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Much concern has been expressed in recent years regarding the practice of using nitrates as meat preservatives. In one study involving possible effects of these chemicals, bacteria cultures were grown in a medium containing nitrates. The rate of uptake of radio-labeled amino acid was then determined for each culture, yielding the following observations: \(\begin{array}{llllllll}7251 & 6871 & 9632 & 6866 & 9094 & 5849 & 8957 & 7978\end{array}\) \(\begin{array}{lllllll}7064 & 7494 & 7883 & 8178 & 7523 & 8724 & 7468\end{array}\) Suppose that it is known that the true average uptake for cultures without nitrates is 8000 . Do the data suggest that the addition of nitrates results in a decrease in the true average uptake? Test the appropriate hypotheses using a significance level of \(.10 .\)

Short Answer

Expert verified
To provide a short answer, one needs to calculate sample mean, sample standard deviation, t-statistic, p-value and then make a decision based on comparison between p-value and significance level. The final answer will depend on the computed p-value.

Step by step solution

01

Calculate the Sample Mean and Sample Standard Deviation

First, you need to find out the sample mean (\( \overline{x} \)) and the sample standard deviation (s). The sample mean is the sum of all the observations divided by the number of observations. The sample standard deviation is a measure of how spread out the numbers in the data are. After the calculation, you can find \(\overline{x}\) and s.
02

Formulate the Hypotheses and Perform the Test

The null hypothesis is that the true average uptake equals to 8000, denoted as \( H_0: \mu = 8000 \), and the alternative hypothesis is that the true average uptake is less than 8000, denoted as \( H_a: \mu < 8000 \). To test these hypotheses, calculate the t-statistic using the formula: \( t = \frac{\overline{x} - \mu_0}{s/\sqrt{n}} \). Here, \( \overline{x} \) is sample mean, \( \mu_0 \) is the population mean, s is the sample standard deviation, and n is the size of the sample.
03

Calculate the P-Value and Make a Decision

Once you have the t-statistic, you can find the corresponding p-value from the t-distribution table or using statistical software. Since the test is a one-tailed test to the left, the p-value is the probability that a t random variable is less than the calculated t-statistic. If the p-value is less than the significance level of 0.10, you reject the null hypothesis and infer that the addition of nitrates results in a decrease in the true average uptake. Otherwise, there is not enough evidence to support the claim.

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Key Concepts

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

Sample Mean
Understanding the sample mean is crucial when dealing with statistical analyses. The sample mean, often represented by the symbol \( \overline{x} \), is the average value of a sample set of numbers. It's calculated by adding up all the values and dividing by the number of observations in the sample. For example, if you are analyzing the effect of nitrates on bacteria cultures, you'd sum all observed rates of amino acid uptake and divide by the total number of cultures. This value serves as an estimate of the population mean and forms the basis from which we can make inferences about the population as a whole.

When we calculate the sample mean, we are essentially looking for the center point of our data. This mean gives us an idea of where most of our values lie, providing a measure that can be compared with known standards or other samples to draw conclusions about trends, effects, or differences within the data.
Sample Standard Deviation
The sample standard deviation (often denoted as s) measures the amount of variation or dispersion within a set of sample data. In other words, it tells us how spread out the values are from the sample mean. A lower standard deviation means that the values are closely clustered around the mean, indicating consistency within the sample. Conversely, a high standard deviation implies a wide range of values and, possibly, outliers or anomalies.

To calculate it, you first find the differences between each data point and the sample mean, square these differences, sum them up, and then divide by one less than the sample size to find the variance. Taking the square root of the variance gives you the sample standard deviation. This statistic is essential for gauging the reliability of the sample mean as an estimate of the population mean, which leads us to evaluate our hypotheses with more confidence.
Null Hypothesis
In statistical hypothesis testing, the null hypothesis (designated as \( H_0 \)) is the statement that there is no effect or no difference, and it serves as a starting assumption. For the exercise on nitrates in meat preservatives, the null hypothesis suggests that there is no decrease in the true average uptake of amino acids in bacteria cultures when nitrates are present, setting this benchmark as 8000 units.

