### Sampling Distribution Analysis: Veterans Day Celebrations in Virginia**Background Context:**The average number of people attending Veterans Day celebrations in Virginia in 2017 is reported to be 182, with a corresponding standard deviation of 37 individuals. Due to a few large celebrations, the data distribution is significantly right-skewed.**Scenario for Analysis:**Consider a random sample consisting of 54 Veterans Day celebrations. We aim to determine the sampling distribution of the mean number of attendees calculated from this sample. Below is a comprehension of the sampling distribution's spread and shape based on this scenario.**Question 24: Spread of the Sampling Distribution**"What is the spread of the sampling distribution?"- 156- 134- 5.035- 4.318- 182- 3.335**Answer Considerations:**The spread (standard error) of the sampling distribution of the sample mean is calculated with the formula: \[ \text{Standard Error (SE)} = \frac{\sigma}{\sqrt{n}} \]where:- $\sigma$ is the population standard deviation (37),- $n$ is the sample size (54).\[ \text{SE} = \frac{37}{\sqrt{54}} ≈ 5.035 \]Hence, the correct answer is **5.035**.**Question 25: Shape of the Sampling Distribution**"What is the shape of the sampling distribution?"- Bimodal- Normal- Slightly Skewed Left- Heavily Skewed Right- Slightly Skewed Right**Answer Considerations:**Given the Central Limit Theorem, for sufficiently large sample sizes, the sampling distribution of the sample mean will approximate a normal distribution, regardless of the population's distribution shape. Here, with a sample size of 54, the shape of the sampling distribution can be approximated to be **Normal**.### Visual Aids*Currently, no diagrams or visual graphs accompany the text. If applicable, consider adding characteristic diagrams demonstrating the skewness of distributions and the transformational effect of large sample sizes towards normality in sampling distributions.*---This analysis aids understanding of key concepts in statistical sampling and distribution for educational purposes.

The question is about sampling dist. Given : Popl. mean no. of people attending all 2017 Veterans…

