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On Estimation of Parameters ACTIVITY 7: On Estimation of Parameters General Instructions. Solve the following as required. Show complete solution whenever possible. 1. A local

On Estimation of Parameters

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ACTIVITY 7: On Estimation of Parameters General Instructions. Solve the following as required. Show complete solution whenever possible. 1. A local government official observes an increase in the number of individuals with cardiovascular and obesity prl.645oblems in his barangay. In order to improve the health conditions of his constituents, he aims to promote an easy and cheap way to reduce weight. It is known that obesity results in risk of having illnesses like diabetes and heart problems. He encouraged his constituents to participate in his Dance for Life project every weekend for three months. To know if the program is effective in reducing weight, he randomly selected 12 participants from the group who completed the program. The weight loss data, in kilograms, of the 12 randomly selected participants after completing the program, are 0.5. 0.7. 0.9, 1.1. 1.2, 1.3, 1.4, 2.0. 2.3, 2.4, 2.7, and 3.0. It is known that the weight loss of those who have completed the dance program follows a normal distribution with variance of 3.24 kg2. a. Construct and interpret a 90% confidence interval for the true mean weight loss of the participants who have completed the dance program. b. Calculate the width of the confidence interval in (a.) c.. How many participants must be sampled in order to be 95% confident that the estimated mean weight loss of the participants will be within 0.5 kg of the true mean? 2. A manufacturer claims that the average content of a particular lubricant is 10 liters. The contents of a random sample of 10 containers were recorded as follows: 10.2, 9.7, 10.1, 10.3. 10.1, 9.8, 9.9, 10.4, 10.3 and 9.8 liters. Assume the distribution of contents is normal. Construct a 99% confidence interval for u. 3. The average entrance exam score of a random sample of 35 first year students of CSM was 89.66 with a standard deviation of 18.28. Find a 96% confidence interval for the true mean score. -end

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