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10.20 A random sample of 64 bags of white Cheddar popcorn weighed, on average, 5.23 ounces with a standard deviation of 0.24 ounces. Test the hypothesis that \(p=5.5\) ounces against the alternative hypothesis, \(\mu<5.5\) ounces at the 0.05 level of significance

Short Answer

Expert verified
The hypothesis that the average weight of a bag of Cheddar popcorn is 5.5 ounces is rejected. There is significant evidence to suggest the average weight is less than 5.5 ounces.

Step by step solution

01

Calculate the z-score

The z-score is calculated with the following formula: \(z = \frac{\overline{x} - \mu}{ \frac{s}{\sqrt{n}}}\). In this case, \(\overline{x} = 5.23\) ounces, \(\mu = 5.5\) ounces, \(s = 0.24\) ounces, and \(n = 64\). After substituting these values into the formula, one gets \(z = \frac{5.23 - 5.5}{ \frac{0.24}{\sqrt{64}}}.\)
02

Perform the calculation

Subtract 5.23 from 5.5 to get -0.27. Then divide 0.24 by the square root of 64, which is 8, to get 0.03. Finally, divide -0.27 by 0.03 to get a z-score of -9.
03

Compare the z-score to the critical value

The calculated z-score of -9 is less than the critical value of -1.645. Therefore, it falls within the critical area.
04

Conclude the test

Since the calculated z-score falls within the critical area, this means that we reject the null hypothesis that the average weight of a bag of popcorn is 5.5 ounces. There is significant evidence to suggest that the average weight is less than 5.5 ounces.

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Key Concepts

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

z-score
The z-score is an essential concept in hypothesis testing, helping us determine how far a sample mean is from the null hypothesis mean, measured in standard deviations. This score provides a way to assess the likelihood of observing a sample mean at least as extreme as the one we have, assuming the null hypothesis is true. To compute the z-score, the formula is used: \[z = \frac{\overline{x} - \mu}{ \frac{s}{\sqrt{n}}}\] where:
  • \(\overline{x}\) is the sample mean
  • \(\mu\) is the population mean from the null hypothesis
  • \(s\) is the sample standard deviation
  • \(n\) is the sample size
In our exercise, the z-score was calculated as \(-9\), indicating that the sample mean is far lower than the hypothesized population mean of 5.5 ounces. This significant difference, highlighted by the high magnitude of the z-score, suggests statistical evidence against the null hypothesis. Understanding z-scores helps us effectively interpret statistical results and draw valid conclusions.
level of significance
The level of significance, often represented as \(\alpha\), is a critical threshold in hypothesis testing that helps us decide when to reject the null hypothesis. This value reflects how much risk of error we are willing to accept when making our decision. Common levels of significance include 0.05, 0.01, and 0.10, with a smaller \(\alpha\) indicating a more stringent test.In hypothesis testing, when the calculated z-score falls within the predefined critical region determined by \(\alpha\), we reject the null hypothesis. For the cheddar popcorn exercise, the level of significance is set at 0.05. This means there's a 5% probability of rejecting the null hypothesis when it is actually true—a phenomenon known as a Type I error. Choosing an appropriate level of significance depends on the context of the test, where we weigh the consequences of making such an error. A smaller \(\alpha\) reduces the chances of committing a Type I error but increases the possibility of a Type II error, which is failing to reject a false null hypothesis.
null hypothesis
The null hypothesis, denoted as \(H_0\), represents a statement of no effect or no difference, which we seek to test. It's the foundation of hypothesis testing, as we use statistical methods to decide whether we have enough evidence against it. In practice, the null hypothesis forms a baseline expectation that the observed data results occur purely by chance.For the popcorn exercise, the null hypothesis states that the average weight of the popcorn bags, \(\mu\), is 5.5 ounces (\(H_0: \mu = 5.5\)). The purpose of the test is to assess this claim against an alternative possibility. We seek evidence that the true average weight is, in fact, less than 5.5 ounces.Rejecting the null hypothesis means that the available data provides enough statistical significance to support the alternative hypothesis. Conversely, if the data doesn't reveal a significant deviation from \(H_0\), we fail to reject it, implying insufficient evidence to support a change or effect.
critical value
The critical value is a crucial component in hypothesis testing, marking the threshold beyond which we reject the null hypothesis. It serves as a boundary for determining whether the observed test statistic is extreme enough to warrant such a decision.When conducting a test, the critical value corresponds to the chosen level of significance and the nature of the alternative hypothesis (one-tailed or two-tailed). For our popcorn exercise with a 0.05 level of significance and a one-tailed test (since we posit that \(\mu < 5.5\)), the critical value is found from z-tables, at approximately \(-1.645\).If our calculated test statistic, like the z-score, falls within or beyond this critical region, we reject the null hypothesis in favor of the alternative. In this scenario, the calculated z-score of \(-9\) is well beyond \(-1.645\), placing it firmly in the critical region. This large deviation signifies that the sample result is statistically improbable under the null hypothesis, supporting the alternative hypothesis.

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Most popular questions from this chapter

Past experience indicates that the time for high school seniors to complete a standardized test is a normal random variable with a mean of 35 minutes. If a random sample of 20 high school seniors took an average of 33.1 minutes to complete this test with a standard deviation of 4.3 minutes, test the hypothesis at the 0.05 level of significance that \(p=35\) minutes against the alternative that \(p<35\) minutes.

Past expericnce indicates that the time required for high school seniors to complete a standardized test is a normal random variable with a standard deviation of 6 minutes. Test the hypothesis that \(0=0\) against the alternative that \(1 \mathrm{~T}<6\) if a random sample of 20 high school seniors has a standard deviations \(s=4.51 .\) Use a 0.05 level of significance.

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