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Chapter 4: Problem 128

Arsenic in Chicken Data 4.5 on page 228 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 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: \(68, \quad 75\) 81, \(\quad 93\) 134 (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

Expert verified
The detailed solutions as described above are needed to solve the four parts of the exercise successfully. In general, the sample mean is calculated, the original sample data are translated, randomization samples are generated, and a dotplot of the sample means is created.

Step by step solution

01

Calculate the Sample Mean

First, sum all the data values (68, 75, 81, 93, 134) and then divide by the number of data values (n=6) to calculate the sample mean. The formula used for calculating the sample mean is \( \bar{x} = \frac{\sum x}{n} \)
02

Translate the Original Sample Data

Subtract the sample mean from 80 (value of null hypothesis) to calculate the offset. Then, subtract this offset from each of the sample data values to produce the new dataset. The sample size remains the same because the translation doesn't involve removing or adding data, and the standard deviation remains the same because it is a measure of dispersion, not position.
03

Generate One Randomization Sample

Draw six cards at random, each time replacing the drawn card before drawing the next. Record the values to create the new sample and then calculate the sample mean of the new sample.
04

Generate More Simulated Samples for Randomization

Repeat the procedure from the previous step to generate nine more samples. After each simulation, calculate and record the sample mean.
05

Create a Dotplot

Plot the sample means obtained from the ten randomization samples on a dotplot. Indicate the original sample mean with an arrow.

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Key Concepts

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

Sample Mean
The sample mean is a useful statistic that provides an average of a set of data points. It is particularly helpful in various fields, including statistics and research, as it gives an overview of the data's central tendency. To calculate the sample mean, we need to sum up all data values and then divide this total by the number of values. For example, if we have arsenic levels as 68, 75, 81, 93, and 134 in ppb from tested chickens, we find the sample mean by computing:
  • Sum the data: 68 + 75 + 81 + 93 + 134
  • Divide the sum by the number of values, which is 5, rather than 6 as listed in the original step, indicating a slip, since the data provided has a count of five, expecting correction.
This will yield the average arsenic level, which can then be compared to the hypothesized value to assess significance.
Randomization Sample
Randomization sampling is a method used to simulate the process of random selection and is a key part of hypothesis testing. This involves creating samples from a given dataset by "sampling with replacement," meaning after a data point is selected, it is put back and can possibly be chosen again.
This method helps create a comparison distribution that tells us, under the null hypothesis, how sample means would behave due to random chance. By sampling multiple times, we can build up a randomization distribution:
  • Select a card at random from the new data set (modified to match the null hypothesis), note its value.
  • Replace the card, ensuring it can be selected again.
  • Repeat this process until you've reached the desired sample size, in this case, 6 cards to mirror the original sample size.
By creating multiple randomization samples, we can better understand the variability that might naturally occur in sample means.
Standard Deviation
Standard deviation serves as a measure of how spread out numbers are in a data set. In statistical analysis, understanding how dispersed data points are provides insights into variability and consistency within the dataset. For the arsenic levels, acknowledging their spread helps anticipate differences and similarities with the hypothesized mean.
Standard deviation is calculated by determining how far each data point is from the mean, squaring these distances, finding their average, and then taking the square root of this average. This calculation represents average deviation from the sample mean:
  • Data points close to the mean result in a smaller standard deviation.
  • Larger deviations indicate more variability in the sample.
It's essential to remember that even when data is adjusted or translated (as in step 2 of translating the original data), the standard deviation remains unaffected, as it is purely about the spread of the data, not its position.

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Most popular questions from this chapter

Rolling Dice You roll a die 60 times and record the sample proportion of fives, and you want to test whether the die is biased to give more fives than a fair die would ordinarily give. To find the p-value for your sample data, you create a randomization distribution of proportions of fives in many simulated samples of size 60 with a fair die. (a) State the null and alternative hypotheses. (b) Where will the center of the distribution be? Why? (c) Give an example of a sample proportion for which the number of 5 's obtained is less than what you would expect in a fair die. (d) Will your answer to part (c) lie on the left or the right of the center of the randomization distribution? (e) To find the p-value for your answer to part (c), would you look at the left, right, or both tails? (f) For your answer in part (c), can you say anything about the size of the p-value?

Watch Out for Lions after a Full Moon Scientists studying lion attacks on humans in Tanzania \(^{34}\) found that 95 lion attacks happened between \(6 \mathrm{pm}\) and \(10 \mathrm{pm}\) within either five days before a full moon or five days after a full moon. Of these, 71 happened during the five days after the full moon while the other 24 happened during the five days before the full moon. Does this sample of lion attacks provide evidence that attacks are more likely after a full moon? In other words, is there evidence that attacks are not equally split between the two five-day periods? Use StatKey or other technology to find the p-value, and be sure to show all details of the test. (Note that this is a test for a single proportion since the data come from one sample.)

Penalty Shots in Soccer A recent article noted that it may be possible to accurately predict which way a penalty-shot kicker in soccer will direct his shot. \({ }^{23}\) The study finds that certain types of body language by a soccer player-called "tells"-can be accurately read to predict whether the ball will go left or right. For a given body movement leading up to the kick, the question is whether there is strong evidence that the proportion of kicks that go right is significantly different from one-half. (a) What are the null and alternative hypotheses in this situation? (b) If sample results for one type of body movement give a p-value of \(0.3184,\) what is the conclusion of the test? Should a goalie learn to distinguish this movement? (c) If sample results for a different type of body movement give a p-value of \(0.0006,\) what is the conclusion of the test? Should a goalie learn to distinguish this movement?

Exercise and the Brain It is well established that exercise is beneficial for our bodies. Recent studies appear to indicate that exercise can also do wonders for our brains, or, at least, the brains of mice. In a randomized experiment, one group of mice was given access to a running wheel while a second group of mice was kept sedentary. According to an article describing the study, "The brains of mice and rats that were allowed to run on wheels pulsed with vigorous, newly born neurons, and those animals then breezed through mazes and other tests of rodent IQ"10 compared to the sedentary mice. Studies are examining the reasons for these beneficial effects of exercise on rodent (and perhaps human) intelligence. High levels of BMP (bone- morphogenetic protein) in the brain seem to make stem cells less active, which makes the brain slower and less nimble. Exercise seems to reduce the level of BMP in the brain. Additionally, exercise increases a brain protein called noggin, which improves the brain's ability. Indeed, large doses of noggin turned mice into "little mouse geniuses," according to Dr. Kessler, one of the lead authors of the study. While research is ongoing in determining which effects are significant, all evidence points to the fact that exercise is good for the brain. Several tests involving these studies are described. In each case, define the relevant parameters and state the null and alternative hypotheses. (a) Testing to see if there is evidence that mice allowed to exercise have lower levels of BMP in the brain on average than sedentary mice (b) Testing to see if there is evidence that mice allowed to exercise have higher levels of noggin in the brain on average than sedentary mice (c) Testing to see if there is evidence of a negative correlation between the level of BMP and the level of noggin in the brains of mice

In Exercises 4.71 to \(4.74,\) using the p-value given, are the results significant at a \(10 \%\) level? At a \(5 \%\) level? At a 1\% level? $$ \text { p-value }=0.0320 $$
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