/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 12 The manufacturer of processed de... [FREE SOLUTION] | 91Ó°ÊÓ

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The manufacturer of processed deli meats reports that the standard deviation of the number of carbohydrates in its smoked turkey breast is 0.5 gram per 2-ounce serving. A dietitian does not believe the manufacturer and randomly selects eighteen 2 -ounce servings of the smoked turkey breast and determines the number of carbohydrates per serving. The standard deviation of the number of carbs is computed to be 0.62 gram per serving. Is there sufficient evidence to indicate that the standard deviation is not 0.5 gram per serving at the \(\alpha=0.05\) level of significance? A normal probability plot indicates that the number of carbohydrates per serving is normally distributed.

Short Answer

Expert verified
No, there is insufficient evidence to indicate that the standard deviation is not 0.5 gram per serving at the 0.05 level of significance.

Step by step solution

01

State the hypotheses

Formulate the null and alternative hypotheses. The null hypothesis (\text{H}_0) states that the standard deviation is 0.5 gram, and the alternative hypothesis (\text{H}_1) states that the standard deviation is not 0.5 gram.\[\text{H}_0: \sigma = 0.5 \text{ grams} \] \[\text{H}_1: \sigma eq 0.5 \text{ grams} \]
02

Determine the test statistic

Use the chi-square test for the standard deviation. The test statistic is calculated using the formula: \[\chi^2 = \frac{(n-1)s^2}{\sigma^2} \] where \(n = 18\) is the sample size, \(s = 0.62\) grams is the sample standard deviation, and \(\sigma = 0.5\) grams is the claimed standard deviation. Thus: \[\chi^2 = \frac{(18-1)(0.62)^2}{(0.5)^2} = \frac{17 \times 0.3844}{0.25} \approx 26.1328 \]
03

Find the critical values

Determine the critical values at the \(\alpha = 0.05\) significance level. For a two-tailed test with \(\alpha = 0.05\) and \(n-1 = 17\) degrees of freedom, the critical values are found from the chi-square distribution table: \[\chi^2_{1 - \frac{\alpha}{2}} = 30.191\], and \[\chi^2_{\frac{\alpha}{2}} = 6.908\]
04

Make a decision

Compare the test statistic to the critical values. If the test statistic falls outside the range of the critical values, reject the null hypothesis. Here, \(\chi^2 = 26.1328\) falls within \(6.908\) and \(30.191\). Thus, we do not reject the null hypothesis.
05

Conclusion

There is not sufficient evidence at the \(\alpha = 0.05\) significance level to conclude that the standard deviation of the number of carbohydrates per serving is different from 0.5 gram.

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Key Concepts

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

Hypothesis Testing
Hypothesis testing is a fundamental method in statistics used to determine whether there is enough evidence to reject a null hypothesis. It involves a few steps to evaluate the validity of a claim statistically. First, you state the null hypothesis (\text{H}_0) and the alternative hypothesis (\text{H}_1). The null hypothesis represents the status quo or a specific claim, and it is assumed to be true unless strong evidence suggests otherwise. In our exercise, the null hypothesis is that the standard deviation of carbohydrates in smoked turkey breast is 0.5 grams.
Then, we select a significance level (typically denoted by \( \alpha \))). This is the probability of rejecting the null hypothesis when it is true (Type I error). Here, the significance level \( \alpha = 0.05 \), meaning there's a 5% risk of concluding that there is an effect when there is none. The next step involves calculating the test statistic using sample data and comparing it to critical values determined by the significance level and degrees of freedom. If the test statistic falls in the critical region, we reject the null hypothesis; otherwise, we fail to reject it. This systematic approach allows us to make data-driven decisions about hypotheses.
Standard Deviation
The standard deviation is a measure of the amount of variation or dispersion in a set of values. It shows how much the individual data points differ from the mean of the dataset. In simple terms, it tells you how spread out the values are. For example, a low standard deviation means that most of the numbers are close to the mean, while a high standard deviation means that the numbers are more spread out.
In our exercise, the manufacturer claims that the standard deviation of carbohydrates in smoked turkey breast is 0.5 grams. The dietitian's sample, however, shows a standard deviation of 0.62 grams. To determine if this difference is statistically significant, we use the chi-square test for the standard deviation. Understanding standard deviation is crucial because it affects many aspects of data analysis, from hypothesis testing to confidence intervals.
Significance Level
The significance level, often denoted as \( \alpha \), is the probability of rejecting the null hypothesis when it is actually true. It is a threshold set by the researcher to decide whether an observed effect is statistically significant or not. Common values for \( \alpha \) are 0.05, 0.01, and 0.10. In our exercise, we use \( \alpha = 0.05 \), meaning there is a 5% risk of concluding that the standard deviation is different from 0.5 grams when it is actually the same.
Choosing the right significance level is important as it balances the risk of making a Type I error (false positive) against the rigor of the test. A lower \( \alpha \) makes the test stricter, reducing the likelihood of a false positive but increasing the chance of a false negative (Type II error). Therefore, researchers must choose a significance level appropriate for their specific study and how much risk they are willing to accept.
Normal Distribution
The normal distribution is a continuous probability distribution that is symmetrical and bell-shaped, describing how the values of a variable are distributed. In a normal distribution, most values cluster around a central peak, with probabilities for values tapering off equally in both directions from the mean. This is why it is often referred to as a 'bell curve'. Many statistical tests, including the chi-square test, assume that the data follow a normal distribution.
In our exercise, it is important to check that the number of carbohydrates per serving is normally distributed before performing the chi-square test. A normal probability plot was used to confirm this assumption. When the data follows a normal distribution, we can more comfortably rely on the statistical test results to make inferences about the population.

