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The mean weight of a package of frozen fish must equal 22 oz. Five independent samples were selected, and the statistical hypotheses about the mean weight were tested. The \(P\) -values that resulted from these tests were \(0.065,0.0924,0.073,0.025,\) and \(0.021 .\) Is there sufficient evidence to conclude that the mean package weight is not equal to 22 oz?

Short Answer

Expert verified
No, there is not sufficient evidence to conclude the mean weight is not 22 oz.

Step by step solution

01

Understand the Hypotheses

We are conducting a hypothesis test where the null hypothesis \(H_0\) is that the mean weight, \( \mu \), is equal to 22 oz. The alternative hypothesis \(H_a\) is that the mean weight is not equal to 22 oz.
02

Evaluate the P-values

The given \(P\)-values are 0.065, 0.0924, 0.073, 0.025, and 0.021. Each \(P\)-value represents the probability of observing a test statistic as extreme as, or more extreme than, the one observed under the null hypothesis.
03

Determine the Significance Level

Typically, a significance level (\( \alpha \)) of 0.05 is used. This means we would reject the null hypothesis if any \(P\)-value is less than 0.05.
04

Compare P-values with the Significance Level

Compare each \(P\)-value with the significance level of 0.05. The \(P\)-values 0.025 and 0.021 are less than 0.05.
05

Make a Decision

Since there are two \(P\)-values (0.025 and 0.021) less than 0.05, this provides some evidence against the null hypothesis, but with multiple tests, the results should be considered collectively.
06

Sufficient Evidence Evaluation

Although there are some \(P\)-values less than 0.05, the majority of the tests have \(P\)-values above 0.05. Therefore, collectively, there is not sufficient evidence to reject the null hypothesis at a 0.05 significance level for the majority of tests.

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Key Concepts

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

Null Hypothesis
In hypothesis testing, the null hypothesis, denoted as \( H_0 \), is the starting point of any statistical test. It is a statement that there is no effect, change, or difference in the context of the experiment or survey. In the exercise we are considering, the null hypothesis is that the mean weight of a package of frozen fish is equal to 22 oz.
The purpose of stating a null hypothesis is to provide a clear criterion for testing the statistics collected from the samples. It sets a benchmark for determining whether there is any significant effect or difference.
  • The null hypothesis assumes no change or effect, which means variables remain consistent with this assumption being true.
  • It is considered true until evidence in the form of data suggests otherwise.
Rejecting the null hypothesis indicates there's sufficient statistical evidence to support an alternative hypothesis. Otherwise, we continue to assume the null hypothesis is true.
P-Value
The p-value in statistical hypothesis testing quantifies how well your experimental data supports a null hypothesis. A low p-value indicates that the observed data is unlikely under the null hypothesis, suggesting the possibility of the alternative hypothesis being true.
In the exercise, the provided p-values are 0.065, 0.0924, 0.073, 0.025, and 0.021. These values indicate the degree of evidence against the null hypothesis that the mean weight equals 22 oz.
  • The smaller the p-value, the stronger the evidence against the null hypothesis.
  • If a p-value is less than the chosen significance level, we reject the null hypothesis.
P-values help us understand the likelihood of getting our observed data if the null hypothesis is true. It is, however, only a measure of evidence and does not prove the null hypothesis wrong by itself.
Significance Level
The significance level, often denoted by \( \alpha \), is a threshold set by researchers to determine whether a p-value indicates enough evidence to reject the null hypothesis. It defines how sure we want to be about rejecting a hypothesis.
Commonly, a significance level of 0.05 is used, implying a 5% risk of concluding that a difference exists when there is none. It essentially controls the probability of a Type I error, which is falsely rejecting the true null hypothesis.
  • A smaller significance level means a stricter criterion for rejecting the null hypothesis.
  • In the exercise, we compare the given p-values against the 0.05 threshold.
Only when p-values fall below this significance level do we consider there is enough evidence to reject the null hypothesis in favor of the alternative.
Alternative Hypothesis
The alternative hypothesis, represented as \( H_a \) or \( H_1 \), is the statement that contradicts the null hypothesis. It represents a new effect or difference, suggesting that something is happening in the data that is statistically significant.
In our exercise, the alternative hypothesis is that the mean weight of the frozen fish package is not equal to 22 oz. This implies that any observed outcome different from 22 oz. could indicate this hypothesis might be true.
  • Acceptance of the alternative hypothesis requires strong evidence against the null hypothesis.
  • It allows researchers to suggest that the effect observed is real and not by random chance.
The difference indicated by the alternative hypothesis must be quantifiable and statistically valid to reject the null hypothesis confidently, determining whether statistical procedures have captured a truly significant result.

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Most popular questions from this chapter

In a random sample of 500 handwritten zip code digits, 466 were read correctly by an optical character recognition (OCR) system operated by the U.S. Postal Service (USPS). USPS would like to know whether the rate is at least \(90 \%\) correct. Do the data provide evidence that the rate is at least \(90 \%\) at \(\alpha=0.05 ?\)

Suppose that eight sets of hypotheses of the form $$H_{0}: \mu=\mu_{0} \quad H_{1}: \mu \neq \mu_{0}$$ have been tested and that the \(P\) -values for these tests are \(0.15,\) \(0.06 .0 .67,0.01,0.04,0.08,0.78,\) and \(0.13 .\) Use Fisher's procedure to combine all of the \(P\) -values. What conclusions can you draw about these hypotheses?

Consider the test of \(H_{0}: \sigma^{2}=5\) against \(H_{1}: \sigma^{2}<5 .\) What are the critical values for the test statistic \(\chi_{0}^{2}\) for the following significance levels and sample sizes? (a) \(\alpha=0.01\) and \(n=20\) (b) \(\alpha=0.05\) and \(n=12\) (c) \(\alpha=0.10\) and \(n=15\)

The mean pull-off force of a connector depends on cure time. (a) State the null and alternative hypotheses used to demonstrate that the pull-off force is below 25 newtons. (b) Assume that the previous test does not reject the null hypothesis. Does this result provide strong evidence that the pulloff force is greater than or equal to 25 newtons? Explain.

A bearing used in an automotive application is supposed to have a nominal inside diameter of 1.5 inches. A random sample of 25 bearings is selected, and the average inside diameter of these bearings is 1.4975 inches. Bearing diameter is known to be normally distributed with standard deviation \(\sigma=0.01\) inch. (a) Test the hypothesis \(H_{0}: \mu=1.5\) versus \(H_{1}: \mu \neq 1.5\) using $$\alpha=0.01$$ (b) What is the \(P\) -value for the test in part (a)? (c) Compute the power of the test if the true mean diameter is 1.495 inches. (d) What sample size would be required to detect a true mean diameter as low as 1.495 inches if you wanted the power of the test to be at least \(0.9 ?\) (e) Explain how the question in part (a) could be answered by constructing a two-sided confidence interval on the mean diameter.

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