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18 December, 21:18

A sample of 64 account balances from a credit company showed an average daily balance of $1,040. The standard deviation of the population is known to be $200. We are interested in determining if the mean of all account balances (i. e., population mean) is significantly different from $1,000.

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  1. 18 December, 22:24
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    With 95% of confidence, there is no statistical evidence to confirm that the mean of all account balances is different from $1,000.

    Explanation:

    Null hypothesis (H0) : μ=1000

    Alternative hypothesis (H1) : μ≠1000

    We must use the z distribution because we know the population standard deviation.

    z-statistic formula:

    z = (xbar-m) / (σ / (sqrt (n)))

    xbar: sample mean

    m: hypothesized value

    σ: population standard deviation

    n: number of observations

    z = (1,040-1,000) / (200/sqrt (64))

    z-statistic = 1.6

    Because the problem do not specify the significance level, we will use 5%

    The critical value from the z distribution with, 64-1 degrees of freedom and 2.5% significance level (because is a two-tail test), is 1,96.

    Because the z-statistic is less than the critical value, there is no statistical evidence to confirm that the mean of all account balances is different from $1,000.
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