A study was conducted in Mandaluyong to determine the proportion of fiber subscribers. From a random sample of 625 homes, 350 are subscribers. Express all proportions in decimal form. If we let x be the number of fiber subscribers in the sample, what is the value of p¯(sample proportion)? Express your answer in two decimal places. Using a 90% confidence level, find the margin of error (two decimal places) for the interval estimate. Formula: E=zα/2 [√p¯(1−p¯) / n] Confidence Level 1−α Significance Level α α/2 zα/2 90% = 0.90 0.10 0.05 1.645 95% = 0.95 0.05 0.025 1.96 98% = 0.98 0.02 0.01 2.33 99% = 0.99 0.01 0.005 2.576                    What is the lower limit(two decimal places) of the 90% confidence interval?

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