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Most popular questions from this chapter

Suppose you wish to determine if the mean IQ of students on your campus is different from the mean IQ in the general population, \(100 .\) To conduct this study, you obtain a simple random sample of 50 students on your campus, administer an IQ test, and record the results. The mean IQ of the sample of 50 students is found to be 107.3 with a standard deviation of \(13.6 .\) (a) Conduct a hypothesis test (preferably using technology) \(H_{0}: \mu=\mu_{0}\) versus \(H_{1}: \mu \neq \mu_{0}\) for \(\mu_{0}=103,104,105,106,107,108,109,110,111,112\) at the \(\alpha=0.05\) level of significance. For which values of \(\mu_{0}\) do you not reject the null hypothesis? (b) Construct a \(95 \%\) confidence interval for the mean IQ of students on your campus. What might you conclude about how the lower and upper bounds of a confidence interval relate to the values for which the null hypothesis is rejected? (c) Suppose you changed the level of significance in conducting the hypothesis test to \(\alpha=0.01\). What would happen to the range of values of \(\mu_{0}\) for which the null hypothesis is not rejected? Why does this make sense?

A simple random sample of size \(n=20\) is drawn from a population that is normally distributed. The sample variance is found to be \(49.3 .\) Determine whether the population variance is less than 95 at the \(\alpha=0.1\) level of significance.

Sports announcers often say, "I wonder which player will show up to play today."This is the announcer's way of saying that the player's performance varies dramatically from game to game. Suppose that the standard deviation of the number of points scored by shooting guards in the NBA is 8.3. A random sample of 25 games played by Derrick Rose results in a sample standard deviation of 6.7 points. Assume that a normal probability plot indicates that the points scored are approximately normally distributed. Is Derrick Rose more consistent than other shooting guards in the NBA at the \(\alpha=0.10\) level of significance?

In \(1994,52 \%\) of parents with children in high school felt it was a serious problem that high school students were not being taught enough math and science. A recent survey found that 256 of 800 parents with children in high school felt it was a serious problem that high school students were not being taught enough math and science. Do parents feel differently today than they did in 1994 ? Use the \(\alpha=0.05\) level of significance? Source: Based on "Reality Check: Are Parents and Students Ready for More Math and Science?" Public Agenda, \(2006 .\)

To test \(H_{0}: p=0.25\) versus \(H_{1}: p \neq 0.25,\) a simple random sample of \(n=350\) individuals is obtained and \(x=74\) successes are observed. (a) What does it mean to make a Type II error for this test? (b) If the researcher decides to test this hypothesis at the \(\alpha=0.05\) level of significance, compute the probability of making a Type II error if the true population proportion is 0.23. What is the power of the test? (c) Redo part (b) if the true population proportion is 0.28 .

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