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The five categories are equally likely to occur, and the category counts are shown in the table. Category Observed Count 49 1 2 65 3 76 State the null and alternative hypotheses. OH: At least one p, is different from 0. Ha: P₁ P₂ = P3 = P4 = P5 = 0 O Ho: P₁ = P₂ = P3 = P4 = P5 = 1 H: At least one p, is different from 1. OH: At least one p, is different from ¹/1/ O Ho: P₁ = P₂ = P3 = P4 = P5 = - =/-/- We are asked to determine whether one or more categories are preferred over another. Our assumption for the null hypothesis will be that the categories occur with equal frequency and the alternative will be that there is a different probability for one or more categories. Enter the probability below as a fraction. P(Category 1) = P(Category 2) == P(Category 5) = H₂: P₁ = P₂ = P3 = P₁ = P5 = = = O Ho: P1 = P₂ = P3 = P4=P5 = 0 H: At least one p, is different from 0. H: At least one p, is different from 4 To determine the probability that one category would occur, we will use the given fact that the five categories are equally likely. We can translate this statement into a probability for the occurrence of each category. Much like a fair six-sided die where there are six possible outcomes with equal probability of occurrence, these five categories translate into five possible outcomes with equal probability. Thus, according to our null hypothesis assumption, the responses are classified with a one in five chance of being placed into any one category. 49 ²/3. 5 61

The five categories are equally likely to occur, and the category counts are shown in the table. Category Observed Count 49 1 2 65 3 76 State the null and alternative hypotheses. OH: At least one p, is different from 0. Ha: P₁ P₂ = P3 = P4 = P5 = 0 O Ho: P₁ = P₂ = P3 = P4 = P5 = 1 H: At least one p, is different from 1. OH: At least one p, is different from ¹/1/ O Ho: P₁ = P₂ = P3 = P4 = P5 = - =/-/- We are asked to determine whether one or more categories are preferred over another. Our assumption for the null hypothesis will be that the categories occur with equal frequency and the alternative will be that there is a different probability for one or more categories. Enter the probability below as a fraction. P(Category 1) = P(Category 2) == P(Category 5) = H₂: P₁ = P₂ = P3 = P₁ = P5 = = = O Ho: P1 = P₂ = P3 = P4=P5 = 0 H: At least one p, is different from 0. H: At least one p, is different from 4 To determine the probability that one category would occur, we will use the given fact that the five categories are equally likely. We can translate this statement into a probability for the occurrence of each category. Much like a fair six-sided die where there are six possible outcomes with equal probability of occurrence, these five categories translate into five possible outcomes with equal probability. Thus, according to our null hypothesis assumption, the responses are classified with a one in five chance of being placed into any one category. 49 ²/3. 5 61

College Algebra
College Algebra
10th Edition
ISBN:9781337282291
Author:Ron Larson
Publisher:Ron Larson
Chapter8: Sequences, Series,and Probability
Section8.7: Probability
Problem 4ECP: Show that the probability of drawing a club at random from a standard deck of 52 playing cards is...
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The five categories are equally likely to occur, and the category counts are shown in the table Category 1 Observed Count 49 2 65 3 O Ho: P₁ P₂ P3 = P4 = P5 = 1 76 We are asked to determine whether one or more categories are preferred over another. Our assumption for the null hypothesis will be that the categories occur with equal frequency and the alternative will be that there is a different probability for one or more categories. State the null and alternative hypotheses. O Ho: At least one p, is different from 0. Ha: P₁ P₂ P3 = P4 = P5 = 0 To determine the probability that one category would occur, we will use the given fact that the five categories are equally likely. We can translate this statement into a probability for the occurrence of each category. Much like a fair six-sided die where there are six possible outcomes with equal probability of occurrence, these five categories translate into five possible outcomes with equal probability. Thus, according to our null hypothesis assumption, the responses are classified with a one in five chance of being placed into any one category. Enter the probability below as a fraction. P(Category 1) = P(Category 2) == P(Category 5) = Ha: P₁ P₂ P3 P4 P5 = O Ho: P₁ = P₂ = P3 = P4 = P5 H: At least one p, is different from 1. O Ho: At least one p, is different from 1 5 O Ho: P₁ = P₂ = P3 = P4 = P5 = 0 H: At least one p, is different from 0. H: At least one p, is different from 4 1 5 5 49 61 //
Transcribed Image Text:The five categories are equally likely to occur, and the category counts are shown in the table Category 1 Observed Count 49 2 65 3 O Ho: P₁ P₂ P3 = P4 = P5 = 1 76 We are asked to determine whether one or more categories are preferred over another. Our assumption for the null hypothesis will be that the categories occur with equal frequency and the alternative will be that there is a different probability for one or more categories. State the null and alternative hypotheses. O Ho: At least one p, is different from 0. Ha: P₁ P₂ P3 = P4 = P5 = 0 To determine the probability that one category would occur, we will use the given fact that the five categories are equally likely. We can translate this statement into a probability for the occurrence of each category. Much like a fair six-sided die where there are six possible outcomes with equal probability of occurrence, these five categories translate into five possible outcomes with equal probability. Thus, according to our null hypothesis assumption, the responses are classified with a one in five chance of being placed into any one category. Enter the probability below as a fraction. P(Category 1) = P(Category 2) == P(Category 5) = Ha: P₁ P₂ P3 P4 P5 = O Ho: P₁ = P₂ = P3 = P4 = P5 H: At least one p, is different from 1. O Ho: At least one p, is different from 1 5 O Ho: P₁ = P₂ = P3 = P4 = P5 = 0 H: At least one p, is different from 0. H: At least one p, is different from 4 1 5 5 49 61 //
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