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 (in dpm, disintegrations per minute) was then determined for each culture, yielding the following observations: \(\begin{array}{llllllll}7251 & 6871 & 9632 & 6866 & 9094 & 5849 & 8957 & 7978 \\\ 7064 & 7494 & 7883 & 8178 & 7523 & 8724 & 7468 & \end{array}\) Suppose that it is known that the mean rate of uptake for cultures without nitrates is 8000 . Do the data suggest that the addition of nitrates results in a decrease in the mean rate of uptake? Test the appropriate hypotheses using a significance level of .10 .

Short answer

To check whether the addition of nitrates decreases the mean rate of uptake, we conduct a one-sample t-test. After calculating the sample mean, standard deviation, the t-value and looking up the critical value for the chosen significance level, we compare these two values. If the t-value is greater than the critical t-value, we reject the null hypothesis and conclude that the addition of nitrates decreases the mean rate of uptake.

Step-by-step solution

Formulate Hypotheses

Set up the null hypothesis (H0) and the alternative hypothesis (HA): \nH0: μ ≥ 8000 \nHA: μ < 8000 \nwhere μ is the population mean rate of uptake with nitrates.

Calculate Sample Statistics

Count the number of observations (n), calculate the sample mean (\(\bar{x}\)) and the sample standard deviation (s).

Calculate Test Statistic

Calculate the t-test statistic using the formula: \n\(t = \frac{\bar{x} - μ}{s / \sqrt{n}}\) \nWhere: \n\(\bar{x}\) is the sample mean \nμ is the population mean hypothesized under the null hypothesis \ns is the sample standard deviation \nn is the number of observations.

Find the Critical Value

Find the critical value for t from the t-distribution table given the significance level (0.10) and the degrees of freedom (n-1).

Make a Decision

Compare the t statistic with the critical value. If the absolute value of t is greater than the critical value, reject the null hypothesis and conclude that the mean rate of uptake is less than 8000.

Key concepts

Understanding the T-Test

The t-test is a fundamental statistical analysis method used to determine if there is a significant difference between the means of two groups. In the context of our exercise, we use a one-sample t-test. This helps us analyze whether the sample mean from a nitrate-treated group significantly differs from a known value, which in this case is the mean rate of uptake for cultures without nitrates.
A t-test accomplishes this by comparing the calculated t value from our sample with a critical value obtained from t-distribution tables. The outcome informs us whether the observed sample results are due to random variation or a significant effect, which might suggest that nitrates impact the bacteria culture's mean rate of uptake.
Performing a t-test involves several steps, including calculating the sample mean, standard deviation, and test statistic, which we will delve into in each subsequent section.

Exploring the Null Hypothesis

The null hypothesis, represented as \(H_0\), serves as the default or baseline assumption in hypothesis testing. In our exercise, the null hypothesis states that the mean rate of uptake with nitrates is at least 8000 disintegrations per minute (dpm).

• Symbolically, this is shown as \(H_0: \mu \geq 8000\).
• The null hypothesis represents the idea that any differences observed in the sample are due to random chance rather than a true effect.

The null hypothesis stands unless evidence suggests otherwise, which means that statistical tests need strong evidence to reject \(H_0\). This rigorous approach prevents false positives, maintaining the integrity of conclusions drawn from the data.

Alternative Hypothesis Explained

In contrast to the null hypothesis, the alternative hypothesis (\(H_A\)) represents the outcome of interest that researchers aim to support. In this exercise:

• The alternative hypothesis proposes that the nitrates indeed cause a decrease, hence \(H_A: \mu < 8000\).

It's important to note that the decision to accept \(H_A\) suggests that external factors, like the nitrates, contribute to a lower mean rate of uptake than expected. The formulation of this hypothesis aligns with the study's objective—testing whether nitrate addition negatively affects bacterial uptake rate. With evidence favoring \(H_A\), scientists can conclude there is an actual decrease in uptake rates when nitrates are introduced.

Understanding Significance Level

A significance level, often denoted as \(\alpha\), is a threshold for determining when to reject the null hypothesis. It represents the probability of committing a Type I error—rejecting \(H_0\) when it's actually true.

• In our study, we use a 0.10 significance level.
• This means there's a 10% risk of incorrectly rejecting the null hypothesis.

Choosing the right significance level is crucial: if it's too high, you risk false discoveries (Type I errors); too low, and you might miss real effects (Type II errors). In scientific studies, a common choice is \(\alpha = 0.05\), but in our case, testing with 0.10 makes evaluations more sensitive to detect possible changes in mean uptake due to nitrates.

Concept of Sample Mean

The sample mean (\(\bar{x}\)) is a key statistic that represents the average of all observations in a sample. It's a critical component in calculating the t-test statistic. The formula for the sample mean is: \[ \bar{x} = \frac{1}{n} \sum_{i=1}^{n}x_i \] where \(x_i\) represents each observation, and \(n\) is the total number of observations.
The sample mean provides an estimate of the population mean. In our nitrate study:

• We calculate the sample mean from the recorded bacterial culture uptake rates.
• This calculated mean is leveraged to compare against the known mean of 8000 dpm without nitrates.

Understanding the sample mean is essential as it underpins the t-test, helping determine whether the observed data suggest a statistically significant effect of the nitrates on bacterial cultures.