You are conducting a multinomial hypothesis test for the claim that the 4 categories occur with thefollowing frequencies:Complete the table.CategoryABCD4032ObservedExpectedSquaredFrequency Frequency Residual10Ho: PA21=0.5; PB = 0.3; pc = 0.1; PD = 0.1What is the chi-square test-statistic for this data?x² =Report all answers accurate to three decimal places.

The data set is,CategoryObserved frequencyProportionA400.5B320.3C100.1D210.1

Answered: You are conducting a multinomial…

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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Professor Nord stated that the mean score on the final exam from all the years he has been teaching is a 79%. Colby was in his most recent class, and his class’s mean score on the final exam was 82%. Colby decided to run a hypothesis test to determine if the mean score of his class was significantly greater than the mean score of the population. α = .01. What is the mean score of the population? What is the mean score of the sample? Is this test one-tailed or two-tailed? Why?
You are conducting a multinomial Goodness of Fit hypothesis test for the claim that the 4 categories occur with the following frequencies: H.: PA = 0.4; PB 0.1; Pc = 0.3; PD %3D0.2 Complete the table. Report all answers accurate to three decimal places. Observed Expected Frequency Frequency Category A 50 24 28 34 What is the chi-square test-statistic for this data? (2 decimal places) x = %3| What is the P-Value? (3 decimal places) P-Value = For significance level alpha 0.1, What would be the conclusion of this hypothesis test? O Fail to reject the Null Hypothesis O Reject the Null Hypothesis
You wish to test the following claim (Hå) at a significance level of a Ho: P₁ = P2 Ha: P₁ P2 0.005. You obtain 10.2% successes in a sample of size ni 518 from the first population. You obtain 15.2% successes in a sample of size n₂ 269 from the second population. Round values for X to the nearest whole number. What is the test statistic for this sample? (Round to three decimal places.) test statistic = What is the p-value for this sample? (Round to four decimal places.) p-value =
Which of the following is not needed to compute a t statistic? Question 18 options: A hypothesized value for the population mean The value of the population variance or standard deviation The value of the sample mean The value of the sample variance or standard deviation
Suppose you are interested in testing whether average monthly gasoline expenditure differs from $50 at the 5% significance level. You take a sample of 30 and calculate a sample mean of 51 and sample standard deviation of 12. Given this information the null and alternative hypotheses are as follows: H0: μ=50 H1: μ≠50 What is the conclusion of this test?
Suppose you are interested in testing whether average monthly gasoline expenditure differs from $50 at the 5% significance level. You take a sample of 30 and calculate a sample mean of 51 and sample standard deviation of 12. Given this information the null and alternative hypotheses are as follows: H0: μ=50 H1: μ≠50 What is the p-value for this test? (Round to four decimal places. )
You wish to test the following claim (HaHa) at a significance level of α=0.05α=0.05. Ho:μ=63.7Ho:μ=63.7 Ha:μ<63.7Ha:μ<63.7You believe the population is normally distributed, but you do not know the standard deviation. You obtain the following sample of data: Column A Column B Column C Column D 66.7 56.1 82.1 74.8 60.4 57.5 55.2 52.4 74.8 56.1 83.6 98 49.5 33.7 83.6 44.9 27.3 29.8 39 92.9 42.2 29.8 25.8 42.7 64.5 69.7 58 38.4 76.6 50.9 57.5 69.7 47 43.9 65 67.8 What is the critical value for this test? (Report answer accurate to three decimal places.)critical value = What is the test statistic for this sample? (Report answer accurate to three decimal places.)test statistic =
Question #3 Samples of body temperatures were collected for a group of women and a group of men. The summary statistics are given below: Women:n1 = 15 sample mean = 98.38 ̊F s1 = 0.45 ̊F Men:n2= 91 sample mean= 98.17 ̊F s2 = 0.65 ̊F 1. Use a 0.01 significance level to test the claim that women and men have different mean body temperatures using the p value method. Show all the steps of your hypothesis test including the p value: initial conclusion: a final conclusion about the original claim:
Clinical tests of the allergy drug Seldane yielded the results summarized in the table below. At the 5%5% significance level, test to determine if there is a relationship between the use of the drug and the occurrence of headaches. Seldane Users Placebo Users Control Total Headache 42 56 60 158 No headache 760 682 635 2077 Total 802 738 695 2235 Perform a Chi-squared test in the following steps: Step 1: State the null and alternative hypotheses: H0:H0:There is between the use of the drug Seldane and the occurrence of headaches. H1:H1: There is between the use of the drug Seldane and the occurrence of headaches. Step 2: Find the critical value(s): State αα-value (in decimals) with the degree(s) of freedom, separated by a comma. State the critical value(s): CV= (Do not round the table value.) Step 3. State the rejection region: Which area(s) is/are rejection region(s)? Reject H0H0 if test statistic value the critical value(s).…
You are conducting a test of independence for the claim that there is a relationship between favorite ice cream flavor and gender. Use a significance level of .01. Vanilla Chocolate Strawberry Male | 32 67 37 Female 24 40 25 The expected observations for this table would be (Round to 3 decimal places.) Vanilla Chocolate Strawberry Male Female At the .01 significance level, what is the critical chi-square for this data? (Round to 3 decimal places.) с What is the chi-square test-statistic for this data? (Round to 3 decimal places.) What can you conclude about Gender and Favorite Ice Cream Flavor? Gender and Ice Cream Flavor are Dependent O Gender and Ice Cream Flavor are Independent
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Question 4 く Test the claim that the mean GPA of night students is larger than the mean GPA of day students at the .01 significance level. The null and alternative hypothesis would be: Ho:PN = PD Ho:HN H1:PN > PD H1:µN > HD H1:PN < PD H1:PN # PD H1:µN < HD H1:µN # µD HD Ho:PN = PD Ho:PN = pD Ho:HN = HD Ho:UN = D The test is: two-tailed left-tailed right-tailed The sample consisted of 70 night students, with a sample mean GPA of 2.22 and a standard deviation of 0.03, and 50 day students, with a sample mean GPA of 2.21 and a standard deviation of 0.02. The test statistic is: (to 2 decimals) The critical value is: (to 2 decimals) Based on this we: O Reject the null hypothesis O Fail to reject the nul hypothesis

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