A set of solar batteries is used in a research satellite. The satellite can run on only one battery, but it runs best if more than one battery is used. The variance ?2 of lifetimes of these batteries affects the useful lifetime of the satellite before it goes dead. If the variance is too small, all the batteries will tend to die at once. Why? If the variance is too large, the batteries are simply not dependable. Why? Engineers have determined that a variance of ?2 = 23 months (squared) is most desirable for these batteries. A random sample of 26 batteries gave a sample variance of 15.8 months (squared). Using a 0.05 level of significance, test the claim that ?2 = 23 against the claim that ?2 is different from 23. (a) What is the level of significance? State the null and alternate hypotheses. H0: ?2 = 23; H1: ?2 ≠ 23H0: ?2 = 23; H1: ?2 > 23 H0: ?2 > 23; H1: ?2 = 23H0: ?2 = 23; H1: ?2 < 23 (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 binomial population distribution.We assume a normal population distribution. We assume a uniform population distribution.We assume a exponential 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 (d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Since the P-value > ?, we fail to reject the null hypothesis.Since the P-value > ?, we reject the null hypothesis. Since the P-value ≤ ?, we reject the null hypothesis.Since the P-value ≤ ?, we fail to reject the null hypothesis. (e) Interpret your conclusion in the context of the application. At the 5% level of significance, there is insufficient evidence to conclude that the variance of battery life is different from 23. At the 5% level of significance, there is sufficient evidence to conclude that the variance of battery life is different from 23.

A set of solar batteries is used in a research satellite. The satellite can run on only one battery, but it runs best if more than one battery is used. The variance ?2 of lifetimes of these batteries affects the useful lifetime of the satellite before it goes dead. If the variance is too small, all the batteries will tend to die at once. Why? If the variance is too large, the batteries are simply not dependable. Why? Engineers have determined that a variance of ?2 = 23 months (squared) is most desirable for these batteries. A random sample of 26 batteries gave a sample variance of 15.8 months (squared). Using a 0.05 level of significance, test the claim that ?2 = 23 against the claim that ?2 is different from 23. (a) What is the level of significance? State the null and alternate hypotheses. H0: ?2 = 23; H1: ?2 ≠ 23H0: ?2 = 23; H1: ?2 > 23 H0: ?2 > 23; H1: ?2 = 23H0: ?2 = 23; H1: ?2 < 23 (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 binomial population distribution.We assume a normal population distribution. We assume a uniform population distribution.We assume a exponential 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 (d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Since the P-value > ?, we fail to reject the null hypothesis.Since the P-value > ?, we reject the null hypothesis. Since the P-value ≤ ?, we reject the null hypothesis.Since the P-value ≤ ?, we fail to reject the null hypothesis. (e) Interpret your conclusion in the context of the application. At the 5% level of significance, there is insufficient evidence to conclude that the variance of battery life is different from 23. At the 5% level of significance, there is sufficient evidence to conclude that the variance of battery life is different from 23.

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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A set of solar batteries is used in a research satellite. The satellite can run on only one battery, but it runs best if more than one battery is used. The variance ?² of lifetimes of these batteries affects the useful lifetime of the satellite before it goes dead. If the variance is too small, all the batteries will tend to die at once. Why? If the variance is too large, the batteries are simply not dependable. Why? Engineers have determined that a variance of ?² = 23 months (squared) is most desirable for these batteries. A random sample of 26 batteries gave a sample variance of 15.8 months (squared). Using a 0.05 level of significance, test the claim that ?² = 23 against the claim that ?² is different from 23.

(a) What is the level of significance?

State the null and alternate hypotheses.

H₀: ?² = 23; H₁: ?² ≠ 23H₀: ?² = 23; H₁: ?² > 23 H₀: ?² > 23; H₁: ?² = 23H₀: ?² = 23; H₁: ?² < 23

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

(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis?

Since the P-value > ?, we fail to reject the null hypothesis.Since the P-value > ?, we reject the null hypothesis. Since the P-value ≤ ?, we reject the null hypothesis.Since the P-value ≤ ?, we fail to reject the null hypothesis.

(e) Interpret your conclusion in the context of the application.

At the 5% level of significance, there is insufficient evidence to conclude that the variance of battery life is different from 23. At the 5% level of significance, there is sufficient evidence to conclude that the variance of battery life is different from 23.

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.

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