The nominal resistance of a wire is 0.15 Ohm. Random testing of the wire stock yields the following resistance data: 0.148 0.147 0.15

Question

The nominal resistance of a wire is 0.15 Ohm. Random testing of the wire stock yields the following resistance data:
0.148 0.147 0.151 0.146 0.151 0.148 0.147 0.152
0.151 0.148 0.149 0.147 0.146 0.149 0.151 0.147
Does the sign test yield the conclusion, at the 5% significance level, that the median resistance is less than 0.15 Ohm?

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Thu Giang 4 years 2021-07-22T00:12:03+00:00 1 Answers 13 views 0

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  1. Cherry
    0
    2021-07-22T00:13:09+00:00
    Reply

    Answer:

    The median resistance of the wire is not small than 0,15 Ω at 0,05 of level of significance

    Step-by-step explanation:

    From data we get    x ( sample mean ) s  ( sample standard deviation) and n size of the sample

    0.148 0.147 0.151 0.146 0.151 0.148 0.147 0.152

    0.151 0.148 0.149 0.147 0.146 0.149 0.151 0.147

    n = 16

    x ≈ 0,1487

    s = 0,00193

    Test Hypothesis

    1.-Null Hypothesis               H₀                x  =  μ₀  = 0,15

    Alternative Hypothesis  Hₐ                 x < 0,15

    We assume data follows normal distribution

    as n = 16 we should use a t-student table

    As the question is : “Is the median resistance less than 0,15 0hm ” we must use  one-tail-test ( to the left)

    Then:

    2.-Significance level  α = 5 %      α = 0,05

    degree of freedom   n = 16     df = n – 1    df = 15

    From t-table we find t(c) = 1,7531      at the left is t(c) = – 1,7531

    3.-t(s) = ( x – 0,15 ) / s / √n

    t(s) = 0,1487 – 0,15 / 0,00193/√16

    t(s) = – 0,0013 * 4 / 0,00193

    t(s) = – 2,69

    4.- Comparing t(s) and t(c)  

    |t(s) | > |t(c)|       2,69 > 1,753

    Then t(s) is in the rejection region

    5.- We reject H₀  . At 95 % of confidence Interval

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