**Understanding Probabilities and Unusual Sample Means**Welcome to our educational resource. In this section, we will guide you in assessing probabilities and determining the unusualness of sample means in a given data set.### Example Problem#### Given Data:- **Population Mean (μ):** 12,750- **Population Standard Deviation (σ):** 1.8- **Sample Size (n):** 38#### Problem Statement:Find the probability that a sample mean is either less than 12,750 or greater than 12,753 and determine if this range is considered unusual.### Steps to Solve:1. **Calculate the Standard Error of the Mean (SEM):**\[ \text{SEM} = \frac{\sigma}{\sqrt{n}} = \frac{1.8}{\sqrt{38}} \]2. **Determine Z-Scores:** - For \( x = 12,750 \): \[ Z = \frac{12,750 - 12,750}{\text{SEM}} = 0 \] - For \( x = 12,753 \): \[ Z = \frac{12,753 - 12,750}{\text{SEM}} \]3. **Find Probabilities:** - Use the Z-scores to find corresponding probabilities from the standard normal distribution table or using technology tools. - Sum the probabilities for \( x < 12,750 \) and \( x > 12,753 \).4. **Determination of Unusualness:** - Compare the calculated probability with the threshold (0.05). - Based on this comparison, decide if the sample mean falls into the unusual range.### Calculated Probability:\[ \text{Probability} = \]_Round your answer to four decimal places as a precise probability measurement is required._### Decision:Would the given sample mean be considered unusual?**Options:**A. The sample mean would not be considered unusual because there is a probability greater than 0.05 of the sample mean being within this range.B. The sample mean would be considered unusual because there is a probability greater than 0.05 of the sample mean being within this range.C. The sample mean would not be considered unusual because there is a probability less than 0.05 of the sample mean being within this range.D. The sample mean would

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Description: **Understanding Probabilities and Unusual Sample Means**Welcome to our educational resource. In this section, we will guide you in assessing probabilities and determining the unusualness of sample means in a given data set.### Example Problem#### Gi

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d to four decimal places as needed.) the given sample mean be considered unusual? The sample mean would not be considered unusual because there is a probability greater than 0.05 of the sample mean beir The sample mean would be considered unusual because there is a probability greater than 0.05 of the sample mean being w The sample mean would not be considered unusual because there is a probability less than 0.05 of the sample mean being

d to four decimal places as needed.) the given sample mean be considered unusual? The sample mean would not be considered unusual because there is a probability greater than 0.05 of the sample mean beir The sample mean would be considered unusual because there is a probability greater than 0.05 of the sample mean being w The sample mean would not be considered unusual because there is a probability less than 0.05 of the sample mean being

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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In this problem, assume that the distribution of differences is approximately normal. Note: For degrees of freedom d.f. not in the Student's t table, use the closest d.f. that is smaller. In some situations, this choice of d.f. may increase the P-value by a small amount and therefore produce a slightly more "conservative" answer.
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In anticipation of an upcoming election, officials in Rising Falls County are looking at the distance from each voter's home to that voter's nearest polling station. Assume that the population of all such distances for voters in Rising Falls County is approximately normally distributed. An article for the newspaper Politically Sound claimed that the mean of this population is 5.15 km. You want to test this claim, so you select a random sample of 23 Rising Falls County voters, and for each you record the distance the voter lives from their nearest polling station. Follow the steps below to construct a 90% confidence interval for the population mean of all the distances voters in Rising Falls County live from their nearest polling station. Then state whether the confidence interval you construct contradicts the reporter's claim. (If necessary, consult a list of formulas.) (a) Click on "Take Sample" to see the results for your random sample. Take Sample Sample size: П Point estimate: 0…
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In a city with three high schools, all the ninth graders took a Standardized Test, with these results: Number of ninth graders (Frequency) Mean Test Score High School Glenwood 82 262 Central 92 311 Lincoln 67 168 The city's PR manager, who never took statistics, claimed the mean score of all ninth graders in the city was the average of 82, 92, and 67, whic incorrect. What is the mean score for all ninth graders in the city? Round to one decimal place 80.3. Of course, that is Mean of all ninth grader's scores *
B Choosing a topic Sk.. For data that is not normally distributed we can't use z-scores. However, there is an equation that works on any distribution. It's called Chebyshev's formula. The formula 1 where p is the minimum percentage of scores that fall within k k2 is p = 1 - standard deviations on both sides of the mean. Use this formula to answer the following questions. a) If you have scores and you don't know if they are normally distributed, find the minimum percentage of scores that fall within 2.2 standard deviations on both sides of the mean? b) If you have scores that are normally distributed, find the percentage of scores that fall within 2.2 standard deviations on both sides of the mean? c) If you have scores and you don't know if they are normally distributed, how many standard deviations on both sides of the mean do we need to go to have 64 percent of the scores? Note: To answer part c you will need to solve the equation for k.
The average number of cavities that thirty-year-old Americans have had in their lifetimes is 4. Do twenty-year-olds have a different number of cavities? The data show the results of a survey of 11 twenty- vear-olds who were asked how many cavities they have had. Assume that the distribution of the population is normal. 2, 5, 3, 2, 5, 3, 5, 2, 2, 2, 4 What can be concluded at the a = 0.05 level of significance? a. For this study, we should use Select an answer b. The null and alternative hypotheses would be: ns Ho: ? vSelect an answer v re H: ? v Select an answer c. The test statistic 2 v = (please show your answer to 3 decimal places.) urse d. The p-value = e. The p-value is ? va f. Based on this, we should Select an answer v the null hypothesis. g. Thus, the final conclusion is that... (Please show your answer to 4 decimal places.) O The data suggest that the population mean number of cavities for twenty-year-olds is not significantly different from 4 at a = 0.05, so there is…
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