We are starting true tests of hypotheses and it is critical to get off on the right foot. I will add some useful info here shortly, but first I am sharing a question on a homework problem that is a good example of what you will see in the second half of the course.

Question 7.2.27 from MyStatlab
The Question




What key information should you glean from reading the problem?

  1. This is a test for a single factor, the population mean, μ, which is a quantitative and not a categorical/qualitative variable.
  2. We are given the Claim that the mean, μ, equals (=) 50. Because the null hypothesis, Ho, always contains a form of equality (≤, =, ≥), the claim is the null hypothesis which is known as the ‘No difference’ hypothesis.
  3. The alternative hypothesis, Ha, is always the complement of the null. This means it always is a form of inequality, (<, ≠, >).
  4. The math operator in the alternative hypothesis always determines the ‘tail’ of the test we run. In this case, because the alternative contains the not = (≠) math operator, we have a two-tailed test.
  5. Because we have a two-tailed test, we must put half of alpha, α, in each tail to find the critical values. In this case, α =.06, α/2 = .03. You use either the tables or technology to find the critical value. I recommend you choose StatCrunch.
  6. Because we know sigma, the population standard deviation, and we are interested in a single quantitative variable, the mean, this is a z-test for the mean.

Now, how do we solve this problem?

  1. Using the StatCrunch normal (z) distribution calculator, and because we have a two-tailed test, we put the value of c = 1 – α = 0.94 in the area field of the “Between” option in the calculator to find the two critical values of z to be +/- 1.88. Note: if the area in the middle is 0.94, then there is 0.03 (α/2) in each tail.
  2. Using the Stat>Z-Stats>One Sample>with Summary (since we do not have raw data) sequence, we get the “One Sample Z Summary” dialog box.
    1. Put the sample mean, x-bar, in the “Sample Mean” box.
    2. Put the population sigma in the “Standard Deviation” box.
    3. Put the sample size n in the “Sample Size” box.
    4. Make sure the Hypothesis test for μ is selected in the “Perform:” area.
    5. Put the null hypothesis value of 50 in the “Ho: μ=” drop down box.
    6. Select the ≠ symbol in the “Ha: μ” box.
    7. Click “Compute!”
  3. The answer box (with the Options button) restates the test we have run. The standardized test statistic, Z-Stat, is -1.01, rounded to two decimal places.
  4. Because the standardized test statistic, -1.01, does not fall within either rejection area, we Fail to Reject the Null hypothesis and conclude “At the 6% significance level, there is not enough evidence to reject the claim.”
  5. Also, because the p-value of 0.314 is greater than the significance level of 0.06, we come to the same conclusion – Fail to Reject the Null, which is also the claim in this problem.

Here is a composite screenshot of the StatCrunch solution:

Using StatCrunch to solve the Problem
Using StatCrunch to Solve the Problem

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