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A company claims that the mean net weight of the contents of its All Taste cereal boxes is at least 18 ounces. Suppose you want to test whether or not the claim of the company is true. Explain briefly how you would conduct this test using a large sample. Assume that \(\sigma=.25\) ounce.

Short Answer

Expert verified
In order to test the company's claim, carry out a one-tailed z-test under the null hypothesis that the mean weight of the cereal boxes is at least 18 ounces. Base on the z-score calculated from the sample data, and compare it to the critical z-score for the chosen significance level. If the calculated z-score is less than the critical z-score, reject the null hypothesis and conclude that the mean weight is less than 18 ounces. Else, fail to reject the null hypothesis and conclude that there is not enough evidence to contradict the company's claim.

Step by step solution

01

Set Up the Hypotheses

The null hypothesis (\(H_0\)) is that the mean weight of the cereal boxes is at least 18 ounces: \(H_0: \mu \geq 18\) ounces. The alternate hypothesis (\(H_1\)) is that the mean weight is less than 18 ounces: \(H_1: \mu < 18\) ounces
02

Conduct the Z-Test

Assuming we have a large sample size (n), we would first calculate the sample mean (\(\bar{x}\)). Then, we can calculate the z-score using the formula: \(Z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}}\), where \(\bar{x}\) is the sample mean, \(\mu\) is the population mean (18 ounces), \(\sigma\) is the standard deviation (0.25 ounce), and n is the sample size.
03

Determine the Critical Region

If we are testing at a significance level of say, 0.05 (5%), the critical z-score for a one-tailed test would be the value that leaves 5% of the distribution in the tail. For a one-tailed test, the critical z-score is typically -1.645. If the calculated z-score falls in the critical region (less than -1.645), we would reject the null hypothesis.
04

Drawing Conclusion

Based on whether the calculated z-score falls in the critical region, we will either reject or fail to reject the null hypothesis. If we reject the null hypothesis, we are saying that the mean weight of the cereal boxes is less than 18 ounces, contradicting the company's claim. If we fail to reject the null hypothesis, we do not have enough evidence to say that the mean weight is less than 18 ounces, and so the company's claim stands.

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Key Concepts

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

Z-Test
When you're faced with the question of whether the mean of a population differs from a known value, the Z-Test is the golden tool of choice. It is especially useful when the sample size is large or the population variance is known. A Z-Test helps you determine if there's enough statistical evidence to support a particular hypothesis about the population mean.

The process begins by calculating the sample mean (\(\bar{x}\)). Then, the Z-score is computed using the formula: \[Z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}}\]where:
  • \(\bar{x}\) = sample mean
  • \(\mu\) = population mean
  • \(\sigma\) = population standard deviation
  • \(n\) = sample size
This formula standardizes our scenario to see how far and in what direction our sample mean deviates from the population mean in units of standard deviation.

A large absolute Z-score indicates a significant difference from the null hypothesis, thus putting it under scrutiny.
Null Hypothesis
In hypothesis testing, the null hypothesis stands as the statement we are trying to disprove or provide evidence against. In simple terms, it's the assumption of "no difference" or "no effect." For the company claiming their cereal boxes have a mean weight of at least 18 ounces, the null hypothesis \(H_0\) would be \(\mu \geq 18\) ounces.

The null hypothesis is crucial because it sets the stage for the test. By establishing this baseline, you can objectively determine whether the results from your data indicate that the status quo is correct or should be rejected.

Rejecting the null hypothesis is akin to saying "there's enough evidence to suggest a deviation from these expectations." Failing to reject it indicates the contrary; not necessarily that it is true, but you don't have enough evidence to prove otherwise.
Significance Level
The significance level, often denoted by \(\alpha\), is the probability threshold set before conducting a hypothesis test. It determines how extreme the test statistic from the Z-Test must be for us to reject the null hypothesis.

Commonly used \(\alpha\) values are 0.05, 0.01, or 0.10, corresponding to 5%, 1%, and 10% levels of significance, respectively. A 5% level means there's a 5% risk of concluding that a difference exists when there is none.

