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Chapter 9: Problem 46

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
We would conduct a one-sample z-test for the hypothesis if the mean weight of the cereal boxes is indeed 18 ounces. However, this step-by-step doesn't lead to a concrete conclusion as we lack the sample mean and size for calculation.

Step by step solution

01

State the Hypotheses

In this case, our null hypothesis (H0) is that the mean net weight of the cereal boxes is 18 ounces. This can be written in statistical terms as \(H0: \mu = 18\). The alternative hypothesis (Ha), which is what we're testing, is if the mean net weight is less than 18 ounces. In statistical terms, this can be written as \(Ha: \mu < 18\) ounces.
02

Calculate the Standard Error

We know that \(\sigma = .25\) ounce, and the sample size (n) is large. The standard error (SE) is given by \(\sigma/\sqrt{n}\). Without a provided sample size, we'll use 'n' in the equation for now.
03

Conduct the Z-Test

The test statistic for a z-test is calculated as \(Z= (X - \mu) / SE\) where X is the sample mean. In our case, we don't have a sample mean, so we'll use 'X' in the equation for now.
04

Find the p-value

After calculating the Z-score, refer to a Z-table or use a statistical calculator to find the p-value associated with that Z-score.
05

Make a Conclusion

If the p-value is less than the chosen significance level (usually 0.05), reject the null hypothesis. This would indicate that the mean weight of the cereal boxes is less than 18 ounces, which would contradict the company's claim. If the p-value is higher than the significance level, do not reject the null hypothesis. The result isn't convincing enough to say the mean weight isn't 18 ounces.

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Most popular questions from this chapter

The Environmental Protection Agency (EPA) recommends that the sodium content in public water supplies should be no more than \(20 \mathrm{mg}\) per liter (http://www.disabledworld.com/artman/publish/ sodiumwatersupply.shtml). Forty samples were taken from a large reservoir, and the amount of sodium in each sample was measured. The sample average was \(23.5 \mathrm{mg}\) per liter. Assume that the population standard deviation is \(5.6 \mathrm{mg}\) per liter. The EPA is interested in knowing whether the average sodium content for the entire reservoir exceeds the recommended level. If so, the communities served by the reservoir will have to be made aware of the violation. a. Find the \(p\) -value for the test of hypothesis. Based on this \(p\) -value, would the communities need to be informed of an excessive average sodium level if the maximum probability of a Type I error is to be \(.05 ?\) What if the maximum probability of a Type I error is to be \(.01 ?\) b. Test the hypothesis of part a using the critical-value approach and \(\alpha=.05 .\) Would the communities need to be notified? What if \(\alpha=.01 ?\) What if \(\alpha\) is zero?

Shulman Steel Corporation makes bearings that are supplied to other companies. One of the machines makes bearings that are supposed to have a diamcter of 4 inches. The bearings that have a diameter of either more or less than 4 inches are considered defective and are discarded. When working properly, the machine does not produce more than \(7 \%\) of bearings that are defective. The quality control inspector selects a sample of 200 bearings each week and inspects them for the size of their diameters. Using the sample proportion, the quality control inspector tests the null hypothesis \(p \leq .07\) against the alternative hypothesis \(p \geq\) 07, where \(p\) is the proportion of bearings that are defective. He always uses a \(2 \%\) significance level. If the null hypothesis is rejected, the machine is stopped to make any necessary adjustments. One sample of 200 bearings taken recently contained 22 defective bearings. a. Using the \(2 \%\) significance level, will you conclude that the machine should be stopped to make necessary adjustments? b. Perform the test of part a using a \(1 \%\) significance level. Is your decision different from the one in part a?

The manufacturer of a certain brand of auto batteries claims that the mean life of these batteries is 45 months. A consumer protection agency that wants to check this claim took a random sample of 24 such batteries and found that the mean life for this sample is \(43.05\) months. The lives of all such batteries have a normal distribution with the population standard deviation of \(4.5\) months. a. Find the \(p\) -value for the test of hypothesis with the alternative hypothesis that the mean life of these batteries is less than 45 months. Will you reject the null hypothesis at \(\alpha=.025\) ? b. Test the hypothesis of part a using the critical-value approach and \(\alpha=.025\).

A real estate agent claims that the mean living area of all single-family homes in his county is at most 2400 square feet. A random sample of 50 such homes selected from this county produced the mean living area of 2540 square feet and a standard deviation of 472 square feet. a. Using \(\alpha=.05\), can you conclude that the real estate agent's claim is true? What will your conclusion be if \(\alpha=.01 ?\)

Consider \(H_{0}: p=.70\) versus \(H_{1}: p \neq .70 .\) 1\. A random sample of 600 observations produced a sample proportion equal to .68. Using \(\alpha=.01\), would you reject the null hypothesis? b. Another random sample of 600 observations taken from the same population produced a sample proportion equal to .76. Using \(\alpha=.01\), would you reject the null hypothesis?
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