Question:

How do you find a critical value when using a 1% significance?

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The data i'm working with is

Sample mean(voltage)=128.25

variance=9/20

sample size=20

SD=3

Z=-2.61

The first part of what I was doing was a hypothesis test that the true voltage is 130 compared to the alternative less than 130. Which I worked out the Z score to be -2.61

What I have to do now is calculate the critical value for this test using a 1% significance level.. I know the answer is 128.44 but for the life of me I can't work out any formulas that will help me... can anyone shed some light??? Thanks :)

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  1. (a) ANSWER: Yes! The "true average" voltage is less than 130.

    Why??

    SINGLE SAMPLE TEST, ONE-TAILED, 7 - Step Procedure for t Distributions,  SMALL SAMPLE SIZE

    1. Parameter of interest: "μ" = population mean voltage

    2. Null hypothesis Ho: μ = 130 volts

    3. Alternative hypothesis Ha: μ < 130 volts ("one-tailed" because "alternative less than 130")

    4. Test statistic formula: t = (x-bar - μ)/(s/SQRT(n))

    x-bar = estimate of the Population Mean (statistical mean of the sample) [128.25]

    n = number of individuals in the sample [20]

    s = sample standard deviation [3]

    Note: Discrepancy "variance=9/20" and "SD=3"

    (Sample) variance = (Sample) SD^2   [9/20 ≠ 3^2]

    μ = Population Mean [130] (used for Test statistic)

    5. Computation of Test statistic formula t = (approx) -2.61

    6. Determination of the P-value  based on n -1 = 19 df (degrees of freedom). Table "look-up" value shows area under the 19 df curve to the left of t = -2.61 is (approximately) 0.00863

    Excel Formula =TDIST(2.61,19,1) which is [0.008629962 ] the value in the "right-tail" is "P-value" [not "Z score"].    Note: Excel Formula requires "positive t" and  t-DISTRIBUTION is symmetrical and that "right-tail" equals "left-tail".

    SMALL SAMPLE SIZE requires "t-DISTRIBUTION", not "Z score" (Normal Distribution).

    7. Conclusion: with significance value α = 0.01 the above shows P-value <= α, [0.00863 <= 0.01]. Null hypothesis Ho: μ = 130 should be rejected. "true average" voltage is less than 130.

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