Answered: lass 1 /Class 2 Sample size 25 /21…

Math

Statistics

lass 1 /Class 2 Sample size 25 /21 Sample mean 82/ 84 Sample standard deviation 6.2/ 7.6

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

See similar textbooks

Similar questions

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
A company would like to know if efficiency can be improved through a training course for employees. A sample of employees is taken. The time taken for an employee to complete a task before training and after training is observed.
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
Given below are the analysis of variance results from a Minitab display. Assume that you want to use a 0.05 significance level in testing the null hypothesis that the different samples come from populations with the same mean. Identify the p-value. Source DF SS MS F Factor 3 30 10.00 1.6 0.264 Error 50 6.25 Total 11 80 O A. 6.25 O B. 1.6 OC. 10.00 O D. 0.264 Click to select your answer. étv MacBook Air F10 F9 F7 30 888 *- F6 F3 esc F1 F2 & @ 7 4 5 2 3.
A real estate agent believes that the mean home price in the northern part of a county is higher than the mean price in the southern part of the county and would like to test the claim. A simple random sample of housing prices is taken from each region. The results are shown below. Mean Variance Observations Southern Northern 155.056 168.889 345.938 560.928 18 Pooled Variance 453.433 Hypothesized Mean Difference df 34 1949 t Stat P(T<=t) one-tail p= Ex: 1.234 0.030 2.441 0.060 t Critical one-tail P(T<=t) two-tail t Critical two-tail 2.728 2. Confidence Level 99% t = Ex: 1.234 n= Ex: 9 Degrees of freedom: df = Point estimate for the southern part of the county: 1 Ex: 1.234 Point estimate for the northern part of the county: 2 0耳_ 国 23 72°F
Does color enhance creativity? Test the indicated claim about the standard deviations or variances of two populations. Subjects are given a creativity exercise on a computer with either a red background or a blue background. The scores are shown in the table. At the 0.05 significance level, test the claim that those tested with red background have creativity scores with a standard deviation equal to the standard deviation for those tested with a blue background. Red Background Blue Background n1 = 39 n2 = 34 xˉx̄1 = 19.6 xˉx̄2 = 20.5 s1 = 0.72 s2 = 0.01 What are the correct hypotheses? (Select the correct symbols and use decimal values not percentages.)H0: Select an answer x̄₂ x̄₁ μ(red) p₂ p μ₁ μ₂ σ₁ s₁² σ(red) p₁ μ p̂₁ ? ≠ < > ≤ ≥ = Select an answer μ₁ μ₂ p₁ μ(blue) p σ₂ p₂ σ(blue) s₁² x̄₁ x̄₂ μ p̂₁ H1: Select an answer μ σ₁ x̄₁ μ(red) p̂₂ p₁ s₂² p₂ μ₂ μ₁ σ(red) x̄₂ p ? ≤ ≥ ≠ > = < Select an answer μ(blue) σ(blue) p μ μ₁ p₁ x̄₂ s₁² x̄₁ σ₂ p̂₁ p₂ μ₂…
An article in a Journal reports that 34% of American fathers take no responsibility for childcare. A researcher claims that the figure is higher for fathers in the town of Littleton. A random sample of 234 fathers from Littleton yielded 96 who did not help with childcare. How many populations? 01 What is the parameter? O Mean O Proportion O Difference between Means O Variance Standard Deviation
Which outcome is expected if the null hypothesis is true for an analysis of variance? a. SSbetween should be about the same size as SStotal. b. SSbetween should be about the same size as SSwithin. c. MSbetween should be about the same size as MStotal. d. MSbetween should be about the same size as MSwithin.
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.
Given below are the analysis of variance results from a Minitab display. Assume that you want to use a 0.05 significance level in testing the null hypothesis that the different samples come from populations with the same mean. Identify the value of the test statistic. Source DF MS F Factor 3 13.500 4.500 5.17 0.011 Error 16 13.925 0.870 Total 19 27.425 Seleccione una: O A. 5.17 О В. 0.011 O C. 4.500 O D. 13.500
Listed below are amounts of arsenic in samples of brown rice from three different states. The amounts are in micrograms of arsenic and all samples have the same serving size. The data are from the Food and Drug Administration. Use a 0.05 significance level to test the claim that the three samples are from populations with the same mean. Assume that the distributions for each state are normal and that all variances are equal. Arkansas 4.8 4.9 5.0 5.4 5.4 5.4 5.6 5.6 5.6 5.9 6.0 6.1 California. 1.5 3.7 4.0 4.5 4.9 5.1 5.3 5.4 5.4 5.5 5.6 5.6 Texas. 5.6 5.8 6.6 6.9. 6.9 6.9 7.1 7.3 7.5 7.6 7.7 7.7
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 =