/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 128 Arsenic in Chicken Data 4.5 on p... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Does Massage Really Help Reduce Inflammation in Muscles? In Exercise 4.132 on page \(279,\) we learn that massage helps reduce levels of the inflammatory cytokine interleukin-6 in muscles when muscle tissue is tested 2.5 hours after massage. The results were significant at the \(5 \%\) level. However, the authors of the study actually performed 42 different tests: They tested for significance with 21 different compounds in muscles and at two different times (right after the massage and 2.5 hours after). (a) Given this new information, should we have less confidence in the one result described in the earlier exercise? Why? (b) Sixteen of the tests done by the authors involved measuring the effects of massage on muscle metabolites. None of these tests were significant. Do you think massage affects muscle metabolites? (c) Eight of the tests done by the authors (including the one described in the earlier exercise) involved measuring the effects of massage on inflammation in the muscle. Four of these tests were significant. Do you think it is safe to conclude that massage really does reduce inflammation?

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?

Describe tests we might conduct based on Data 2.3 , introduced on page \(66 .\) This dataset, stored in ICUAdmissions, contains information about a sample of patients admitted to a hospital Intensive Care Unit (ICU). For each of the research questions below, define any relevant parameters and state the appropriate null and alternative hypotheses. Is the average age of ICU patients at this hospital greater than \(50 ?\)

Multiple Sclerosis and Sunlight It is believed that sunlight offers some protection against multiple sclerosis (MS) since the disease is rare near the equator and more prevalent at high latitudes. What is it about sunlight that offers this protection? To find out, researchers \({ }^{15}\) injected mice with proteins that induce a condition in mice comparable to MS in humans. The control mice got only the injection, while a second group of mice were exposed to UV light before and after the injection, and a third group of mice received vitamin D supplements before and after the injection. In the test comparing UV light to the control group, evidence was found that the mice exposed to UV suppressed the MS-like disease significantly better than the control mice. In the test comparing mice getting vitamin D supplements to the control group, the mice given the vitamin \(\mathrm{D}\) did not fare significantly better than the control group. If the p-values for the two tests are 0.472 and 0.002 , which p-value goes with which test?

Beer and Mosquitoes Does consuming beer attract mosquitoes? A study done in Burkino Faso, Africa, about the spread of malaria investigated the connection between beer consumption and mosquito attraction. \(^{9}\) In the experiment, 25 volunteers consumed a liter of beer while 18 volunteers consumed a liter of water. The volunteers \({ }^{8}\) Bouchard, M., Bellinger, D., Wright, \(\mathrm{R}\), and Weisskopf, M. "Attention-Deficit/Hyperactivity Disorder and Urinary Metabolites of Organophosphate Pesticides," Pediatrics, \(2010 ; 125:\) e1270-e1277. \({ }^{9}\) Lefvre, T., et al., "Beer Consumption Increases Human Attractiveness to Malaria Mosquitoes," PLoS ONE, 2010; 5(3): e9546.were assigned to the two groups randomly. The attractiveness to mosquitoes of each volunteer was tested twice: before the beer or water and after. Mosquitoes were released and caught in traps as they approached the volunteers. For the beer group, the total number of mosquitoes caught in the traps before consumption was 434 and the total was 590 after consumption. For the water group, the total was 337 before and 345 after. (a) Define the relevant parameter(s) and state the null and alternative hypotheses for a test to see if, after consumption, the average number of mosquitoes is higher for the volunteers who drank beer. (b) Compute the average number of mosquitoes per volunteer before consumption for each group and compare the results. Are the two sample means different? Do you expect that this difference is just the result of random chance? (c) Compute the average number of mosquitoes per volunteer after consumption for each group and compare the results. Are the two sample means different? Do you expect that this difference is just the result of random chance? (d) If the difference in part (c) is unlikely to happen by random chance, what can we conclude about beer consumption and mosquitoes? (e) If the difference in part (c) is statistically significant, do we have evidence that beer consumption increases mosquito attraction? Why or why not?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.