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    Statistical Confidence Intervals and One-sample tests

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    8.28
    Errors made: 2, 3, 5, 3 and 5

    a. How many different samples of 2 tellers are possible?
    Answer: 3 different samples of 2 tellers are possible: (2,3), (2,5) and (3,5)

    b. List all possible samples of size 2 and compute the mean of each.
    (2,3), (2, 5), (3,5)
    Samples mean
    x1_bar = (2+3)/2 = 2.5
    x2_bar = (2+5)/2 = 3.5
    x3_bar = (3+5)/2 = 4.0

    c. Compute the mean of the sample means and compare it to the population mean.
    Mean of samples mean
    X_bar = (x1_bar + x2_bar + x3_bar)/3 = (2.5+3.5+4.0)/3 = 10.0/3 = 3.33

    Population mean,
    mu = (2+3+5+3+5)/5 = 18/5 = 3.6

    Comparison
    X_bar - mu = 3.33 - 3.60 = -0.27

    Mean of samples mean is less than population mean by 0.27.

    8.31
    mean, mu = 35 hours
    standard deviation, sd = 5.5 hours
    sample size, n = 25

    a. What can you say about the shape of the distribution of the sample mean?
    The sample mean is Normal distributed, --Answer
    with average value, X_bar = mu = 35 hours --Answer
    and variance, V = sd^2/n = 5.5^2/25 = 1.21

    b. What is the standard error of the distribution of the sample mean?
    Standard error, SE = sqrt(V) = sqrt(1.21) = 1.1 --Answer

    c. What proportion of the samples will have a mean useful life of more than 36 hours?
    x = 36
    P(X >= x) == P(Z >= (x - X_bar)/SE)
    = P(Z >= (36 - 35.5)/1.1)
    = P(Z >= 0.45)
    = 0.5000 - 0.1736 [From N-table with upper half]
    = 0.3264

    Hence, 0.3264 proportion or 32.64% of the ...

    Solution Summary

    A few problems related to probability and confidence intervals are solved here.

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