What conclusion would be appropriate for an upper-tailed chi-squared test in each of the following situations?USE SALT(a) a = 0.05, df = 4, x² = 12.75O We would reject Ho.O We would fail to reject Ho.(b) a = 0.01, df = 3, x² = 8.59O We would reject Ho.O We would fail to reject Ho.(c) a = 0.10, df = 2, x² = 4.36We would reject HoO We would fail to reject Ho.(d) a = 0.01, k = 6, x² = 11.40O We would reject Ho.We would fail to reject Ho.

Given Upper tailed chi squared test Note: According to Bartleby guidelines expert solve only one…

Answered: What conclusion would be appropriate…

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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You are conducting a test of independence for the claim that two different populations have the same proportions of the following two characteristics. Here is the sample data. Category A B A Population Population #1 #2 B 9 57 The expected observations for this table would be Population Population Category 21 119 What is the chi-square test-statistic for this data? 2 x² Report all answers accurate to three decimal places.
What conclusion would be appropriate for an upper-tailed chi-squared test in each of the following situations? LUSE SALT (a) a=0.05, df = 4, x² = 11.25 O We would reject Ho O We would fail to reject Ho. (b) a = 0.01, df = 3, x² = 8.52 O We would reject Ho O We would fail to reject Ho (c) a = 0.10, df -2, x² - 4.06 = O We would reject Ho O We would fail to reject Ho (d) a = 0.01, k = 6, x² = 11.80 O We would reject Ho- O We would fail to reject Ho
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The degrees of freedom is critical to computing the p-value for the chi-square goodness of fit. We will discuss the p-value in the next lab. The degrees of freedom for the Chi-Square goodness of fit is equal to (the number of non-zero rows in the table) - 1. The number of non-zero rows is the number of expected values that are non-zero. For example: Phenotype ... Expected ... Disease ... 234 ... Wild-type ... 6743 ... has two rows where the Expected value is non-zero. So the degrees of freedom is 2-1 = 1. Another example: Phenotype ... Expected ... Disease ... 2374 ... Wild-type ... 0 ... has one row where the Expected value is non-zero. So the degrees of freedom is 1-1 = 0. With 0 degrees of freedom, the chi-square test cannot be calculated. A third example (sex-linked): Phenotype ... Expected ... Disease-Male ... 2374 ... Disease-Female ... 9456 Wild-type-Male ... 1001 ... Wild-type-Female ... 235 ... has four rows where the Expected values are…
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You are conducting a test of independence for the daim that there is an association between morning beverage and age. Juice Coffee Pop Tea 5 32 | 43 less than 25 19 18 25 or older 27 30 2 What is the chi-square test-statistic and the corresponding p-value for this data? [three decimal accuracy] p-value = [three decimal accuracy]
What is the critical t score if you are testing a one-tailed hypothesis, α level of .05 with df = 36? a. 1.671 b. 2.026 c. 2.028 d. 1.697
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You are performing a left-tailed test with test statistic z=−2.04, find the p-value to 4 decimal places.Note: You will need to use the z and t sheet to get this answer.
The mean and the variance of a sample of size 16 from normal population are 32 and 8, respectively. For testing the hypotheses H,:0%=46vsH, : 0> # 4.6 thetest statistic is: Select one: O a.22.83 O b.19.57 () ¢.29.35 O d.26.09

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