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28 April, 20:29

Supposethat z is a normally distributed variable with variance 4. You collect a sample of size n for which you get a sample average value of 3. You are asked to test the null hypothesis that the mean is smaller than 2. Find the sample size n above which you can reject the null hypothesiswith a 95% of confidence.

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  1. 29 April, 00:02
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    The sample size is 4

    Step-by-step explanation:

    Null hypothesis: The mean is 2

    Alternate hypothesis: The mean is less than 2

    Mean = 3

    sd = sqrt (variance) = sqrt (4) = 2

    At 95% confidence level, t-value is 1.960

    Assuming the lower bound of the mean is 1.04

    Lower bound = mean - (t*sd/√n)

    1.04 = 3 - (1.96*2/√n)

    3.92/√n = 3 - 1.04

    3.92/√n = 1.96

    √n = 3.92/1.96

    √n = 2

    n = 2^2

    n = 4
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