Answered: Suppose the mean number of people…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Fran is training for her first marathon, and she wants to know if there is a significant difference between the mean number of miles run each week by group runners and individual runners who are training for marathons. She interviews 37 randomly selected people who train in groups, and finds that they run a mean of 47.7 miles per week. Assume that the population standard deviation for group runners is known to be 3.3 miles per week. She also interviews a random sample of 49 people who train on their own and finds that they run a mean of 49.4 miles per week. Assume that the population standard deviation for people who run by themselves is 4.4 miles per week. Test the claim at the 0.10 level of significance. Let group runners training for marathons be Population 1 and let individual runners training for marathons be Population 2. Step 2 of 3 : Compute the value of the test statistic. Round your answer to two decimal places.
Suppose Professor Alpha and Professor Omega each teach Introductory Biology. You need to decide which professor to take the class from and have just completed your Introductory Statistics course. Records obtained from past students indicate that students in Professor Alpha's class have a mean score of 80% with a standard deviation of 5%, while past students in Professor Omega's Class have a mean score of 80% with a standard deviation of 10%. Decide which instructor to take for Introductory Biology using a statistical argument.
A taxi company operates in two different regions, and it is analyzing the prices of rides in those regions. It has compiled the prices (in dollars) of 101 rides in each of the two regions during the last week. The histograms below show the distributions of the two data sets. Each histogram shows price (in dollars) on the horizontal axis and the number of taxi rides on the vertical axis. The means and standard deviations for the data sets are also given. Region A 25- 20+ 15+ 10+ 5+ 1₂ 0.5 5.5 10.5 15.5 20.5 25.5 30.5 35.5 40.5 45.5 50.5 Region A mean: $25.28 Region A standard deviation: $7.95 25 Expl 20- 15+ 10+ neck 5+ 0A 0.5 5.5 5.5 10.5 15.5 20.5 25.5 30.5 35.5 40.5 45.5 50.5 (a) Identify the data set for which it is appropriate to use the Empirical Rule. It is appropriate to use the Empirical Rule for the (Choose one) data set. The company wants to use the Empirical Rule to make some approximations about both data sets. Unfortunately, it is appropriate to use the Empirical Rule on…
A researcher would like to estimate the mean amount of money that a family spends to within $75. If the amount spent by a family is considered to be normally distributed with a standard deviation of $412, what sample size would be necessary to be 93% confident?
Students preparing for comprehensive exams usually spend many hours studying. Suppose it is known that the time students spend studying for comprehensive exams has a distribution that is skewed heavily to the right with a mean of 41.3 hours and a standard deviation of 6.8 hours. If a simple random sample of 72 students is selected and the amount of time each spent studying for the comprehensive exam is determined. State and check the two assumptions.
A certain prescription medicine is supposed to contain an average of 250 parts per million (ppm) of a certain chemical. If the concentration is higher than this, the drug may cause harmful side effects; if it is lower, the drug may be ineffective. The manufacturer runs a check to see if the mean concentration in a large shipment conforms to the target level of 250 ppm or not. A simple random sample of 100 portions is tested, and the sample mean concentration is found to be 247 ppm. The sample concentration standard deviation is s = 12 ppm. What are the appropriate null and alternative hypotheses? Group of answer choices H 0: x̄ = 250 vs. H a: x̄ < 250 H 0: x̄ = 250 vs. H a: x̄ ≠ 250 H 0: x̄ = 250 vs. H a: x̄ > 250 H 0: μ = 250 vs. H a:μ < 250 H 0: μ = 250 vs. H a: μ ≠ 250 H 0: μ = 250 vs. H a: μ > 250
Fran is training for her first marathon, and she wants to know if there is a significant difference between the mean number of miles run each week by group runners and individual runners who are training for marathons. She interviews 42 randomly selected people who train in groups and finds that they run a mean of 47.1 miles per week. Assume that the population standard deviation for group runners is known to be 4.4 miles per week. She also interviews a random sample of 47 people who train on their own and finds that they run a mean of 48.5 miles per week. Assume that the population standard deviation for people who run by themselves is 1.8 miles per week. Test the claim at the 0.01 level of significance. Let group runners training for marathons be Population 1 and let individual runners training for marathons be Population 2. Step 2 of 3 : Compute the value of the test statistic. Round your answer to two decimal places.
Suppose that on a certain section of I-95, with a posted speed limit of 65 miles per hour, the speeds of all vehicles have a bell-shaped distribution with a mean of 72 mph and a standard deviation of 3 mph. Find the percentage of vehicles with the following speeds on this section of I-95: greater than 75 mph.
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An adventure company runs two obstacle courses, Fundash and Coolsprint. The designer of the courses suspects that the mean completion time of Fundash is not equal to the mean completion time of Coolsprint. To test this, she selects 240 Fundash runners and 200 Coolsprint runners. (Consider these as random samples of the Fundash and Coolspring runners.) The 240 Fundash runners complete the course with a mean time of 76.7 minutes and a standard deviation of 5.2 minutes. The 200 Coolsprint runners complete the course with a mean time of 77.5 minutes and a standard deviation of 5.4 minutes. Assume that the population standard deviations of the completion times can be estimated to be the sample standard deviations, since the samples that are used to compute them are quite large. At the 0.05 level of significance, is there enough evidence to support the claim that the mean completion time, μ1 , of Fundash is not equal to the mean completion time, μ2 , of Coolsprint?…
A recent survey reported that 47% of 18- to 29-year-olds in a certain country own tablets. Use binomial distribution. What are the mean and standard deviation of the number of 18 to 29 year old who will own a tablet in a survey of six?
The first experiment I created 200 samples with 10 students in each and took each sample's mean. The mean of this distribution was about 10.7 miles with a standard deviation of 3.7 miles. (See video) Next I did 200 samples with 20 students each, then I did 200 samples with 50 students each and finally I did 200 samples with 100 students each. Watch the following video and answer the following questions about the sampling distributions being created. 1) For the samples with 100 students each, when the sampling is complete, what will the mean be close to or equal to? 2) For the samples with 100 students each, when the sampling is complete, will the standard deviation be smaller than the standard deviation for the samples with 50 students each? 3) What is the mean of the population of students that I am randomly sampling from? 4) What is the name of the important theorem that I am demonstrating right now.?

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