The null and alternative hypotheses for a hypothesis test of the difference in twopopulation means are:Null Hypothesis:Alternative Hypothesis: 1 ₂Notice that the alternative hypothesis is a two-tailed test. Suppose ttest ind methodfrom scipy module is used to perform the test and the output is (-1.99, 0.0512)What is the P-value for this hypothesis test? Select one.1.99μ=1₂1.990.02560.0512

given data null hypothesis : μ1 = μ2alternative hypotheis : μ1 ≠ μ2two tailed test

Answered: The null and alternative hypotheses for…

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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You are conducting a multinomial hypothesis test (a = 0.05) for the claim that all 5 categories are equally likely to be selected. Complete the table. |Category Frequency Frequency Observed Expected A 10 В 9. 11 17 E 12 Report all answers accurate to three decimal places. But retain unrounded numbers for future calculations. What is the chi-square test-statistic for this data? (Report answer accurate to three decimal places, and remember to use the unrounded Pearson residuals in your calculations.) What are the degrees of freedom for this test? d.f.= What is the p-value for this sample? (Report answer accurate to four decimal places.) p-value = Is the p-value less than a? O yes O no This test statistic leads to a decision to... O reject the null accept the null fail to reject the null accept the alternative As such, the final conclusion is that... O There is sufficient evidence to warrant rejection of the claim that all 5 categories are equally likely to be selected. O There is not…
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Could you help find the test statistic p-value and if the H0 should be reject or fails to reject
The display provided from technology available below results from using data for a smartphone carrier's data speeds at airports to test the claim that they are from a population having a mean less than 6.00 Mbps. Conduct the hypothesis test using these results. Use a 0.05 significance level. Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim. T-Test μ<6.00 t=−0.198213 p=0.421896 x=5.83 Sx=5.753382 n=45 Assuming all conditions for conducting a hypothesis test are met, what are the null and alternative hypotheses? identify the test statistic pvalue conclusion
Hello, I could really use some help, I thought I had a handle on this and I just do not, I am mentally exhausted.
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I need a hypothesis, description and interpretation please!! Thanks
T-Test μ<4.00 t=−3.803793 p=0.000218 x=3.33 Sx=1.181583 n= 45 The display provided from technology available below results from using data for a smartphone carrier's data speeds at airports to test the claim that they are from a population having a mean less than 4.00 Mbps. Conduct the hypothesis test using these results. Use a 0.05 significance level. Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim. Identify the test statistic: (Round to two decimal places as needed.) Identify the P-value. (Round to three decimal places as needed.) State the final conclusion that addresses the original claim: Choose the correct answers in parentheses. (Reject/Fail to reject) H0. There is not (sufficient/not sufficient) evidence to support the claim that the sample is from a population with a mean less than 4.00 Mbps.
ter question 13 A repeated-measures study with a sample of n = 6 participants produces a mean difference of MD = 4 with SS= 30. Use a two-tailed hypothesis test !! with a = .05 to determine whether this sample provides evidence of a significant treatment effect. t-critical + 2.571 2.000 t Distribution Degrees of Freedom = 21 5000 5000 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 0.000
Assume that random variable X distributed normal with mean u and variance o?. Use following summary statistics to answer below questions. (use 2 digits after decimal point) n=8 ΣΧ=393 x2 = 21027 >(X-Xbar )2 = SSX = 1720.88 Sample mean = Sample standard deviation = 95% confidence interval for the population mean is 95% confidence interval for the population variance is between | to
What is the value of the test statistic for a test with a null hypothesis that mu = 250, Xbar = 260 the standard error = 5 and n = 100?

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