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Math Help - Population Problems

  1. #1
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    Question Population Problems

    1.) A recent study of 100 individuals revealed that in the US the mean amount of money spent on gifts for Mother’s Day for a given individual is $38. The population standard deviation is not known but is estimated to be $12.
    a.) Using a 90% confidence, what is the margin of error in estimating the mean?
    b.) Using your results from part (a), what is the 90% confidence interval for the mean? Interpret your results.
    c.) How many individuals need to be sampled in order to have a margin of error of $1.2 at 90% confidence

    2.) Suppose a survey reveals that 12 out of 1500 individuals are a twin.
    a.) What is the point estimate of the population proportion?
    b.) Using a 95% confidence, what is the margin of error in estimating the population proportion?
    c.) Using your results from part (b), what is the 95% confidence interval for the population proportion?

    3.) In the United States, the mean monthly electric bill is $250 per household. A sample of 100 households is taken and showed a sample mean of $247.60. Suppose the population standard deviation is $10.00. Suppose you wish to determine if the mean monthly electric bill is under $250 per household. Assume the probability of a Type I error is 0.05.
    a.) State the null and alternative hypotheses.
    b.) What is the value of the test statistic?
    c.) What is the p-value?
    d.) State your conclusions.
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  2. #2
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    Exclamation

    Here's what I've got, is it right?

    1.) A recent study of 100 individuals revealed that in the US the mean amount of money spent on gifts for Mother’s Day for a given individual is $38. The population standard deviation is not known but is estimated to be $12.
    a.) Using a 90% confidence, what is the margin of error in estimating the mean?
    • α = 1 – 0.90 = 0.10
    • p* = 1 – (0.10/2) = 0.95 ---> critical value of: 1.65
    • Margin of Error: 1.65 * 12/sqrt(100) = 1.98

    b.) Using your results from part (a), what is the 90% confidence interval for the mean? Interpret your results.
    • 38 +/- 1.98 or 36.02 to 39.98 or 36 to 40

    c.) How many individuals need to be sampled in order to have a margin of error of $1.2 at 90% confidence?
    • About 270




    2.) Suppose a survey reveals that 12 out of 1500 individuals are a twin.
    a.) What is the point estimate of the population proportion?
    • 12 / 15000 = .008 or .8%

    b.) Using a 95% confidence, what is the margin of error in estimating the population proportion?
    • α = 1 – 0.95 = 0.05
    • 1.96 x sqrt([(0.008)(0.992)]/1500) = 1.96 x 0.0023 = 0.004508

    c.) Using your results from part (b), what is the 95% confidence interval for the population proportion?
    • .008 +/- 0.004508 or .003492 to .012508

    3.) In the United States, the mean monthly electric bill is $250 per household. A sample of 100 households is taken and showed a sample mean of $247.60. Suppose the population standard deviation is $10.00. Suppose you wish to determine if the mean monthly electric bill is under $250 per household. Assume the probability of a Type I error is 0.05.
    a.) State the null and alternative hypotheses.
    • H0: μ < 250
    • H1: μ > 250
    b.) What is the value of the test statistic?
    • z = [250 – 247.60] / [10/sqrt(100)] = 2.4
    c.) What is the p-value?
    • For z = 2.4, Cumulative prob = .9932
    • 1-.9932 = .0068
    d.) State your conclusions.
    • Since H0 = .0068 < 0.05, we reject H0
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