conclusion for part c

With a p-value of 0.8849, which is much greater than \( \alpha = 0.01 \), we fail to reject the null hypothesis. There is not enough evidence to support \( H_a \).

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Key Concepts

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

Understanding the p-value in Hypothesis Testing
When conducting a hypothesis test, the p-value is a crucial element. It helps us decide whether to reject the null hypothesis, \( H_0 \). The p-value is the probability of obtaining a test statistic that is as extreme as, or more extreme than, the observed one if the null hypothesis is true. It's a measure of the evidence against \( H_0 \). A smaller p-value indicates stronger evidence against the null hypothesis.

Think of the p-value like a scale:
  • If the p-value is less than a predefined significance level (\( \alpha \)), we reject \( H_0 \) because the evidence is strong.
  • If it's greater, we fail to reject \( H_0 \) since there isn't enough evidence.
In our exercise, with \( \alpha = 0.01 \):
  • For \( \bar{x} = 44 \), the p-value is approximately 0.1251.
  • For \( \bar{x} = 43 \), the p-value is approximately 0.0046.
  • For \( \bar{x} = 46 \), the p-value is approximately 0.8849.
By comparing these to \( \alpha \), we determine whether there's enough evidence to support the alternative hypothesis \( H_a \).
Exploring the Z-Score
The z-score helps us understand how a sample mean compares to the hypothesized population mean. In essence, it tells us how many standard deviations away our sample mean is from the population mean under the null hypothesis.

The z-score formula is:
\[z = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}}\]
where:
  • \( \bar{x} \) is the sample mean.
  • \( \mu_0 \) is the hypothesized population mean.
  • \( s \) is the sample standard deviation.
  • \( n \) is the sample size.
Calculating the z-score for each situation in the exercise helps us find the corresponding p-value and determine the statistical significance of our results. Simply put, the z-score is your guide to converting your sample data into information about population hypotheses.
Significance Level Decoded
Significance levels, often denoted as \( \alpha \), are pre-set thresholds that help measure how strong the evidence against the null hypothesis needs to be. In general, a common choice is \( \alpha = 0.05 \), but stricter tests, like the one in our example, use \( \alpha = 0.01 \).

Here's how to think about significance levels:
  • A significance level of 0.01 implies that there is only a 1% risk of rejecting the null hypothesis when it is actually true. This is known as a Type I error.
  • The lower the significance level, the stronger the evidence must be to reject the null.
Using a significance level allows researchers to have a guide for decision-making, providing a standard by which p-values are judged. In our solution, this threshold determines when the evidence (p-value) is considered strong enough to reject \( H_0 \) and accept \( H_a \). By setting this level, we manage the risk of making incorrect conclusions while interpreting results.

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Most popular questions from this chapter

Carpetland salespersons average \(\$ 8000\) per week in sales. Steve Contois, the firm's vice president, proposes a compensation plan with new selling incentives. Steve hopes that the results of a trial selling period will enable him to conclude that the compensation plan increases the average sales per salesperson. a. Develop the appropriate null and alternative hypotheses. b. What is the Type I error in this situation? What are the consequences of making this error? c. What is the Type II error in this situation? What are the consequences of making this error?

The Coca-Cola Company reported that the mean per capita annual sales of its beverages in the United States was 423 eight-ounce servings (Coca-Cola Company website, February 3 2009 ). Suppose you are curious whether the consumption of Coca-Cola beverages is higher in Atlanta, Georgia, the location of Coca-Cola's corporate headquarters. A sample of 36 individuals from the Atlanta area showed a sample mean annual consumption of 460.4 eight-ounce servings with a standard deviation of \(s=101.9\) ounces. Using \(\alpha=.05\) do the sample results support the conclusion that mean annual consumption of Coca-Cola beverage products is higher in Atlanta?

Consider the following hypothesis test: \\[ \begin{array}{l} H_{0}: p=.20 \\ H_{\mathrm{a}}: p \neq .20 \end{array} \\] A sample of 400 provided a sample proportion \(\bar{p}=.175\) a. Compute the value of the test statistic. b. What is the \(p\) -value? c. \(\quad\) At \(\alpha=.05,\) what is your conclusion? d. What is the rejection rule using the critical value? What is your conclusion?

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.

Consider the following hypothesis test: \\[ \begin{array}{l} H_{0}: \mu \leq 50 \\ H_{\mathrm{a}}: \mu>50 \end{array} \\] A sample of 60 is used and the population standard deviation is \(8 .\) Use the critical value approach to state your conclusion for each of the following sample results, Use \\[\alpha=.05\\] a. \(\quad \bar{x}=52.5\) b. \(\bar{x}=51\) c. \(\quad \bar{x}=51.8\)

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