1. As part of their requirement for graduation, a group of BSIT-Automotive students of CSCST-MC randomly...

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1. As part of their requirement for graduation, a group of BSIT-Automotive students of CSCST-MC randomly selected 217 cars owned by students in all of the universities in Cebu City. They found that mean age of cars is 7.89 years with a corresponding standard deviation of 3.67 years. They also randomly selected 152 faculty cars of the same school and found that the mean age of their cars is 5.99 years with a corresponding standard deviation of 3.65 years. At 5% level of significance, test whether the students' cars are older than those of the faculty's. Solutions: Let μ₁ be the mean age of the cars of the students. μ₂ be the mean age of the cars of the faculty members. STEP 1. Statement of Hypotheses Ho: H₁: STEP 2: Level of Significance a = 0.05(one-tailed) STEP 3: Test Statistic te te (x₁ - x₂) - (1₁-1₂). n₁ m (7.89-5.99)-0 13.40866 STEP 4. Critical Value Pls provide: STEP 5. Decision therefore 1 1 + 217 152 STEP 6. Conclusion (217-1)3.67² + (152-1)3.65² 367 Reject Ho if |tc|2|ta, dfl. Since|tc|=4.91 is Ho. 1.9 13.40866 (369/32984) At 5% level of significance,. = 13.40866 1.9 0.38731 -= 4.91 than Ita, df = 1. As part of their requirement for graduation, a group of BSIT-Automotive students of CSCST-MC randomly selected 217 cars owned by students in all of the universities in Cebu City. They found that mean age of cars is 7.89 years with a corresponding standard deviation of 3.67 years. They also randomly selected 152 faculty cars of the same school and found that the mean age of their cars is 5.99 years with a corresponding standard deviation of 3.65 years. At 5% level of significance, test whether the students' cars are older than those of the faculty's. Solutions: Let μ₁ be the mean age of the cars of the students. μ₂ be the mean age of the cars of the faculty members. STEP 1. Statement of Hypotheses Ho: H₁: STEP 2: Level of Significance a = 0.05(one-tailed) STEP 3: Test Statistic te te (x₁ - x₂) - (1₁-1₂). n₁ m (7.89-5.99)-0 13.40866 STEP 4. Critical Value Pls provide: STEP 5. Decision therefore 1 1 + 217 152 STEP 6. Conclusion (217-1)3.67² + (152-1)3.65² 367 Reject Ho if |tc|2|ta, dfl. Since|tc|=4.91 is Ho. 1.9 13.40866 (369/32984) At 5% level of significance,. = 13.40866 1.9 0.38731 -= 4.91 than Ita, df =