3. Using data in question 6, determine whether there is enough evidence to conclude that the type II hashigher strength than type I? Use a pooled t-test with a = 0.05.1. Identify the parameters2. Null Hypothesis3. Alternative Hypothesis4. Test Statistic5. Rejection Region6. Conclusion 6. The maintenance cost (y in dollars) and the age of (x in years) of nine cash registers in a department storeis given in the following table.sum71354292181234012635867251Xya. Complete the above table and find the sum of each row.b. Draw the scatter plot. Is there a linear trend?200150с.Find the S, Sw, and Swy.у 1005010d. Find the Pearson's correlation coefficient.

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Description: 3. Using data in question 6, determine whether there is enough evidence to conclude that the type II hashigher strength than type I? Use a pooled t-test with a = 0.05.1. Identify the parameters2. Null Hypothesis3. Alternative Hypothesis4. Test Stati

Given Information : The maintenance cost (y in dollars) and age of (x in years) of nine cash…

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3. Using data in question 6, determine whether there is enough evidence to conclude that the type II has higher strength than type I? Use a pooled t-test with a = 0.05. 1. Identify the parameters 2. Null Hypothesis 3. Alternative Hypothesis 4. Test Statistic 5. Rejection Region

3. Using data in question 6, determine whether there is enough evidence to conclude that the type II has higher strength than type I? Use a pooled t-test with a = 0.05. 1. Identify the parameters 2. Null Hypothesis 3. Alternative Hypothesis 4. Test Statistic 5. Rejection Region

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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Sample 2 Assume that both populations are normally distributed. a) Test whether µ, # µ, at the a = 0.05 level of significance for the given sample data. b) Construct a 95% confidence interval about u, - H2. Sample 1 17 13 17 14.3 3.5 3.1 E Click the icon to view the Student t-distribution table. A. Ho: H1 = H2, Hq: Hq # Hz O B. Ho: H1 = H2, H,: Hy H2 Determine the test statistic. t= - 1.15 (Round to two decimal places as needed.) Determine the critical value(s). Select the correct choice below and fill in the answer box(es) within your choice. (Round to three decimal places as needed.) O A. The critical value is O B. The lower critical value is -2.040. The upper critical value is 2.040. Should the hypothesis be rejected? Do not reject the null hypothesis because the test statistic is not in the critical region. b) Construct a 95% confidence interval about u, - H2. The confidence interval is the range from to| (Round to two decimal places as needed. Use ascending order.)
Q2 part 2
I need a hypothesis, description and interpretation, please! Thank you!!
Dr. V. noticed that the more frequently a student is late or absent from class the worse he or she performs on the final exam. He decided to investigate. Dr. V. collected a random sample of 22 students. This sample includes the number of times a student is absent and their grades on the final exam. Discuss the implications of the calculated Statistical Power and the probability of a Type II Error. Significance Level: Use a 5% significance level. What is/are the Critical Value(s)? Calculate the coefficient of determination, r2, using Excel’s RSQ function. Explain what it means.
Use a 0.05 significance level to test the claim that the different samples come from populations with the same mean. Use the traditional method the null and alternative hypothesis, the critical value, the test statistic, Draw the graph and label the test statistic, critical value and rejection regions. Reject or fail to reject? State the conclusion as outlined in the template or book ANOVA Source of Variation SS df MS F P-value Factor 13.500 3 4.500 5.171 0.0011 Error 13.925 16 0.870 Total 27.425 19
Test the hypothesis that the three class sections (section) differed in their performance on Quiz #3 (quiz3). 2 Before doing this, indicate what type of ANOVA you will perform (i.e., Independent or Dependent), state your null and alternative hypotheses, and the alpha level. Type of ANOVA: Step 1: H1: Ho: Step 2: a = 0.01 Step 3: Descriptives quiz3 95% Confidence Interval for Mean Mean N. Std. Deviation Lower Bound Upper Bound Std. Error Minimum Maximum 33 9.00 2.062 9.73 8.27 359 10 7.62 39 2.098 6.94 336 8.30 10 33 7.61 2.549 6.70 1444 8.51 10 0. Total 8.05 2.310 105 225 7.60 8,49 10 ANOVA quiz3 Sum of Squares Mean Square df Sig 21.826 4.356 015 43.652 Between Groups 5.011 102 511.110 Within Groups 104 554.762 Total
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Determine Test statistic. Determine P-value. Reject or fail to reject?
1. True or False: The normality assumption was satisfied at 5% level of significance. 2. True or False: To answer Eriel’s objective, one-sample t-test is the appropriate procedure to use. 3. True or False: The decision is to reject Ho. 4: True or False: There is no sufficient evidence to say that the average lifespan of cats is beyond 15 years.
24. Question content area top Part 1 Find the critical value(s) and rejection region(s) for the indicated t-test, level of significance α, and sample size n. Right-tailed test, α=0.10, n=10 LOADING... Click the icon to view the t-distribution table. Question content area bottom Part 1 The critical value(s) is/are enter your response here . (Round to the nearest thousandth as needed. Use a comma to separate answers as needed.) Part 2 Determine the rejection region(s). Select the correct choice below and fill in the answer box(es) within your choice. (Round to the nearest thousandth as needed.) A. t>enter your response here B. t<enter your response here and t>enter your response here C. enter your response here <t<enter your response here D. t<enter your response here
The student population at the state college is made up of 60% females and 40% males. The college theater department recently staged a production of a modern musical. A researcher recorded the gender of each student entering the theater and found a total of 37 females and 18 males. Is the gender distribution for theater goers significantly different from the distribution for the general college? Test at the .05 level of significance. Chi-Square Distribution Degrees of Freedom = 16 0123456789101112131415161718192021222324252627282930Χ² χ² critical value = (to three decimal places) χ² = (to two decimal places) Conclusion: Fail to reject the null hypothesis; the distribution of gender for theater goers is not significantly different from that of the general college. Fail to reject the null hypothesis; the distribution of gender for theater goers is significantly different from that of the general college. Reject…
What is the observed value? What is the p-value? What is the rejection region?