Statistics - estimate of population mean, margin of error

Related BrainMass Solutions

• QMTH 205Lab 3:Probability Distribution & Interval Estimation

  Construct a 95% confidence interval estimate of population mean (confidence level 1- = 95%): enter Excel formulas to compute sample mean score, sample size, margin of error, lower and upper limits (Upper Limit = the sample mean + the margin error

• Statistics: Margin of Error, Sample Sizes and Confidence Intervals

  40947 Statistics: Margin of Error, Sample Sizes and Confidence Intervals Question: Give an estimate of the sample size needed to obtain the specified margin of error for a 95% confidence interval.

• Statistics Questions Related to Sampling

  Assume that a random sample is used to estimate a population p. Find the margin of error E that corresponds to the given statistics and confidence level. N=550, x=220, 90% confidence.

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  165054 Minimum Sample Required in Statistics 1.) With a margin of error of 0.050; confidence level of 99% ^p and ^q unknown, what is the minimum sample size required to estimate a population proportion or percentage? 2.)

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  If we want to estimate the population mean travel-to-work time for San Francisco residents with a margin of error of 2 minutes, what sample size should be used?

• Confidence interval

  There are also other population parameters that we could estimate with confidence interval. Similar to what we described above, they are found by finding a sample statistic and a margin of error that depends on the level of confidence.

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  Assume that a random sample is used to estimate a population p. Find the margin of error E that corresponds to the given statistics and confidence level. N=550, x=275, 90% confidence. The margin of error E=____ 3.

• All kinds of statistics problems

  Problem 6) A government survey conducted to estimate the mean price of houses in a metropolitan area is designed to have a margin of error of $10,000. Pilot studies suggest that the population standard deviation is $70,000.

• Confidence interval for Smith Travel Research

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• Statistics - Confidence Interval

  s = 12 SE = s/sqrt n= 12/sqrt n Margin of error = 2 The z value corresponding to the 95% CI is 1.96 Margin of error = z * SE 2 = 1.96 * 12/sqrt n On solving, we get n = 138.30 The sample size required is 139.