Question

Much concern has been expressed 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: \(\begin{array}{llllll}7,251 & 6,871 & 9,632 & 6,866 & 9,094 & 5,849 \\ 8,957 & 7,978 & 7,064 & 7,494 & 7,883 & 8,178 \\ 7,523 & 8,724 & 7,468 & & & \end{array}\) Suppose that it is known that the true average uptake for cultures without nitrates is \(8,000 .\) Do these data suggest that the addition of nitrates results in a decrease in the mean uptake? Test the appropriate hypotheses using a significance level of 0.10

Short answer

The conclusion regarding whether the addition of nitrates results in a decrease in the mean uptake depends upon the calculated p-value and the obtained t-value, to be compared with the significance level and critical t-value, respectively. A less than 0.10 p-value or a lesser obtained t-value than the critical t-value would result in rejection of the null hypothesis. The exact results can only be determined once the calculations are made.

Step-by-step solution

Formulate the Hypotheses

The null hypothesis (\(H_0\)): The mean uptake with nitrates (\(μ\)) equals 8000. So, \(H_0 : μ = 8000\).\nThe alternative hypothesis (\(H_a\)): The mean uptake with nitrates (\(μ\)) is less than 8000. So, \(H_a : μ < 8000\). The usage of a < sign instead of ≠ implies that this is a one-tailed test.

Compute Sample Mean and Standard Deviation

Compute the sample mean (x̅) and the sample standard deviation (s) using the provided data. x̅ will be the sum of all values divided by the count of values, and s will be the square root of variance, where variance can be obtained by squaring the difference between each value and the mean, totaling these, and dividing by the count of values minus 1.

Compute Test Statistic

Once the sample mean and standard deviation are obtained, calculate the test statistic (t) using the formula: t = (x̅ - μ) / (s / √n), where μ is the value from the null hypothesis (8000), x̅ is the sample mean, s is the standard deviation, and n is the sample size.

Determine the P-value and Make a Decision

After computing the test statistic, refer to the t-distribution table for the critical t-value corresponding to a significance level of 0.10 and the appropriate degrees of freedom (n-1). The p-value can then be calculated. If the p-value is less than the significance level (0.10), reject the null hypothesis in favor of the alternative. An obtained t-value lower than the critical t-value would also lead to rejection of the null hypothesis. If not, do not reject the null hypothesis.

Key concepts

Null and Alternative Hypotheses

Understanding the null and alternative hypotheses is essential to hypothesis testing. The null hypothesis \(H_0\) is a statement that there is no effect or no difference, and it serves as the starting assumption for statistical significance testing. In our exercise, the null hypothesis posits that the mean uptake with nitrates \(\mu\) is equal to 8000.
In contrast, the alternative hypothesis \(H_a\) suggests that there is an effect, or a difference exists. It is what we would consider if the null hypothesis is rejected. Here the alternative hypothesis is that the mean uptake with nitrates \(\mu\) is less than 8000, indicating a decrease due to the nitrates.
It's important to frame these hypothesis correctly as they determine the structure of the test. This particular test is one-tailed because the alternative hypothesis specifically contends a decrease rather than any difference from the null. This affects how we interpret the test statistic and p-value.

Significance Level

The significance level, often denoted by \(\alpha\), is the threshold used to judge whether a test statistic is sufficiently extreme to warrant rejecting the null hypothesis. It represents the probability of making the error of rejecting the null hypothesis when it is true (Type I error).
For the given exercise, a significance level of 0.10 has been chosen, which implies a 10% risk of concluding that a difference exists when there actually is no difference. This is higher than the commonly used levels of 0.05 or 0.01, denoting a less conservative approach to testing and higher tolerance for Type I error. The choice of significance level affects the critical values and the p-value required to reject the null hypothesis.

Test Statistic

The test statistic is a standardized value calculated from sample data during a hypothesis test. It measures how far the sample statistic such as the sample mean deviates from the null hypothesis. In this exercise, we use a t-test statistic calculated as \(t = (\bar{x} - \mu) / (s / \sqrt{n})\), where \(\bar{x}\) is the sample mean, \(s\) is the sample standard deviation, and \(n\) is the sample size.
This calculated t-value then helps us determine whether the observed data are statistically significantly different from the null hypothesis. It's essentially the signal-to-noise ratio; a larger absolute value of the test statistic indicates a greater difference between the sample statistic and the null hypothesis.

P-value

The p-value is the probability of obtaining test results at least as extreme as the observed data, assuming that the null hypothesis is true. Put simply, it quantifies how extreme the observed data are. If this p-value is smaller than the significance level, the null hypothesis is rejected. In the context of our exercise, the p-value would tell us the probability of observing a mean uptake less than or equal to the one observed if the true mean uptake were actually 8000 (as stated in the null hypothesis).
A low p-value in this case (below 0.10) suggests that the nitrates do indeed have a significant effect on the mean uptake, leading to the rejection of the null hypothesis. The careful interpretation of the p-value, in conjunction with the significance level, is crucial for drawing a valid conclusion from the hypothesis test.