Let x represent the average annual salary of college and university professors (in thousands of dollars) in the United States. For all colleges and universities in the United States, the population variance of x is approximately ?2 = 47.1. However, a random sample of 18 colleges and universities in Kansas showed that x has a sample variance s2 = 86.1. Use a 5% level of significance to test the claim that the variance for colleges and universities in Kansas is greater than 47.1. Find a 95% confidence interval for the population variance. (a) What is the level of significance? State the null and alternate hypotheses. Ho: ?2 = 47.1; H1: ?2 < 47.1Ho: ?2 = 47.1; H1: ?2 > 47.1 Ho: ?2 = 47.1; H1: ?2 ≠ 47.1Ho: ?2 < 47.1; H1: ?2 = 47.1 (b) Find the value of the chi-square statistic for the sample. (Round your answer to two decimal places.) What are the degrees of freedom? What assumptions are you making about the original distribution? We assume a uniform population distribution.We assume a normal population distribution. We assume a exponential population distribution.We assume a binomial population distribution. (c) Find or estimate the P-value of the sample test statistic. P-value > 0.1000.050 < P-value < 0.100 0.025 < P-value < 0.0500.010 < P-value < 0.0250.005 < P-value < 0.010P-value < 0.005

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

See similar textbooks

Related questions

Question

100%

Let x represent the average annual salary of college and university professors (in thousands of dollars) in the United States. For all colleges and universities in the United States, the population variance of x is approximately ?² = 47.1. However, a random sample of 18 colleges and universities in Kansas showed that x has a sample variance s² = 86.1. Use a 5% level of significance to test the claim that the variance for colleges and universities in Kansas is greater than 47.1. Find a 95% confidence interval for the population variance.

(a) What is the level of significance?

State the null and alternate hypotheses.

H_o: ?² = 47.1; H₁: ?² < 47.1H_o: ?² = 47.1; H₁: ?² > 47.1 H_o: ?² = 47.1; H₁: ?² ≠ 47.1H_o: ?² < 47.1; H₁: ?² = 47.1

(b) Find the value of the chi-square statistic for the sample. (Round your answer to two decimal places.)

What are the degrees of freedom?

What assumptions are you making about the original distribution?

We assume a uniform population distribution.We assume a normal population distribution. We assume a exponential population distribution.We assume a binomial population distribution.

P-value > 0.1000.050 < P-value < 0.100 0.025 < P-value < 0.0500.010 < P-value < 0.0250.005 < P-value < 0.010P-value < 0.005

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Expert Solution