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The type of household for the U.S. population and for a random sample of 411 households from a community in Montana are shown below. Type of Household Married with children Married, no children Single parent One person Other (e.g., roommates, siblings) Percent of U.S. Households 26% 29% 9% 25% 11% Observed Number of Households in the Community 90 129 29 98 65 [ L

The type of household for the U.S. population and for a random sample of 411 households from a community in Montana are shown below. Type of Household Married with children Married, no children Single parent One person Other (e.g., roommates, siblings) Percent of U.S. Households 26% 29% 9% 25% 11% Observed Number of Households in the Community 90 129 29 98 65 [ L

MATLAB: An Introduction with Applications
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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7 The type of household for the U.S. population and for a random sample of 411 households from a community in Montana are shown below. Type of Household Married with children Married, no children Single parent One person Other (e.g., roommates, siblings) Percent of U.S. Households 26% 29% 9% 25% 11% Observed Number of Households in the Community 90 129 29 98 65
Transcribed Image Text:7 The type of household for the U.S. population and for a random sample of 411 households from a community in Montana are shown below. Type of Household Married with children Married, no children Single parent One person Other (e.g., roommates, siblings) Percent of U.S. Households 26% 29% 9% 25% 11% Observed Number of Households in the Community 90 129 29 98 65
(b) Find the value of the chi-square statistic for the sample. (Round the expected frequencies to two decimal places. Round the test statistic to three decimal places.) Are all the expected frequencies greater than 5? Yes No What sampling distribution will you use? uniform binomial chi-square Student's What are the degrees of freedom? normal (c) Find or estimate the P-value of the sample test statistic. (Round your answer to three decimal places.) (d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis that the population fits the specified distribution of categories? Since the P-value > x, we fail to reject the null hypothesis. Since the P-value > xx, we reject the null hypothesis. Since the P-values ∞, we reject the null hypothesis. Since the P-values x, we fail to reject the null hypothesis. (e) Interpret your conclusion in the context of the application. At the 5% level of significance, the evidence is sufficient to conclude that the community household distribution does not fit the general U.S. household distribution. At the 5% level of significance, the evidence is insufficient to conclude that the community household distribution does not fit the general U.S. household distribution.
Transcribed Image Text:(b) Find the value of the chi-square statistic for the sample. (Round the expected frequencies to two decimal places. Round the test statistic to three decimal places.) Are all the expected frequencies greater than 5? Yes No What sampling distribution will you use? uniform binomial chi-square Student's What are the degrees of freedom? normal (c) Find or estimate the P-value of the sample test statistic. (Round your answer to three decimal places.) (d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis that the population fits the specified distribution of categories? Since the P-value > x, we fail to reject the null hypothesis. Since the P-value > xx, we reject the null hypothesis. Since the P-values ∞, we reject the null hypothesis. Since the P-values x, we fail to reject the null hypothesis. (e) Interpret your conclusion in the context of the application. At the 5% level of significance, the evidence is sufficient to conclude that the community household distribution does not fit the general U.S. household distribution. At the 5% level of significance, the evidence is insufficient to conclude that the community household distribution does not fit the general U.S. household distribution.
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