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    Finding Sample Amount for Population Mean

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    In a study, the carapace lengths (in mm) of Thenus lobster caught near Singapore measured: 78,66,65,63,60,60,58,56,52 and 50 ( the population data is approximately normal).
    Find:
    (round to the nearest tenth)

    a. The sample mean
    b. The 80% confidence intervals for the population mean=
    c. The 50% confidence intervals for the population mean=
    d. The 98% confidence intervals for the population mean=
    e. How many would they need to sample to be 95% sure to be within 0.2 of the population mean?
    f. How many more (than the original 10) would they need to sample to be 86% sure to be withing .3 of the population mean?

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    https://brainmass.com/statistics/confidence-interval/finding-sample-amount-population-mean-533698

    Solution Preview

    Means and confidence intervals

    Answers
    a. The sample mean = 60.8
    Hint:

    X X - X̅ (X - X̅)^2
    78 17.2 295.84
    66 5.2 27.04
    65 4.2 17.64
    63 2.2 4.84
    60 -0.8 0.64
    60 -0.8 0.64
    58 -2.8 7.84
    56 -4.8 23.04
    52 -8.8 77.44
    50 -10.8 116.64
    608 571.61

    Sample mean, x̅ = sum of X/n = 608/10 = 60.8

    Sample standard deviation = √ 1/n-1 sum of (X_i -x̅)^2 = √571.6/10-1 = 7.969385868

    b. The 80% confidence intervals for the population mean = (57.3, 64.3)

    Hint:
    The 95% confidence interval is given by (x̅ - t_a/2, n-1 S /√n, x̅ + t_a/2, n-1 S/√N), where x̅ = 60.8, s = 7.969385868, n = 10, t_a/2 = 1.383028739
    That is, (60.8 - 1.383028739 * 7.969385868/ √10, 60.8 + 1.383028739 * 7.969385868/√10)
    = (57.3, 64.3)
    Thus with 80% confidence we can claim that population mean is within (57.3, 64.3).
    Details
    Confidence Interval Estimate for the Mean

    Data
    Sample Standard Deviation 7.969385868
    Sample Mean 60.8
    Sample Size 10
    Confidence Level 80%

    Intermediate Calculations
    Standard Error of the Mean 2.52014109
    Degrees of ...

    Solution Summary

    The expert finds the sample amount for population means.

    $2.19

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