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14 January, 12:41

The electric cooperative needs to know the mean household usage of electricity by its non-commercial customers in kWh per day. Assume that the population standard deviation is 1.8 kWh. The mean electricity usage per family was found to be 15.8 kWh per day for a sample of 872 families. Construct the 98% confidence interval for the mean usage of electricity. Round your answers to one decimal place.

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  1. 14 January, 13:08
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    Answer: (14.4, 17.2)

    Step-by-step explanation: We are to construct a 98% confidence interval for mean household usage of electricity.

    We have been given that

    Sample size (n) = 872

    Sample mean (x) = 15.8

    Population standard deviation (σ) = 1.8

    The formulae that defines the 98% confidence interval for mean is given below as

    u = x + Zα/2 * σ/√n ... For the upper limit

    u = x - Zα/2 * σ/√n ... For the lower limit

    Zα/2 = critical value for a 2% level of significance in a two tailed test = 2.33

    By substituting the parameters we have that

    For upper tailed

    u = 15.8 + 2.33 * (1.8/√872)

    u = 15.8 + 2.33 (0.6096)

    u = 15.8 + 1.4203

    u = 17.2

    For lower tailed

    u = 15.8 - 2.33 * (1.8/√872)

    u = 15.8 - 2.33 (0.6096)

    u = 15.8 - 1.4203

    u = 14.4

    Hence the 98% confidence interval for population mean usage of electricity is (14.4kwh, 17.2kwh)
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