13.27 Consider the following 2 x 3 (i.e., r = 2 and c = 3) con-tingency table:Column123Row1293453163025a. Specify the null and alternative hypotheses that shouldbe used in testing the independence of the row and col-umn classifications. Ho: Row and Column are independentb. Specify the test statistic and the rejection region thatshould be used in conducting the hypothesis test of parta. Use a .01. >9.21034c. Assuming that the row classification and the columnclassification are independent, find estimates for the ex-pected cell counts.d. Conduct the hypothesis test of part a. Interpret yourresult.8.71

Answered: 13.27 Consider the following 2 x 3…

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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The Batteries sheet of the Data Excel file shows the results of two random samples that measured the average number of minutes per charge for AA Lithium-ion (Li-ion) rechargeable batteries versus Nickel-Metal Hydride (NiMH) rechargeable batteries. Perform a hypothesis test using significance level (α) = 0.05 to determine if the true average number of minutes per charge for NiMH batteries is smaller than that for Li-ion batteries. Let:µLi-ion be the true average number of minutes per charge for Li-ion batteries µNiMH be the true average number of minutes per charge for NiMH batteries. t-Test: Two-Sample Assuming Unequal Variances NiMH Li-ion Mean 89.35714 95 Variance 3.93956 59.75 Observations 14 17 Hypothesized Mean Difference 0 df 19 t Stat -2.89621 P(T<=t) one-tail 0.004628 t Critical one-tail 1.729133 P(T<=t) two-tail 0.009255 t Critical two-tail 2.093024 Based on the…
If a hypothesis test produces a z-score in the critical region, which decision should be made? a. fail to reject the alternative hypothesis b. fail to reject the null hypothesis c. reject the alternative hypothesis d. reject the null hypothesis
The following MINITAB output presents the results of a hypothesis test for the difference Hx - Hy between two population means: Two-sample T for X vs Y N Mean StDev SE Mean 135 3.94 2.65 0.23 180 4.43 2.38 0.18 Difference = mu (X) – mu (Y) Estimate for difference: -0.484442 95% upper bound for difference: -0.007380 T-Test of difference = 0 (vs <): T-Value = -1.68 P-Value = 0.047 DF = 270 Is this a one-tailed or two-tailed test? a. b. What is the null hypothesis? Can H, be rejected at the 5% level? How can you tell? C. d. The output presents a Student's t test. Compute the P-value using a z test. Are the two results similar? Use the output and an appropriate table to compute a 99% confidence interval for Hy- Hy based on the z statistic. e.
A major oil company has developed a new gasoline additive that is supposed to increase mileage. To test this hypothesis, ten cars are randomly selected. The cars are driven both with and without the additive, The results are displayed in the following table. Can it be concluded, from the data, that the gasoline additive does significantly increase mileage? Let d = (gas mileage with additive)-(gas mileage without additive). Use a significance level of a = 0.05 for the test. Assume that the gas mileages are normally distributed for the population of all cars both with and without the additive, Car 1 2 3 4 6 7 9 10 Without additive 22.9 20.9 11.3 27.1 9.5 19.4 16.8 16.8 20.1 27.2 With additive 25.6 23.6 13.3 28.7 13. 22 17 18.2 21 29.6 Copy Data Step 2 of 5: Find the value of the standard deviation of the paired differences. Round your answer to two decimal places. Answer E Tables E Keypad How to enter your answer Keyboard Shortcuts Previous Step Answers
An article in the newspaper claims that 60% of registered drivers have the organ donor designation on their driver's license. A researcher at a hospital decides to investigate this claim. The researcher selects a random sample of 50 registered drivers and finds that 44.0% have the organ donor designation on their driver's license. Do the data provide convincing evidence at the ?α = 0.05 level that fewer than 60% of registered drivers have the organ donor designation on their license?
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A manufacturer of nickel-hydrogen batteries randomly selects 100 nickel plates for test cells, cycles them a specified number of times, and determines that 13 of the plates have blistered. n USE SALT (a) Does this provide compelling evidence for concluding that more than 10% of all plates blister under such circumstances? State and test the appropriate hypotheses using a significance level of 0.05. Но: р %3D 0.10 Ha: p + 0.10 Ho: p = 0.10 Ha: p > 0.10 Ho: p = 0.10 Ha: p 0.10 Ha: p = 0.10 Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to four decimal places.) z = 1.333 P-value = 0.0913 State the conclusion in the problem context. Reject the null hypothesis. There is sufficient evidence that more than 10% of plates blister under the experimental conditions. o Do not reject the null hypothesis. There is sufficient evidence that more than 10% of plates blister under the experimental conditions. Do not reject the…
A marketing professor has surveyed the students at her university to better understand attitudes towards PPT usage for higher education. To be able to make inferences to the entire student body, the sample drawn needs to represent the university’s student population on all key characteristics. The table below shows the five key student demographic variables. The professor found the breakdown of the overall student body in the university’s fact book posted online. A non-parametric chi-square test was used to test the sample demographics against the population percentages shown in the table above. Review the output for the five chi-square tests on the following pages and answer the five questions: Based on the chi-square test, which sample variables adequately represent the university’s student population and which ones do not? Support your answer by providing the p-value of the chi-square test and explaining what it means. Using the results from Question 1, make recommendation for…

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