Selecting a significance level is a trade-off between what risk of error is acceptable and how strong a claim the research findings can support. A smaller \(\alpha\) level means stricter criteria for rejection, thereby lowering the chances of false positives (type I errors).

In our example with the cereal company, by choosing 0.05 as the significance level, it means we're tolerating a 5% chance that we'll incorrectly reject \(H_0\) when it is true.
Critical Region
The critical region in hypothesis testing is the set of all outcomes which, if occurred, would lead us to reject the null hypothesis. It's the "danger zone" where evidence is against the null. To determine it, we use the significance level and, for a Z-Test, critical Z-score values.

If testing at a 5% level of significance for a one-tailed test, you might find the critical Z-score is -1.645. If your computed Z-score falls beyond -1.645, it lands within the critical region, suggesting that the sample mean is significantly lower than the hypothesized population mean.

Thus, in our cereal box example, if your calculated Z-value is less than -1.645, the results fall in the critical region, leading to a rejection of the null hypothesis. This means the evidence shows the mean weight is likely less than 18 ounces.

Determining the critical region correctly ensures that our decision in hypothesis testing remains objective and statistically supported.

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Most popular questions from this chapter

Consider \(H_{0}=\mu=40\) versus \(H_{1}: \mu>40 .\) a. A random sample of 64 observations taken from this population produced a sample mean of 43 and a standard deviation of \(5 .\) Using \(\alpha=.025\), would you reject the null hypothesis? b. Another random sample of 64 observations taken from the same population produced a sample mean of 41 and a standard deviation of \(7 .\) Using \(\alpha=.025\), would you reject the null hypothesis? Comment on the results of parts a and \(\mathrm{b}\).

A random sample of 18 observations produced a sample mean of \(9.24\). Find the critical and observed values of \(z\) for each of the following tests of hypothesis using \(\alpha=.05 .\) The population standard deviation is known to be \(5.40\) and the population distribution is normal. a. \(H_{0}: \mu=8.5\) versus \(H_{1}: \mu \neq 8.5\) b. \(H_{0}=\mu=8.5\) versus \(H_{1}: \mu>8.5\)

A study claims that all adults spend an average of 14 hours or more on chores during a weekend. A researcher wanted to check if this claim is true. A random sample of 200 adults taken by this researcher showed that these adults spend an average of \(14.65\) hours on chores during a weekend. The population standard deviation is known to be \(3.0\) hours. a. Find the \(p\) -value for the hypothesis test with the alternative hypothesis that all adults spend more than 14 hours on chores during a weekend. Will you reject the null hypothesis at \(\alpha=.01 ?\) b. Test the hypothesis of part a using the critical-value approach and \(\alpha=.01\).

Make the following tests of hypotheses. a. \(H_{0}: \mu=80, \quad H_{1}: \mu \neq 80, \quad n=33, \quad \bar{x}=76.5, \quad \sigma=15, \quad \alpha=.10\) b. \(H_{0}=\mu=32, \quad H_{1}: \mu<32, \quad n=75, \quad \bar{x}=26.5, \quad \sigma=7.4, \quad \alpha=.01\) c. \(H_{0}=\mu=55, \quad H_{1}: \mu>55, \quad n=40, \bar{x}=60.5, \quad \sigma=4, \quad \alpha=.05\)

The manager of a service station claims that the mean amount spent on gas by its customers is $$\$ 15.90$$ per visit. You want to test if the mean amount spent on gas at this station is different from $$\$ 15.90$$ per visit. Briefly explain how you would conduct this test when \(\sigma\) is not known. \(9.73\) A tool manufacturing company claims that its top-of-the-line machine that is used to manufacture bolts produces an average of 88 or more bolts per hour. A company that is interested in buying this machine wants to check this claim. Suppose you are asked to conduct this test. Briefly explain how you would do so when \(\sigma\) is not known.

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