Consider the following information for questions 24-26. A student organization a

Question

Consider the following information for questions 24-26. A student organization at a university is interested in estimating the proportion of students in favor of showing movies biweekly instead of monthly. A SRS of 423 students revealed that 180 are in favor of showing movies biweekly. State the point estimate. (Round to two decimal places.) Determine the margin of error for a 90% level of confidence. (State the z-value used. Round to two decimal places.) Construct a 90% confidence interval for the true proportion of student in favor of showing movies biweekly instead of monthly. (Round to two decimal places.)

Explanation / Answer

point estimate = 180 / 423 = 0.43

margin of error : alpha / 2 = 0.05 Z= 1.64

margin of error : 1.64 * SRQT ( 0.43 * 0.57 / 423 )

margin of error : 0.0395

confidence interval

0.43 +/- 0.0395

0.3905 < p < 0.3395

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