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Homework Help: Hypothesis Testing

  1. Apr 11, 2013 #1
    1. The problem statement, all variables and given/known data

    I have a few questions i have to solve. They are quite similar.

    1. Given the following, draw conclusions about the 10% and 5% levels of significance. Then find a p-value.

    [itex]n = 10,
    mean: 8.179,
    std. dev: 0.02,
    |mean - μ| = 0.021,
    z_{0.05} = 1.645,
    z_{0.025} = 1.96,
    H_{0}: μ = 8.2

    Using the formula from my textbook, i compute a value to compare to [itex]z_{0.05}[/itex] and end up with 3.3203. Since this is greater than 1.645, we reject the null hypothesis. This value is also greater than 1.96 so we reject the null hypothesis in both cases.

    Now, i have to compute a p-value. This is where it gets confusing because the textbook doesn't explain how i determine what sign ( ≤, ≥, >, < ) to use when i write it out. It seems to be completely arbitrary for all example questions i find. Is it based on the sign used in the null hypothesis?

    This is my best guess:

    P( mean ≤ 8.179 ) = [itex]\Phi(-3.32)[/itex] = 0.05%

    2. Given the following, draw conclusions about the 1% and 5% levels of significance. Then find a p-value.

    [itex]n = 64,
    mean: 756.4,
    std. dev: 33.9,
    |mean - μ| = 8.9,
    z_{0.05} = 1.645,
    z_{0.01} = 2.33,
    H_{0}: μ ≤ 747.5

    I compute the value to compare to 1.645 and i get 2.1, which is greater so we reject the null hypothesis. At the 1% significance level, we see that 2.1 is less than 2.33 so we accept the null hypothesis.

    For the p-value: P( mean ≥ 2.1 ) = [itex]\Phi(-2.1)[/itex] = 1.79%

    Again, this is just going off of a similar example that used ≥ for whatever unexplained reason. Possibly because my null hypothesis uses a ≤?

    3. Given the following, decide whether you accept or reject the null hypothesis.

    [itex]n = 10,
    mean = 26.4,
    S^{2} = 12.2667,
    std. dev ≈ S = 3.5,
    |mean - μ| = 3.6,
    H_{0}: μ ≥ 30

    Using the same formula from the prior questions we end up with 3.25. We aren't given a level of significance so i guess I have to compute a p-value then come up with a level of significance on our own?
    Last edited: Apr 11, 2013
  2. jcsd
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