Sample of 35 gas grills has a mean price of $637.50 and a standard deviation of $59.30?

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W asked:


you are given the sample mean and the sample standard deviation. use this information to construct the 90% and 95% confidence intervals for the population mean.

a sample of 35 gas grills has a mean price of $637.50 and a standard deviation of $59.30.

the 90% confidence interval is ___________ ______________

the 95% confidence interval is ___________ ______________

Metal Kitchen Cabinets

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    • This entry was posted on Saturday, July 17th, 2010 at 4:53 am and is filed under gas grills. You can follow any responses to this entry through the RSS 2.0 feed. Both comments and pings are currently closed.

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    No Responses to “Sample of 35 gas grills has a mean price of $637.50 and a standard deviation of $59.30?”

    1. M Says:
      July 19th, 2010 at 1:04 pm

      ANSWER: 90% Resulting Confidence Interval for ‘true mean’: = [$620.56, $654.44]

      Why???

      SMALL SAMPLE, CONFIDENCE INTERVAL, NORMAL POPULATION DISTRIBUTION
      x-bar = Sample mean 637.5
      s = Sample standard deviation59.3
      n = Number of samples 35
      df = degrees of freedom 34

      significant digits2

      Confidence Level90
      “Look-up” Table ‘t-critical value’1.69
      Look-up Table of t critical values for confidence and prediction intervals. Central two-sided area = 90%
      with df = 34. Another Look-up method is to utilize Microsoft Excel function:
      TINV(probability,degrees_freedom) Returns the inverse of the Student’s t-distribution

      90% Resulting Confidence Interval for ‘true mean’:
      x-bar +/- (’t critical value’) * s/SQRT(n)= 637.5 +/- 1.69 * 59.3/SQRT(35) = [$620.56, $654.44]

      ANSWER: 95% Resulting Confidence Interval for ‘true mean’: = [$617.15, $657.85]

      Why???

      SMALL SAMPLE, CONFIDENCE INTERVAL, NORMAL POPULATION DISTRIBUTION
      x-bar = Sample mean 637.5
      s = Sample standard deviation59.3
      n = Number of samples 35
      df = degrees of freedom 34

      significant digits2

      Confidence Level95
      “Look-up” Table ‘t-critical value’2.03
      Look-up Table of t critical values for confidence and prediction intervals. Central two-sided area = 95%
      with df = 34. Another Look-up method is to utilize Microsoft Excel function:
      TINV(probability,degrees_freedom) Returns the inverse of the Student’s t-distribution

      95% Resulting Confidence Interval for ‘true mean’:
      x-bar +/- (’t critical value’) * s/SQRT(n)= 637.5 +/- 2.03 * 59.3/SQRT(35) = [$617.15, $657.85]