Assuming that there is a 5-point difference between the two sample means, which set of samplecharacteristics is most likely to produce a significant value for the independent-measures tstatistic?Question 23 options:Large sample sizes and small sample variancesLarge sample sizes and large sample variancesSmall sample sizes and small sample variancesSmall sample sizes and large sample variances

We have given statement Assuming that there is a difference between the two sample means which set…

Answered: Assuming that there is a 5-point…

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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A local bank claims that the waiting time for its customers to be served is the lowest in the area. A competitor bank checks the waiting times at both banks. The sample statistics are listed below. Test the local bank's claim. Use the information given below. Use a significance level of .05 and assume the variances are equal. Sample statistics for a local bank and a competitor's bank Local Bank Competitor Bank Sample size n1 = 46 n2 = 50 Average waiting time in minutes for each sample X, = 2.3 mins. X2 = 2.6 Sample Standard Deviation of each Sample = 1.1 mins 1.0 mins. S1 S2 1. Are the samples dependent or independent? 2. State your Null/Alternative hypotheses 3. What is the test-statistic? 4. What is the p-value? 5. What are the critical values? 6. Does the test-statistic lie in the rejection region? 7. Interpret the Result? Does the result change for a different value of alpha? Explain?
A two-sample hypothesis test is comparing the averages of the following populations. Population 1 is the salary of female employees and Population 2 is the salary of male employees. A hypothesis test was conducted to see if the average salary of males is greater than the average salary of females. Using the following as the output of the test, state the P-value and interpret the results in context to the problem. Use a significance level of 5% Mean Variance Observations Hypothesized Mean Difference df t Stat P(T<=t) one-tail t Critical one-tail P(T<=t) two-tail t Critical two-tail Females Males 65,852.00 89,562.00 1,003.00 1,025.00 45 30 0 37 -1.26 0.1200 1.3300 0.2400 1.2800 For the toolbar, press ALT+F10 (PC) or ALT+FN+F10 (Mac). BIUS Paragraph Arial 10pt EEVA V Tx XQ5 二三 ||||| WORDS POWERED BY TINY
Violation of which assumption below for the two-factor ANVOA is not a cause for concern with large sample sizes? a. The populations from which the samples are selected must have equal variances. b. The populations from which the samples are selected must be normal. c. A violation of any assumption below would be a concern, even with large sample sizes. d. The observations within each sample must be independent.
What are the values for SS and variance for the following sample of n = 3 scores? Sample: 8, 5, 2
When using an Analysis of Variance test, in order to find out which specific means are different from one another which test is used? a. t-test b. post hoc test c. Cohen’s D test d. Chi-square test
1. Conduct a two-sample t-test at a 1% level of significance to find out if the U.S. stock returns significantly outperform the U.S. corporate bond returns using the monthly data covering the sample period 1980-2019. Please explain the hypothesis and how to use the summary to answer the question.
Class 1 Final Sample Size: 30 Sample Variance: 80.55 Sample Mean: 44.2917 Class 2 Final Sample Size:30 Sample Variance: 55.29 Sample Mean: 36.4833 The assignment says to use the Z test for means and assume the hypothesized difference is 0. 1. Is there a significant difference in the means of the two finals at the .01 level of significance? 2. Why or why not?
Consider the following data: Step 1 of 3: Calculate the value of the sample variance. Round your answer to one decimal place. Answer How to enter your answer (opens in new window) stical Tables nomial Probabilities nomial Cumulative Poisson Probabilities Poisson Cumulative andard Normal Table t Table - Chi Square F Table - Pearson Correlation Coefficient Nonparametrics - Misc. q Table -2, -2, -8,-8, -2, -8, 12 N ww 3.1 3.0 2.9 2.8 2.7 2.6 2.5 Copy Data Tables Keypad Keyboard Shortcuts 0.07 0.08 0.09 3.3 3.2 Standard Normal Probabilities (z to ∞) 0.00 0.01 0.02 0.03 0.04 0.05 0.06 3.4 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.000 0.0005 0.0005 0.0005 0.0004 0.0004 0.0004 0.0004 0.0004 0.0004 0.000 0.0007 0.0007 0.0006 0.0006 0.0006 0.0006 0.0006 0.0005 0.0005 0.000 0.0010 0.0009 0.0009 0.0009 0.0008 0.0008 0.0008 0.0008 0.0007 0.000 0.0013 0.0013 0.0013 0.0012 0.0012 0.0011 0.0011 0.0011 0.0010 0.001 0.0019 0.0018 0.0018 0.0017 0.0016 0.0016 0.0015 0.0015 0.0014…
Given below are the analysis of variance results from a stat software display comparing sample data for the meanmileage for 4 different types of cars. Assume that you want to use a 0.05 significance level in testing the null hypothesisthat the different samples come from cars with the same mileage. What can you conclude about the equality of the population means?Source DF SS MS F pFactor 3 13.500 4.500 5.17 0.011Error 16 13.925 0.870Total 19 27.425 12)What is the null and alternative hypothesis for this scenari?Based on this One Way Anova software analysis, would you reject of fail to reject the nul hypothesis?What does this result mean in context?
Which of the following factors have each of the advantages listed when using analysis of variance? Multiple answers can be chosen. 1. Makes ANOVA more robust to departures from the assumption of normality. 2. Makes ANOVA more robust to departures from the assumption of equal variance. 3. Increases the power of the test. Options: Large sample size. Balanced design.
Perform an analysis of variance (ANOVA) to determine if the differences in the percentage butterfat of the different breeds of cows is statistically significant. Guernsey Jersey Holstein 4.51 4.59 3.43 5.26 6.67 3.68 5.81 5.39 3.72 5.14 4.38 3.76 4.55 4.99 4.43 4.86 5.49 3.84 4.69 5.43 3.54 4.12 5.02 4.18 5.44 5.16 3.4 4.69 5.15 3.77 5.24 5.76 4.03 4.62 5.34 3.63 4.61 5.11 3.37 5.02 4.75 3.54 4.67 5.07 3.64 4.59 4.42 3.98 4.92 6.13 5.58 4.6 5.6 5.83 1. Compute the treatment sum of squares (SSTr) and the error sum of squares (SSE) and use them to complete the following ANOVA table. (Round your answers to 4 decimal places). Source S.S. df M.S. F Treatment Error Total 2. Compute the p-value. (Round your answer to 4 decimal places.)p-value =

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