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13 March, 15:02

Here are summary statistics for randomly selected weights of newborn girls: nequals161 , x overbarequals32.8 hg, sequals7.2 hg. construct a confidence interval estimate of the mean. use a 90 % confidence level. are these results very different from the confidence interval 31.7 hgless thanmuless than34.5 hg with only 12 sample values, x overbarequals33.1 hg, and sequals2.7 hg? what is the confidence interval for the population mean mu ? 31.6 hgless thanmuless than 34.6 hg (round to one decimal place as needed.)

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  1. 13 March, 15:12
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    We have:

    n = 161

    xbar = 32.8hg

    s = 7.2hg

    cl = 90% = 0.90

    Since we have only the sample standard deviation s, we need to use the t-distribution.

    The degrees of freedom DF = 161 - 1 = 160

    α = (1 - 0.90) / 2 = 0.05

    If you look at these values in a t-distribution table you find t = 1.65

    Now we can build the confidence interval:

    xbar + / - (t · s / √n) = 32.8 + / - (1.65 · 7.2 / √161)

    Therefore:

    31.86 < μ < 33.74

    In order to understand if these values are different from the ones you are given, let's calculate the confidence interval for the latest:

    n = 12

    xbar = 33.1hg

    s = 2.7hg

    31.7 < μ < 34.5

    Calculate the error: (34.5 - 33.1) = (33.1 - 31.7) = 1.4

    We know t · s / √n = 1.4

    and we can solve for t:

    t = 1.4 · √n / s = 1.4 · √12 / 2.7 = 1.7962

    Looking at a t-distribution table, we find α = 0.05

    which brings to a confidence level of 90%, which is the same for the previous part.

    Since the sample size is big enough, we can use the normal distribution and the z-score:

    Looking at a normal distribution table, we find z-score = 1.645, which is very similar to the t-value found previously. We don't know the population standard deviation, but for such a big sample the sample standard deviation is a good estimate, therefore:

    xbar + / - (z * · s / √n) = 32.8 + / - (1.645 · 7.2 / √161)

    31.8 < μ < 33.7
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