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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 \(\$ 3.084\) ?

Short Answer

Expert verified
To provide the final answer, compute the t-value, find the corresponding critical value, and make a comparison. The answer depends on the outcome of these calculations.

Step by step solution

01

State the Hypotheses

The null hypothesis \((H_0)\) is that the current average price of milk is the same as the April 2009 average (mean = \$3.084). The alternative hypothesis \((H_a)\) is that the current average price is less than the April 2009 average (mean < \$3.084).
02

Calculate the Test Statistic

We calculate the t-score using the formula: \(t = \frac{\bar{x} - \mu }{(s/\sqrt{n})}\), where \(\bar{x}\) is the sample mean, \(\mu\) is the population mean, \(s\) is the standard deviation, and \(n\) is the number of samples. Substituting the given figures, we have \(t = \frac{3.022 - 3.084 }{(.274/\sqrt{80})}\).
03

Determine the Critical Value

The critical value for a one-tailed t-test at the 1% level and 79 degrees of freedom (n-1) can be found from the t-distribution table. Let's call it \(t_{critical}\).
04

Make the Decision

Compare the calculated t-value with the critical value. If \(|t| > |t_{critical}|\), reject the null hypothesis as there is significant evidence that the average price of milk is lower than the April 2009 average.

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Key Concepts

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

t-score
The t-score is a statistic used in hypothesis testing to determine how much a sample mean deviates from the population mean when standard deviation and sample size are known. In essence, it's a ratio reflecting the difference between the observed data and the null hypothesis, considering the variability in the data.

To compute the t-score, we use the formula:
  • \( t = \frac{\bar{x} - \mu}{(s/\sqrt{n})} \)
where:
  • \(\bar{x}\) is the sample mean, which in this case is \\(3.022.
  • \(\mu\) is the population mean, here being the historical mean of \\)3.084.
  • \(s\) represents the standard deviation, given as \$0.274.
  • \(n\) is the number of observations, which is 80 in this scenario.
Plugging in these numbers helps us to conclude logically if the difference in prices is statistically significant, beyond just incidental variations.
null hypothesis
The null hypothesis, denoted as \(H_0\), is a foundational component in hypothesis testing. It presents a statement of no effect or no difference, serving as a starting point for statistical testing.

In this milk price analysis, the null hypothesis states that there is no change in the milk price from the historical average of \$3.084. In symbols, \(H_0: \mu = 3.084\). The role of the null hypothesis is essentially to act as a 'default' assumption that researchers aim to verify or refute through data analysis.

The null hypothesis is presumed true until evidence suggests otherwise. Thus, when testing a hypothesis, you are generally trying to prove that the null hypothesis is false, implying the alternative hypothesis might be true.
alternative hypothesis
The alternative hypothesis, denoted as \(H_a\), provides a statement that proposes a potential difference or effect in a given situation. It is what researchers hope to support through their study.

In the milk price scenario, the alternative hypothesis suggests that the current average price has indeed decreased from the historical average of \$3.084. Expressed mathematically, it is \(H_a: \mu < 3.084\). This indicates a one-tailed t-test, as we're only looking for evidence of a decrease, not simply a change.

The essence of hypothesis testing is to weigh evidence against the null hypothesis to potentially favor the alternative hypothesis. Thus, if data shows the t-score falls in the extreme end of the t-distribution, far enough from the mean, the null hypothesis can be rejected, thereby supporting the alternative.
t-distribution
The t-distribution is a probability distribution that is based on the degree of variability (variance) in smaller sample sizes, notably when the population standard deviation isn't known.

This distribution resembles a normal distribution but has thicker tails, indicating a higher likelihood of values farther from the mean, which becomes important when dealing with sample sizes like that of our 80 milk retailers.

In our particular analysis, this distribution helps us to determine the critical t-value, giving us a benchmark for statistical significance. The properties of the t-distribution are such that as sample size increases, it approaches a normal distribution.

T-distributions are indexed by the degrees of freedom, calculated as the sample size minus one (n-1), which substantially guides the critical value for hypothesis testing.
significance level
In hypothesis testing, the significance level, often denoted as \( \alpha \), represents a threshold probability used to decide whether to reject the null hypothesis. It quantifies the risk of rejecting a true null hypothesis, known as Type I error.

In our case, the chosen significance level is 1% or 0.01, which is quite stringent. This means we need very strong evidence to reject \(H_0\). Essentially, with a 1% level, there's only a 1% probability of concluding that the average milk price has decreased when it hasn't.

Choosing a lower significance level demands compelling evidence, situated in the critical region of the t-distribution, to support claims against the null hypothesis. Thus, setting the significance level is critical as it directly influences the robustness and reliability of the test outcomes.

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Most popular questions from this chapter

Alpha Airlines claims that only \(15 \%\) of its flights arrive more than 10 minutes late. Let \(p\) be the proportion of all of Alpha's flights that arrive more than 10 minutes late. Consider the hypothesis test $$ H_{0}: p \leq .15 \text { versus } H_{1}: p>.15 $$ Suppose we take a random sample of 50 flights by Alpha Airlines and agree to reject \(H_{0}\) if 9 or more of them arrive late. Find the significance level for this test.

Briefly explain the conditions that must hold true to use the \(t\) distribution to make a test of hypothesis about the population mean.

A 2008 study performed by careerbuilder.com entitled \(\mathrm{No}\), Really, Your Excuse is Totally Believable! notes that \(11 \%\) of workers who call in sick do so to catch up on housework. Suppose that in a survey of 675 male workers who have called in sick, 61 did so to have time to catch up on housework. At the \(2 \%\) significance level, can you conclude that the proportion of all male workers who call in sick do so to catch up on housework is different from \(11 \%\) ?

Consider the null hypothesis \(H_{0}: \mu=12.80 .\) A random sample of 58 observations is taken from this population to perform this test. Using \(\alpha=.05\), show the rejection and nonrejection regions on the sampling distribution curve of the sample mean and find the critical value(s) of \(t\) for the following. a. a right-tailed test b. a left-tailed test c. a two-tailed test

At Farmer's Dairy, a machine is set to fill 32 -ounce milk cartons. However, this machine does not put exactly 32 ounces of milk into each carton; the amount varies slightly from carton to carton but has a normal distribution. It is known that when the machine is working properly, the mean net weight of these cartons is 32 ounces. The standard deviation of the milk in all such cartons is always equal to \(.15\) ounce. The quality control inspector at this company takes a sample of 25 such cartons every week, calculates the mean net weight of these cartons, and tests the null hypothesis, \(\mu=32\) ounces, against the alternative hypothesis, \(\mu \neq 32\) ounces. If the null hypothesis is rejected, the machine is stopped and adjusted. A recent sample of 25 such cartons produced a mean net weight of \(31.93\) ounces a. Calculate the \(p\) -value for this test of hypothesis. Based on this \(p\) -value, will the quality control inspector decide to stop the machine and readjust it if she chooses the maximum probability of a Type I error to be \(.01 ?\) What if the maximum probability of a Type I error is .05? b. Test the hypothesis of part a using the critical-value approach and \(\alpha=.01\). Does the machine need to be adjusted? What if \(\alpha=.05 ?\)

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