Let x = age in years of a rural Quebec woman at the time of her first marriage. In the year 1941, the population variance of x was approximately σ2 = 5.1. Suppose a recent study of age at first marriage for a random sample of 51 women in rural Quebec gave a sample variance s2 = 3.4. Use a 5% level of significance to test the claim that the current variance is less than 5.1. (a) What is the level of significance? State the null and alternate hypotheses. H0: σ2 < 5.1; H1: σ2 = 5.1H0: σ2 = 5.1; H1: σ2 < 5.1 H0: σ2 = 5.1; H1: σ2 > 5.1H0: σ2 = 5.1; H1: σ2 ≠ 5.1 (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 exponential population distribution.We assume a normal population distribution. We assume a uniform 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 (d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Since the P-value > α, we fail to reject the null hypothesis.Since the P-value > α, we reject the null hypothesis. Since the P-value ≤ α, we reject the null hypothesis.Since the P-value ≤ α, 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 age at first marriage is less than 5.1.At the 5% level of significance, there is sufficient evidence to conclude that the variance of age at first marriage is less than 5.1.

Name: Let x = age in years of a rural Quebec woman at the time of her first
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Description: Let x = age in years of a rural Quebec woman at the time of her first marriage. In the year 1941, the population variance of x was approximately σ2 = 5.1. Suppose a recent study of age at first marriage for a random sample of 51 women in rural Quebec

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

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Let x = age in years of a rural Quebec woman at the time of her first marriage. In the year 1941, the population variance of x was approximately σ² = 5.1. Suppose a recent study of age at first marriage for a random sample of 51 women in rural Quebec gave a sample variance s² = 3.4. Use a 5% level of significance to test the claim that the current variance is less than 5.1.

(a) What is the level of significance?

State the null and alternate hypotheses.

H₀: σ² < 5.1; H₁: σ² = 5.1H₀: σ² = 5.1; H₁: σ² < 5.1 H₀: σ² = 5.1; H₁: σ² > 5.1H₀: σ² = 5.1; H₁: σ² ≠ 5.1

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

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

Since the P-value > α, we fail to reject the null hypothesis.Since the P-value > α, we reject the null hypothesis. Since the P-value ≤ α, we reject the null hypothesis.Since the P-value ≤ α, 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 age at first marriage is less than 5.1.At the 5% level of significance, there is sufficient evidence to conclude that the variance of age at first marriage is less than 5.1.

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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