4. Two types of plastic are suitable for use by an electronic calculator manufacturer. The breaking...
4. Two types of plastic are suitable for use by an electronic calculator manufacturer. The breaking strength of the plastic is important. From random samples of n 10 and n2- 12 batches the respective means were obtained as yı. 152.5 and 2. 153.7. The total sum of square for the data was SSTotal31.865 (a) Complete the following ANOVA table. Source SSdf MSF Treatment Error Total 31.865 (b) Calculate the t statistic for the appropriate two sample t test for checking...
Two types of plastic are suitable for use by an electronic calculator manufacturer. The breaking strength of the plastic is important. From random samples of ni- 10 and n2 12 batches the respective means were obtained as -152.5 and 2 153.7. The total sum of square for the data was SSTotal -31.865. (a) Complete the following ANOVA table. Source SS dfMS F Treatment Error Total 31.865 (b) Calculate the t statistic for the appropriate two sample t test for checking...
Two types of plastic are suitable for use by an electronic calculator manufacturer. The breaking strength of the plastic is important. From random samples of ni- 10 and n2 -12 batches the respective means were obtained as ji. -152.5 and j2.-153.7. The total sum of square for the data was SSTotal 31.865. (a) Complete the following ANOVA table. Source Treatment Error Total SSdf MS F 31.865 (b) Calculate the t statistic for the appropriate two sample t test for checking...
7. Two types of plastic are suitable for use by an electronic calculator manufacturer. The breaking strength of this plastic is important. It is known that σ-σ,-1.0 psi. From random samples of n1-10 and n2-12 we obtain У1-162.5 and у,-155.0. The company will not adopt plastic 1 unless its breaking strength exceeds that of plastic 2 by at least 10 psi. Based on the sample information, should they use plastic !? Set up and test appropriate hypotheses using α= 0.01.
Q4- Two types of plastic are suitable for use by an electronics component manufacturer. The breaking strength of this plastic is important. It is known that the distribution is normal and a = 02 = 1.0 psi. From a random sample of size n = 10 and n2 = 12, we obtain x1 = 162.5 and X2 = 155. The company will not adopt plastic 1 unless its mean breaking strength exceeds that of plastic 2 by at least 10...
2.25. Two types of plastic are suitable for use by an elec- tronic calculator manufacturer. The breaking strength of this plastic is important. It is known that ( σ,-1.0 psi. From random samples of n,-10 and n2= 12 we obtain yi = 162.5 and y2 155.0. The company will not adopt plastic 1 unless its breaking strength exceeds that of plastic 2 by at least 10 psi. Based on the sample information, should they use plastic 1? In answering this...
Question 5 1 pts Two types of plastics are suitable for an electronics component manufacturer to use. The breaking strength of this plastic is important. It is known that the standard deviations of the two types of plastics are the same, with a value of 1.0 psi. From a random sample of 10 and 12 for type 1 and type 2 plastics, respectively, we obtain sample means of 162.5 and 155. The company will not adopt plastic 1 unless its...
Treatment 1 Treatment 2 Treatment 3 0 1 6 1 4 5 0 1 6 3 2 3 T = 4 T = 8 T = 20 SS = 6 SS = 6 SS = 6 N = 12 G = 32 ƩX2= 138 1a. Conduct a single-factor independent-measures ANOVA to test the hypothesis that there are significant differences in the mean scores among the three treatment conditions. Use α = .01. The alternative hypothesis is Group of answer choices...
4. Measuring effect size for two-factor ANOVA Аа Аа It is projected that approximately 580,000 veterans will take advantage of the GI Bill for the 21st Century. Boots to Books is a course for all veterans, current military members, and their family members, friends, and supporters. The goal of Boots to Books is to assist deployed, postdeployed, and veteran students in making a positive transition from military to civilian life or from deployment to postdeployment life, including the acquisition of...
QUESTION 2 An ANOVA is to be performed comparing 4 means. Using the information provided, calculates *1 = 12, X2 = 13, X3 = 16,4 = 15. Note: The overall mean is 7 = 14. Round your answer to two decimal places. QUESTION 3 A comparison of 4 means is to be performed using ANOVA. Given the following information, calculate the value of the F-Test statistic. Round your answer to two decimal places. sî - 3,53 - 11,5 - 14,54...