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Day 5: Drills. (10 pts) Sample Size Determination. Show complete solutions. Express your final answer in the appropriate units of measurement (i.e.: people, bags, etc.) (1 pt each) Ferdie observed that the average yield of palay per sack follows a normal distribution with a standard deviation of 25 grams per sack. How large should the sample be for him to be 95% confident that the estimate for the true average yield of palay will differ by at most 5 grams per sack? a) b) Luke makes some of the best Leche Flan in town. The average amount of Leche Flan he makes per container is normally distributed with a standard deviation of 2 grams per container. How large a sample should he take for him to be 90% confident that the estimate for the true average amount of Leche Flan will differ by no more than 1.5 grams? c) Julie observed that the average amount of coffee dispensed per cup by a certain coffee dispenser is normally distributed with a standard deviation of 2mL per cup. How large a sample should be taken if she wants to be 99% confident that her estimate for the true mean amount of coffee dispensed per cup will differ by no more than 1mL per cup? d) An independent polling group conducted a survey two years ago showing that 38% of individuals in Pasig are in favor of the current administration. They would like to repeat the same survey once again to see if there were any changes in this proportion given the policies implemented by the administration. How large a sample should take for them to be 90% confident that their estimate of the true proportion of individuals in favor of the administration will differ by no more than 0.08? e) Dr. Walter wants to determine the proportion of mothers in a certain barangay in Makati who had a caesarian section. The results of a pilot test he conducted beforehand showed that 62.8% of mothers had a caesarian section. How large should the sample size be for er to be 95% confident that the estimated proportion of mothers who had a caesarian section will differ by no more than 0.12?

ANSWER

(a) difference in mean = 5 and s = 25
\alpha = 0.05
n = ?
P(Z<(5/(25//n0.5) = 0.95
1.6449 = 5/(25/n0.5)
n = 68

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