Statistics: 1001 Practice Problems For Dummies (+ Free Online Practice)
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For these three sample questions, consider that: A researcher conducted an Internet survey of 300 students at a particular college to estimate the average amount of money students spend on groceries per week. The researcher knows that the population standard deviation of weekly spending is $25. The mean of the sample is $85.

The following table provides the z*- values for selected (percentage) confidence levels.

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Sample questions

  1. What is the margin of error if the researcher wants to be 99% confident of the result?

    Answer: plus/minus $3.72

    The formula for margin of error when estimating a population mean is

    image1.jpg

    where z* is the value from the table for a given confidence level (99% in this case, or 2.58),

    image2.jpg

    is the standard deviation ($25), and n is the sample size (300).

    Now, substitute the values into the formula and solve:

    image3.jpg

    The margin of error for a 99% confidence interval for the population mean is plus/minus $3.72.

  2. What is the margin of error if the researcher wants to be 95% confident in the result?

    Answer: plus/minus $2.83

    The formula for margin of error when estimating a population mean is

    image4.jpg

    where z* is the value from the table for a given confidence level (95% in this case, or 1.96),

    image5.jpg

    is the standard deviation ($25), and n is the sample size (300).

    Now, substitute the values into the formula and solve:

    image6.jpg

    The margin of error for a 95% confidence interval for the population mean is plus/minus $2.83.

  3. What is the lower limit of an 80% confidence interval for the population mean, based on this data?

    Answer: $83.15

    To find the lower limit for the 80% confidence interval, you first have to find the margin of error. The formula for the margin of error when estimating a population mean is

    image7.jpg

    where z* is the value from the table for a given confidence level (80% in this case, or 1.28),

    image8.jpg

    is the standard deviation ($25), and n is the sample size (300).

    Now, substitute the values into the formula and solve:

    image9.jpg

    Next, subtract the MOE from the sample mean to find the lower limit: $85.00 – $1.85 = $83.15.

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