Question

Data 4.3 on page 265 introduces a situation in which a restaurant chain is measuring the levels of arsenic in chicken from its suppliers. The question is whether there is evidence that the mean level of arsenic is greater than \(80 \mathrm{ppb},\) so we are testing \(H_{0}: \mu=80\) vs \(H_{a}:\) \(\mu>80,\) where \(\mu\) represents the average level of arsenic in all chicken from a certain supplier. It takes money and time to test for arsenic, so samples are often small. Suppose \(n=6\) chickens from one supplier are tested, and the levels of arsenic (in ppb) are: \(\begin{array}{llllll}68, & 75, & 81, & 93, & 95, & 134\end{array}\) (a) What is the sample mean for the data? (b) Translate the original sample data by the appropriate amount to create a new dataset in which the null hypothesis is true. How do the sample size and standard deviation of this new dataset compare to the sample size and standard deviation of the original dataset? (c) Write the six new data values from part (b) on six cards. Sample from these cards with replacement to generate one randomization sample. (Select a card at random, record the value, put it back, select another at random, until you have a sample of size \(6,\) to match the original sample size.) List the values in the sample and give the sample mean. (d) Generate 9 more simulated samples, for a total of 10 samples for a randomization distribution. Give the sample mean in each case and create a small dotplot. Use an arrow to locate the original sample mean on your dotplot.

Short answer

The exercise first requires the calculation of the original sample mean, which equals to 91. A hypothesis arbitrary set to 80 is examined by creating new data sets and generating repeated random samples. The variations in these means are then displayed on a dotplot where the original sample mean is indicated.

Step-by-step solution

Calculating Sample Mean

First, calculate the sample mean by adding all the given values and dividing by the total number of values, which is 6 in this case. The data provided includes the arsenic levels of 6 chickens: 68, 75, 81, 93, 95, 134. Hence, the sample mean \(\bar{x} = \frac{(68 + 75 + 81 + 93 + 95 + 134)}{6} = 91\).

Generate New Dataset

Translate the original sample data to create a new dataset where the null hypothesis is true (\(\mu = 80\)). This implies subtracting the original sample mean (91) from each data point and then adding the hypothesized mean (80). The new dataset then becomes: 57, 64, 70, 82, 84, 123. The sample size remains the same (n=6) as we are not adding or removing any data points. The standard deviation remains the same because shifting data (adding or subtracting a constant from each data point) does not affect the spread of data.

Generate Randomization Sample

For a randomization sample, select 6 data points from the new dataset with replacement (meaning a value can be picked more than once). Let's say the randomly selected observations are 57, 70, 123, 70, 84, 123. Then, calculate the sample mean of this random sample.

Repeat Random Sampling

Repeat step 3 nine more times to have a total of ten samples for a randomization distribution. For each of these nine new samples, calculate the sample mean.

Create a Dotplot

A dotplot can be used to visually display the variations in the sample means for the ten random samples. On this dotplot, use an arrow to locate the original sample mean (91).

Key concepts

Randomization Test

Understanding the randomization test is crucial for anyone grappling with statistical hypothesis testing. In essence, a randomization test, also known as a permutation test, is a non-parametric method used to test the hypothesis. This technique involves creating numerous simulated samples by randomly reshuffling your dataset. By comparing these shuffled samples against the original data, the randomization test evaluates if the observed data might occur by chance.

Imagine shuffling a deck of cards representing your data points, where the null hypothesis holds true—every shuffle leads to a new possible world under the null hypothesis. By comparing sufficient shuffles against your actual data, you get a sense of whether your original dataset stands out as unusual, or if it's just another possible outcome. This approach doesn't assume a particular distribution, making it versatile and robust, especially with small sample sizes or non-normal data as seen in the chicken arsenic levels case.

Sample Mean Calculation

The sample mean, symbolized as \(\bar{x}\), is a central concept in statistics and represents the average of a set of observations. It is calculated by summing all the values in a sample and dividing by the number of observations in that sample. For instance, with the chicken supplier's arsenic levels, we calculate the sample mean by adding \(68 + 75 + 81 + 93 + 95 + 134\) and dividing by 6, leading to a sample mean of 91 ppb.

This calculation is a straightforward but fundamental step in many statistical analyses, including hypothesis testing. The sample mean serves as the observed estimate of the population mean \(\mu\), which is under scrutiny when testing the null hypothesis \(H_0: \mu = 80\) vs \(H_a: \mu > 80\). The sample mean acts as a pivot point from which the randomization test will generate simulated samples to evaluate the null hypothesis.

Standard Deviation

Standard deviation is a measure of the dispersion or spread of a set of data points in a sample. It reveals how much variation there is from the average (mean). A low standard deviation means that data points are generally close to the mean, while a high standard deviation indicates that the data points are spread out over a wider range of values.

When we manipulate a dataset by adding or subtracting a constant to implement the null hypothesis, this does not affect the standard deviation. The operation shifts all data points uniformly, maintaining their relative distances from each other. Therefore, in the arsenic level example, even after adjusting the data to reflect the null hypothesis, the spread of the original data (as measured by the standard deviation) stays constant. This consistency is key in our statistical toolkit, ensuring that the variability observed in our sample is faithfully represented throughout the randomization test.

Null Hypothesis

At the heart of hypothesis testing lies the null hypothesis, denoted as \(H_0\). It is a statement of no effect or no difference that serves as a starting point for statistical significance testing. In our example, the null hypothesis posits that the mean level of arsenic in the chicken sample is 80 ppb (\(H_0: \mu = 80\)).

The null hypothesis is vital because it allows us to calculate the probability of observing a test statistic at least as extreme as the one we observed, given that the null hypothesis is true. If this probability (p-value) is very low, we have evidence against \(H_0\) and may reject it in favor of the alternative hypothesis, \(H_a\), suggesting a new effect or difference. In the context of the chicken arsenic levels, if the sample mean calculation is significantly higher than 80 ppb, we would consider the evidence against the null hypothesis that the mean arsenic level is not greater than 80 ppb.