The null hypothesis is critical because it sets the stage for statistical testing. By attempting to disprove or fail to disprove it, researchers can make inferences about the probability of an alternative hypothesis (representing an effect or difference) being true. Our analysis then becomes a process of examining whether the data we have is strong enough to reject this null hypothesis.
t-Statistic
The t-statistic is a ratio that compares the difference between the sample mean and the population mean relative to the variation in the sample data. Its calculation is based on your sample data, specifically the sample mean, the population mean hypothesized under the null hypothesis (\( \mu_0 \)), the sample standard deviation, and the sample size \( n \). The formula is \( t = \frac{\overline{x} - \mu_0}{s/\sqrt{n}} \).

The computed t-statistic is then used to assess the plausibility of the null hypothesis. If the t-statistic falls far from zero (either positive or negative, depending on the hypothesis), it suggests that the sample mean is significantly different from the population mean hypothesized under the null hypothesis, indicating that the sample provides sufficient evidence to reject the null hypothesis.
p-Value
The p-value is a probability that measures the evidence against the null hypothesis. It tells us how likely it is to obtain a sample statistic as extreme as the one we got if the null hypothesis were true. In simpler terms, it's the probability of seeing the effect observed in your study (or more extreme) just by chance if there were no real effect.

When the p-value is low, this indicates that such an extreme observed result would be very unusual under the null hypothesis. Consequently, a low p-value, generally below the chosen significance level, suggests that you should reject the null hypothesis in favor of the alternative hypothesis. In the context of our nitrate study, the p-value would help us decide whether the observed decrease in average uptake is likely due to the presence of nitrates or just random chance.
Significance Level
The significance level, often denoted by \( \alpha \), is a threshold set by the researcher to determine when to reject the null hypothesis. It is typically set before any data is viewed and is used as a benchmark for comparing the p-value. Common levels of significance include 0.05, 0.01, or, as in the case of our nitrate study, 0.10.

If the p-value is less than or equal to the significance level, it suggests that the observed data is sufficiently inconsistent with the null hypothesis, and thus, the null hypothesis is rejected. It's important to note, however, that choosing a higher significance level (like 0.10) increases the chance of making a Type I error—rejecting a true null hypothesis, which should be considered when making inferences.

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Most popular questions from this chapter

To determine whether the pipe welds in a nuclear power plant meet specifications, a random sample of welds is selected and tests are conducted on each weld in the sample. Weld strength is measured as the force required to break the weld. Suppose that the specifications state that the mean strength of welds should exceed 100 \(\mathrm{lb} / \mathrm{in} .^{2}\). The inspection team decides to test \(H_{0}: \mu=100\) versus \(H_{a}: \mu>100 .\) Explain why this alternative hypothesis was chosen rather than \(\mu<100\).

Newly purchased automobile tires of a certain type are supposed to be filled to a pressure of 30 psi. Let \(\mu\) denote the true average pressure. Find the \(P\) -value associated with each of the following given \(z\) statistic values for testing \(H_{0}: \mu=30\) versus \(H_{a}: \mu \neq 30\) when \(\sigma\) is known: a. \(2.10\) d. \(1.44\) b. \(-1.75\) e. \(-5.00\) c. \(0.58\)

A random sample of \(n=44\) individuals with a B.S. degree in accounting who started with a Big Eight accounting firm and subsequently changed jobs resulted in a sample mean time to change of \(35.02\) months and a sample standard deviation of \(18.94\) months ("The Debate over Post-Baccalaureate Education: One University's Experience," Issues in Accounting Education [1992]: 18-36). Can it be concluded that the true average time to change exceeds 2 years? Test the appropriate hypotheses using a significance level of \(.01\).

The success of the U.S. census depends on people filling out and returning census forms. Despite extensive advertising, many Americans are skeptical about claims that the Census Bureau will guard the information it collects from other government agencies. In a USA Today poll (March 13,2000 ), only 432 of 1004 adults surveyed said that they believe the Census Bureau when it says the information you give about yourself is kept confidential. Is there convincing evidence that, despite the advertising campaign, fewer than half of U.S. adults believe the Census Bureau will keep information confidential? Use a significance level of \(.01\).

A credit bureau analysis of undergraduate students credit records found that the average number of credit cards in an undergraduate's wallet was \(4.09\) ("Undergraduate Students and Credit Cards in 2004," Nellie Mae, May 2005). It was also reported that in a random sample of 132 undergraduates, the sample mean number of credit cards carried was 2.6. The sample standard deviation was not reported, but for purposes of this exercise, suppose that it was \(1.2\). Is there convincing evidence that the mean number of credit cards that undergraduates report carrying is less than the credit bureau's figure of \(4.09\) ?

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