### Constructing a 95% Confidence Interval for the Population Mean#### Problem Statement:Use the standard normal distribution or the t-distribution to construct a 95% confidence interval for the population mean. Justify your decision. If neither distribution can be used, explain why.#### Scenario:In a recent season, the population standard deviation of the yards per carry for all running backs was 1.32. The yards per carry of 25 randomly selected running backs are shown below. Assume the yards per carry are normally distributed.| 2.3 | 6.4 | 6.8 | 5.7 ||-----|-----|-----|-----|| 5.8 | 5.9 | 5.3 | 3.3 || 5.7 | 2.2 | 2.2 | 3.9 || 2.9 | 2.6 | 3.9 | 4.8 || 5.8 | 4.6 | 7.1 | 4.3 || 5.8 | 3.2 | 6.9 | 4.6 |#### Question:Which distribution should be used to construct the confidence interval?#### Options:A. Use a normal distribution because σ is known and the data are normally distributed. B. Use a t-distribution because n < 30 and σ is unknown. C. Use a normal distribution because n < 30, the miles per gallon are normally distributed and σ is unknown. D. Use a t-distribution because n < 30 and σ is known. E. Cannot use the standard normal distribution or the t-distribution because σ is unknown, n < 30, and the yards are not normally distributed. ### Analysis:Given the provided data and conditions:- The sample size \( n = 25 \), which is less than 30.- The population standard deviation \( \sigma = 1.32 \) is known.- The data are assumed to be normally distributed.Based on these factors, **Option A** is the most appropriate choice. #### Conclusion:**A. Use a normal distribution because σ is known and the data are normally distributed.** By knowing the population standard deviation and assuming the data are normally distributed, the normal distribution is suitable even when the sample size is small.

It is given that the data of size 25 randomly selected of the yards per carry for all running backs.…

Answered: Which distribution should be used to…

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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The finite population correction is most useful for which of the following? a. It adjusts the estimated mean when the sample size is a large fraction of the population size. b. It adjusts the standard error of the mean when the sample size is a small fraction of the population size. c. It adjusts the estimated mean when the sample size is a small fraction of the population size. d. It adjusts the standard error of the mean when the sample size is a large fraction of the population size.
Q1. Which of these is not a parameter of the population distribution of X? A. Sample Mean of X. B. Variance of X. C. Std. deviation of X. D. The Expected value of X, E(X). O A. A В. Ь О С. с O D. d
Assume y follows theX distribution. Assume c₁ and c₂ satisfy P(Y> c₁) = 0.05 and P(Y < c₂) (b). = 0.05. Then C₁ = (a) and C2 =
C. Compute the test value D. Make a decision
Which of the statement is NOT true of F distributions. a. There is only one F distribution for each pair of degrees of freedom. b. The F distribution is always positively skewed. c. The exact shape of the distribution is going to be determined by two degrees of freedom. d. Under a specific circumstance, the F distribution would be normal.
The percent of fat calories that a person in America consumes each day is normally distributed with a mean of about 36 and a standard deviation of 10. Suppose that one individual is randomly chosen. Let X=percent of fat calories. Round all answers to 4 decimal places if where possible a. What is the distribution of X? X N( b. Find the probability that a randomly selected fat calorie percent is more than 40. c. Find the minimum number for the lower quarter of percent of fat calories.
There are frogs and koi in a pond, and the number of frogs and the number of koi in the pond are independent. Let X represent the number of frogs in any given week, and let Y represent the number of koi in any given week. X has a mean of 28 with a standard deviation of 2.7, and Y has a mean of 15 with a standard deviation of 1.6. Which answer choice correctly calculates and interprets the standard deviation of the difference, D = X - Y? = 1.05; this pond can expect the difference of frogs and koi to vary by approximately 1.05 from the mean. = 1.1; this pond can expect the difference of frogs and koi to vary by approximately 1.1 from the mean. = 3.1; this pond can expect the difference of frogs and koi to vary by approximately 3.1 from the mean. = 13; this pond can expect the difference of frogs and koi to vary by approximately 1.05 from the mean.
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