E.D.Species Sum of Sample ValuesAΣi Yai = 42.3BCΣi-1 YBi = 44.1ΣΥΒYci = 48.6Calculate the sum of squares for treatment in a one-way ANOVA withspecies as the factor.Jade asked her colleague to perform a two-way ANOVA on the samesample data with food type and species as the two factors. Her colleagueperformed the two-way ANOVA (assuming any required assumptions weresatisfied) but unfortunately misplaced the results. All that he could remember wasthat there was an interaction between food type and species at the 5%significance level but not at the 2.5% significance level. Calculate the two valuesbetween which the sum of squares for interaction in this two-way ANOVA must lie.

Given the sample statistics for adult wingspan (Y) asFood TypeSum of Sample ValuesSum of Squared…

Answered: E. D. A Species Sum of Sample Values…

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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Is the proportion of wildfires caused by humans in the south higher than the proportion of wildfires caused by humans in the west? 356 of the 579 randomly selected wildfires looked at in the south were caused by humans while 322 of the 597 randomly selected wildfires looked at the west were caused by humans. What can be concluded at the a = 0.01 level of significance? a. For this study, we should use z-test for the difference between two population proportions b. The null and alternative hypotheses would be: Ho: p1 p2 v (please enter a decimal) H1: p1 V p2 v (Please enter a decimal) c. The test statistic z v = 2.620 (please show your answer to 3 decimal places.) d. The p-value = 0.0213 e. The p-value is >v a f. Based on this, we should reject g. Thus, the final conclusion is that ... (Please show your answer to 4 decimal places.) |the null hypothesis. O The results are statistically significant at a = 0.01, so there is sufficient evidence to conclude that the proportion of the 579…
A researcher wants to estimate a population mean. The estimated average population variance from a pilot sample is 1.5 for the most important questions in the study. However, of the 2 most important questions, one had a variance of 1.4 and the other had a variance of 2.4. What is the desired sample size? O a. 922 QUESTION 20 O b. 538 OC 470 Od.576
Independent random samples were selected from populations 1 and 2. Thesample sizes, means, and variances are as follows: Population 1 2 Sample Size 35 49 Sample Mean 12.7 7.4 Sample Variance 1.38 4.14 Find a 95% confidence interval for estimating the difference in thepopulation means (μ1 - μ1)
Two different box-filling machines are used on an assembly line. The critical measurement influenced by these machines is the weight of the product in the boxes. Engineers are quite certain that the variance of the weight of product is o² =3 ounces. Experiments are conducted using both machines with sample sizes of 81 each. The sample averages for machines A and B are XA = 12.1 ounces and = 12.3 ounces. Engineers are surprised that the two sample averages for the filling machines are so different. Complete parts (a) XB = and (b) below. Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table. (a) Use the Central Limit Theorem to determine P(XB-XA20.2) under the condition that HA = HB. P(XB-XA20.2) = 4 Get more help - Clear all Check answer (1)
I need help solving part a and b of the following question attached.
A sample has a minimum value of 7 and a maximum value of 58 .Estimate its variance.
Pls show complete solution
Three different methods for assembling a product were proposed by an industrial engineer. To investigate the number of units assembled correctly with each method, 30 employees were randomly selected and randomly assigned to the three proposed methods in such a way that each method was used by 10 workers. The number of units assembled correctly was recorded, and the analysis of variance procedure was applied to the resulting data set. The following results were obtained: SST =10,870; SSTR =4580. a. Set up the ANOVA table for this problem (to 2 decimals but p-value to 4 decimals, if necessary). Source of Variation Sum of Squares Degrees of Freedom Mean Square F p-value Treatments 4580 Error 6290 Total 10870 b.Use a = 0.05 to test for any significant difference in the means for the three assembly methods. Calculate the value of the test statistic (to 2 decimals). The p-value is - Select your answer - What is your conclusion? Select your answer -
Two different formulas of an oxygenated motor fuel are being tested to study their road octane numbers. The variance of road octane number for formula 1 is o? 1.5, and for formula 2 it is o, = 1.2. Two random samples of size n1 15and n2 20 are tested, and the mean octane numbers observed are x 89.5 fluid ounces and X2 93.1 fluid ounces. Assume normality. (a) Test the hypotheses Ho : µ1 = H2 versus Hị :µ1 < µz using a = 0.05. Round your answer to three decimal places (e.g. 98.765). Zo = Но- (b) Calculate a 95% two-sided confidence interval on the mean difference road octane number, x1 – X2. Round your answer to three decimal places (e.g. 98.765). < H1 - H2 < (c) What sample size would be required in each population if you wanted to be 95% confident that the error in estimating the difference in mean road octane number is less than 1? ni = n2 = i . Round your answer up to the nearest integer.
Several methods of estimating the number of seeds in soil samples have been developed by ecologists. An article gave the accompanying data on the number of seeds detected by the direct method and by the stratified method for 27 soil specimens. Specimen 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Direct 21 33 0 60 20 61 40 8 13 93 1 68 77 21 Stratified 8 37 8 57 53 61 28 8 8 100 0 57 68 53 Specimen 15 Find the test statistic. (Round your answer to two decimal places.) t = 16 17 18 19 20 21 22 23 24 25 26 27 Direct Find the df. (Round your answer down to the nearest whole number.) df = 33 0 37 17 93 1 40 21 0 8 13 Use technology to find the P-value. (Round your answer to four decimal places.) P-value = 17 40 Stratified 28 0 37 13 93 13 48 21 0 13 40 Do the data provide sufficient evidence to conclude that the mean number of seeds detected differs for the two methods? Test the relevant hypotheses using a = 0.05. (Use direct stratified.) 13 77
In an industrial experiment, a job was performed by 10 workers using Method I, and by 7 workers using Method II. The experiment yielded the following data. The unit of measurement is minute. A statistician wants to test for the equality of two variances and the difference in the mean times to complete the job using the two methods. (This is a case of two independent samples drawn from two normal populations). Method 1 Method 2 12221915101615222613 61071261012 n1=10 Mean=17 Variance=26 n2=7 Mean=9 Variance=7 (a) Test at the 5% significance level whether the two variances are equal. (Give null and the alternative hypotheses, test statistic, rejection rule, conclusion) (b) Estimate with 95% confidence the ratio of the two population variances. Briefly describe what the interval estimate tells you. (c) Briefly explain how to use the interval estimate in part (b) to test the hypotheses in (a).
A retail designer wants to assess the mannequin proportions to which the public will respond most positively. She e-mails one of six pictures of mannequins of different sizes to 500 retail clothing customers from different parts of the country. The dependent variable is how pleasing and likable the customers find each of the mannequins. Which of these most accurately states the test for assumptions and whether the ANOVA can be completed? A. The independence of cases assumption has been violated, which is not robust. Thus, the ANOVA cannot be completed. B. The normality assumption has been violated, which is not robust. Thus, the ANOVA cannot be completed. C. The homogeneity of variances assumption has been violated, which is not robust. Thus, the ANOVA cannot be completed. D. None of the assumptions has been violated and so the ANOVA can be completed.

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