1. Use the following methods to test Hoμ₁ = μ2 versus H₁ μ₁ μ2 for the data in worksheet 9.1 ofthe Excel data file:a. Tukey's Quick Testb. A boxplot slippage testc. The two-sample t testd. The Mann-Whitney test (use Stat> Nonparametrics> Mann-Whitney) x1x213181919111814158201816161521

Do not reject H0There is enough evidence to support the claim that there is difference in two…

Answered: 1. Use the following methods to test…

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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Suppose there is sufficient evidence to reject Ho: ₁ = ₂ = μ3 using a one-way ANOVA. The mean square error from the ANOVA is determined to be 22.8. The sample means are x₁ = 4.6, x₂ = 10.7, and x3 = 19.6, with n₁ = n₂ = n3 = 7. Use Tukey's test to determine which pairwise means are significantly different using a familywise error rate of α=0.01. Click here to view page 1 of the table of critical values for Tukey's Test for an alpha of 0.05. Click here to view page 2 of the table of critical values for Tukey's Test for an alpha of 0.05. Click here to view page 1 of the table of critical values for Tukey's Test for an alpha of 0.01. Click here to view page 2 of the table of critical values for Tukey's Test for an alpha of 0.01. Calculate the test statistic qo for each pairwise difference. 912 913 923 (Round to three decimal places as needed.) The critical value is. (Round to three decimal places as needed.) Based on this data, the hypothesis Ho: #₁ =1₂₂ the hypothesis Ho: #₂ = 13- the…
Listed in the data table are amounts of strontium-90 (in millibecquerels, or mBq, per gram of calcium) in a simple random sample of baby teeth obtained from residents in two cities. Assume that the two samples are independent simple random samples selected from normally distributed populations. Do not assume that the population standard deviations are equal. Use a 0.05 significance level to test the claim that the mean amount of strontium-90 from city #1 residents is greater than the mean amount from city #2 residents. Click the icon to view the data table of strontium-90 amounts. What are the null and alternative hypotheses? Assume that population 1 consists of amounts from city #1 levels and population 2 consists of amounts from city #2. OA. Ho: H₁ H₂ H₁: Hy > H₂ OC. Ho: H₁ H₂ H₁: H₁ H₂ More Info City #1 106 86 121 118 101 20 104 213 130 290 100 277 145 Print City #2 117 63 100 85 87 107 110 111 143 133 101 209 Done ~ X OB. Ho: ₁₂ H₁: Hy > H₂ OD. Ho Hi#H H₁: H₁ H₂
You wish to test the following claim (HaHa) at a significance level of α=0.01. For the context of this problem, μd=PostTest−PreTest where the first data set represents a pre-test and the second data set represents a post-test. (Each row represents the pre and post test scores for an individual. Be careful when you enter your data and specify what your μ1 and μ2 are so that the differences are computed correctly.) Ho:μd=0 Ha:μd≠0You believe the population of difference scores is normally distributed, but you do not know the standard deviation. You obtain the following sample of data: pre-test post-test 58.5 -47 63.7 56.5 55.4 48.2 59.1 95.5 56.3 -30.6 62.3 21 51.1 11.8 57.5 -0.2 60.9 15.6 57.5 43 54.3 5.8 60.2 44.7 62.8 31.1 47.7 62.6 65 101.4 65 86.1 62 68.7 What is the test statistic for this sample?test statistic = (Report answer accurate to 4 decimal places.)What is the p-value for this sample?p-value = (Report answer accurate to…
10.44
The accompanying summary data on skeletal-muscle CS activity (nmol/min/mg) appeared in an article. Old Old Young Sedentary Active Sample size 10 8. 10 Sample mean 46.28 47.41 58.34 Sample sd 7.16 5.59 8.43 Carry out a test at level a = 0.05 to decide whether true average activity differs for the three groups. Suppose that all required ANOVA assumptions are satisfied. If appropriate, investigate differences amongst the means with a multiple comparisons method. (Use i = 1, 2, and 3 respectively.) State the appropriate hypotheses. Ho: H1 = H2 = H3 H: all three H's are unequal Ho: H1 = H2 = H3 Ha: : at least two u's are unequal O Ho: H1 # H2 + H3 all three u;'s are equal Ha Ho: H1 + H2# H3 H: at least two u's are equal Calculate the test statistic. (Round your answer to two decimal places.) f =
9.1/5 Identify the test statistic p-value
Listed in the data table are amounts of strontium-90 (in millibecquerels, or mBq, per gram of calcium) in a simple random sample of baby teeth obtained from residents in two cities. Assume that the two samples are independent simple random samples selected from normally distributed populations. Do not assume that the population standard deviations are equal. Use a 0.05 significance level to test the claim that the mean amount of strontium-90 from city #1 residents is greater than the mean amount from city #2 residents. City_#1 City_#2108 11786 53121 100116 85101 83104 107213 110113 111290 142100 133266 101145 209 What are the null and alternative hypotheses? Assume that population 1 consists of amounts from city #1 levels and population 2 consists of amounts from city #2? The test statistic is? The P-value is? State the conclusion for the test?
Consider the following sample observations on stabilized viscosity of asphalt specimens. 2,762 2,907 3,012 2,813 2,876 Suppose that for a particular application, it is required that true average viscosity be 3,000. Does this requirement appear to have been satisfied? State the appropriate hypotheses. (Use ? = 0.05.) H0: ? > 3,000 Ha: ? < 3,000 H0: ? = 3,000 Ha: ? ≠ 3,000 H0: ? ≠ 3,000 Ha: ? = 3,000 H0: ? < 3,000 Ha: ? = 3,000 Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to three decimal places.) t = P-value = What can you conclude? Reject the null hypothesis. There is sufficient evidence to conclude that the true average is viscosity differs from 3000.Do not reject the null hypothesis. There is sufficient evidence to conclude that the true average is viscosity differs from 3000. Reject the null hypothesis. There is not sufficient evidence to conclude that…
11.33 Osteoporosis. Osteoporosis is a condition in which the bones become brittle due to loss of minerals. To diagnose os- teoporosis, an elaborate apparatus measures bone mineral den sity (BMD). BMD is usually reported in standardized form The standardization is based on a population of healthy young adults. The World Health Organization (WHO) criterion for osteoporosis is a BMD 2.5 or more standard deviations below the mean for young adults. BMD measurements in a popula- tion of people who are similar in age and sex roughly follow a Normal distribution.? (a) What percent of healthy young adults have osteoporosis by the WHO criterion? (b) Women aged 70 to 79 are, of course, not young adults. The mean BMD in this age is about -2 on the standard scale for young adults. Suppose that the standard deviation is the same as for young adults. What percent of this older population has osteoporosis!
a) Test the following claim. Make sure to include the critical value, a sketch of the critical value, and the test statistic. A simple random sample of weights of 19 green M&Ms has a mean of .8635 g. Assume that o =.0565 g. Use a .05 significance level to test the claim that the mean weight of all green M&Ms is less than 8435g, which is the mean weight printed on the label.

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