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14 October, 01:02

In a survey, 27 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $31 and standard deviation of $3. Construct a confidence interval at a 99% confidence level. Give your answers to one decimal place.

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  1. 14 October, 01:05
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    CI (29.4, 32.6)

    Step-by-step explanation:

    Given

    n = 27 (Size)

    X = 31 (Mean)

    sd = 3 (Standard deviation)

    Required to construct 99% Confidence Interval CI

    X±z*б/√n

    What we do have is the z value and to find it we need degrees of freedom and the significance level

    DF = 27-1 = 26

    SL = 100%-99% = 1%

    The z value in a two tailed test with (26,1%) = 2.779

    Back to formula

    Lower limit = 31 - 2.779 * 3/√27 = 29.4

    Upper limit = 32.6
  2. 14 October, 01:11
    0
    Answer: (29.4, 32.6)

    Step-by-step explanation:

    From the question, we know that

    Sample size (n) = 27

    Sample mean (x) = $31

    Sample standard deviation (s) = $3

    We are to construct a 99% confidence interval interval for average amount spent on gift.

    The formulae for constructing a 99% confidence interval for population mean is given as

    u = x + tα/2 * s/√n ... For upper limit

    u = x - tα/2 * s/√n ... For lower limit

    tα/2 is the critical value for a 2 tailed test. This value of gotten from a t distribution table by checking the degree of freedom (27 - 1 = 26) against the level of significance (100% - 99% = 1%).

    Hence tα/2 = 2.779

    Let us substitute our parameters and solve

    For lower limit

    u = 31 - 2.779 * 3/√27

    u = 31 - 2.779 (0.5773)

    u = 31 - 1.6045

    u = 29.4

    For upper limit

    u = 31 + 2.779 * 3/√27

    u = 31 + 2.779 (0.5773)

    u = 31 + 1.6045

    u = 32.6

    Hence the 99% confidence interval for population mean is given as 29.4, 32.6
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