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Chapter 9: Problem 117

More and more people are abandoning national brand products and buying store brand products to save money. The president of a company that produces national brand coffee claims that \(40 \%\) of the people prefer to buy national brand coffee. A random sample of 700 people who buy coffee showed that 259 of them buy national brand coffee. Using \(\alpha=.01\), can you conclude that the percentage of people who buy national brand coffee is different from \(40 \%\) ? Use both approaches to make the test.

Short Answer

Expert verified
No, we can not conclude that the percentage of people who buy national brand coffee is different from 40%. The test results fail to reject the null hypothesis at the 0.01 level of significance, which suggests that the population proportion for those preferring the national brand coffee could be 40%.

Step by step solution

01

State the Null and Alternative Hypotheses

The null hypothesis \(H_0\) would be that the population proportion \(p\) equals the claimed proportion. The alternative hypothesis \(H_A\) is the opposite of the null hypothesis. In this case, \n\n\(H_0 : p = 0.40\) (The proportion of people who prefer national brand coffee is 40%)\n\(H_A : p \neq 0.40\) (The proportion of people who prefer national brand coffee is not equal to 40%).
02

Compute the Sample Proportion

The sample proportion (\(\hat{p}\)) is computed as \(x/n\) where \(x\) is the number of successes (those who prefer national brand coffee) and \(n\) is the total numbers in the sample. So, \(\hat{p} = 259/700 = 0.37\).
03

Compute the Standardized Test Statistic (Z)

The test statistic \(Z\) for testing a hypothesis about a proportion is calculated using the formula: \(Z = (\hat{p} - p_0) / 鈭歔p_0(1-p_0)/n]\), where \(p_0 = 0.40\) (the hypothesized population proportion), and \(n = 700\) (the sample size). Substituting these values, we get \(Z = (0.37 - 0.40) / 鈭歔0.40(0.60)/700]\) which evaluates to \(Z = -1.71\).
04

Determine the Critical Z-values and Make a Decision

Since this is a two-tailed test (we are checking if the proportion is not equal to 40%), and the level of significance \(\alpha = .01\), the critical values for Z from the standard normal distribution table for an \(\alpha/2=.005\) in each tail are -2.58 and +2.58. Since our computed Z value of -1.71 does not lie beyond these critical values, we fail to reject the null hypothesis.
05

Calculate the P-value and Make a Decision

The two-tailed P-value can be found by looking up the Z-value in a standard normal table which represents the probability that a value is found at least as extreme as our test statistic. For \(Z = -1.71\), the P-value is approximately 0.087. As this P-value (0.087) is larger than our significance level of 0.01, we fail to reject the null hypothesis.

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Most popular questions from this chapter

According to the University of Wisconsin Dairy Marketing and Risk Management Program (http:// future.aae.wisc.edu/index.html), the average retail price of 1 gallon of whole milk in the United States for April 2009 was \(\$ 3.084\). A recent random sample of 80 retailers in the United States produced an average milk price of \(\$ 3.022\) per gallon, with a standard deviation of \(\$ .274 .\) Do the data provide significant evidence at the \(1 \%\) level to conclude that the current average price of 1 gallon of milk in the United States is lower than the April 2009 average of \(\$ 3.084\) ?

A company claims that its 8 -ounce low-fat yogurt cups contain, on average, at most 150 calories per cup. A consumer agency wanted to check whether or not this claim is true. A random sample of 10 such cups produced the following data on calories. 147 159 1 153 146 144 161 163 153 143 158 Test at the \(2.5 \%\) significance level whether the company's claim is true. Assume that the numbers of calories for such cups of yogurt produced by this company have an approximately normal distribution.

According to the American Diabetes Association (www.diabetes.org), \(23.1 \%\) of Americans aged 60 years or older had diabetes in 2007. A recent random sample of 200 Americans aged 60 years or older showed that 52 of them have diabetes. Using a \(5 \%\) significance level, perform a test of hypothesis to determine if the current percentage of Americans aged 60 years or older who have diabetes is higher than that in 2007 . Use both the \(p\) -value and the critical-value approaches.

Perform the following tests of hypotheses for data coming from a normal distribution. a. \(H_{0}: \mu=94.80, H_{1}: \mu<94.80, n=12, \quad \bar{x}=92.87, s=5.34, \quad \alpha=.10\) b. \(H_{0}: \mu=18.70, \quad H_{1}: \mu \neq 18.70, n=25, \quad \bar{x}=20.05, s=2.99, \quad \alpha=.05\) c. \(H_{0}: \mu=59, \quad H_{1}: \mu>59, \quad n=7, \quad \bar{x}=59.42, s=.418, \quad \alpha=.01\)

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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