A hypothesis test for independence has produced a randomisation distribution with the quantiles in the belowtable. If the test statistic is 99 and alpha = 0.05, what is the outcome of the test?l aQuantile Value120.010|| 30.000.025|| 35.0000.050 40.0000.100 50.0000.900| 90.0000.950| 95.0000.975||100.000180.990|110.000Reject the Alternative HypothesisAccept the Alternative HypothesisAccept the Null HypothesisReject the Null HypothesisDo not reject the Alternative HypothesisDo not reject the Null Hypothesis

Name: A hypothesis test for independence has produced a randomisation distr
Uploaded: 2024-03-20T07:07:12.000Z
Channel: Bartleby
Description: A hypothesis test for independence has produced a randomisation distribution with the quantiles in the belowtable. If the test statistic is 99 and alpha = 0.05, what is the outcome of the test?l aQuantile Value120.010|| 30.000.025|| 35.0000.050 40.0

The conclusion of a hypothesis test is always written in terms of the null hypothesis. If a test is…

Math

Probability

A hypothesis test for independence has produced a randomisation distribution with the quantiles in the below table. If the test statistic is 99 and alpha = 0.05, what is the outcome of the test? Quantile Value 0.010 30.000 1 0.025 35.000 0.050 40.000 4 5 3 0.100 50.000 0.900 90.000 0.950 95.000 6 0.975100.000 0.990 110.000 Reject the Alternative Hypothesis Ассеpt the Alternative Hypothesis Accept the Null Hypothesis Reject the Null Hypothesis Do not reject the Alternative Hypothesis Do not reject the Null Hypothesis ㅇ

A hypothesis test for independence has produced a randomisation distribution with the quantiles in the below table. If the test statistic is 99 and alpha = 0.05, what is the outcome of the test? Quantile Value 0.010 30.000 1 0.025 35.000 0.050 40.000 4 5 3 0.100 50.000 0.900 90.000 0.950 95.000 6 0.975100.000 0.990 110.000 Reject the Alternative Hypothesis Ассеpt the Alternative Hypothesis Accept the Null Hypothesis Reject the Null Hypothesis Do not reject the Alternative Hypothesis Do not reject the Null Hypothesis ㅇ

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...

See similar textbooks

Similar questions

A clinical researcher performs a clinical trial on 6 patients to determine whether a drug treatment has an effect on serum glucose. The sample mean glucose of the patients before and after the treatment are summarized in the following table. The sample standard deviation of the differences was 9. Sample mean glucose (mg/dL) Before treatment 96 After treatment 87 What type of hypothesis test should be performed? What is the test statistic? What is the number of degrees of freedom? Does sufficient evidence exist to support the claim that the drug treatment has an effect on serum glucose at the α=0.01 significance level?
In the airline business, "on-time" flight arrival is important for connecting flights and general customer satisfaction. Is there a difference between summer and winter average on-time flight arrivals? Let x1 be a random variable that represents percentage of on-time arrivals at major airports in the summer. Let x2 be a random variable that represents percentage of on-time arrivals at major airports in the winter. A random sample of n1 = 16 major airports showed that x1 = 74.9%, with s1 = 5.3%. A random sample of n2 = 18 major airports showed that x2 = 70.2%, with s2 = 8.6%. Note: For degrees of freedom d.f. not in the Student's t table, use the closest d.f. that is smaller. In some situations, this choice of d.f. may increase the P-value a small amount and thereby produce a slightly more "conservative" answer. (a) Does this information indicate a difference (either way) in the population mean percentage of on-time arrivals for summer compared to winter? Use α = 0.05. (i) What is the…
A discount airline allows staff 36.0 minutes between flights to clean the aircraft interior. The aircraft workers union claims that that average time that it takes to this cleaning is more than 36.0 minutes. They want to carry out a hypothesis test. They take a random sample n = 40 flights and time how long it takes to carry out the cleaning. They find that the sample mean is 36.66 minutes and the sample standard deviation is 5.56 minutes. Calculate the test statistic for this test and enter your answer below. Give your answer to 2 decimal places.
A clinical researcher performs a clinical trial on 12 patients to determine whether a drug treatment has an effect on serum glucose. The sample mean glucose of the patients before and after the treatment are summarized in the following table. The sample standard deviation of the differences was 6. Sample mean glucose (mg/dL) Before treatment 114 After treatment 108 What type of hypothesis test should be performed?Select: Left-tailed z-test Unpaired t-test Two-tailed z-test Paired t-test What is the test statistic? What is the number of degrees of freedom? Does sufficient evidence exist to support the claim that the drug treatment has an effect on serum glucose at the α=0.05 significance level? Select: _____Yes OR _____No
In a hypothesis test for the population variance, the alternate hypothesis is the population variance doesn't equal 17.0. The significance level to be used is 0.05 and the sample size to be taken is 25. The table value(s) to use from the chi-square distribution is/are 12.401 and 39.364. 13.181 39.364 40.671
n engineer has designed a valve that will regulate water pressure on an automobile engine. The valve was tested on 160 engines and the mean pressure was 7.7 pounds/square inch (psi). Assume the population variance is 0.36. If the valve was designed to produce a mean pressure of 7.8 psi, is there sufficient evidence at the 0.01 level that the valve performs below the specifications? Step 2 of 3: Find the P-value for the hypothesis test. Round your answer to four decimal places.
In a completely randomized design, 12 experimental units were used for the first treatment, 15 for the second treatment, and 20 for the third treatment. Complete the following analysis of variance (to 2 decimals, if necessary). If the answer is zero enter "0". Source of Variation Sum of Squares Degrees of Freedom Mean Square F p-value Treatments 1,100 Error Total 1,700 At a .05 level of significance, is there a significant difference between the treatments? The -value is - Select your answer - less than .01 between .01 and .025 between .025 and .05 between .05 and .10 greater than .10 What is your conclusion? - Select your answer - Cannot conclude that the means of the three treatments are equal Conclude that the means of the three treatments are equal