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Question 736:

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We can use a 1-sample z-test. We have a criterion against which we're using a sample to test a hypothesis and we know the population standard deviation.

 

  1. The Null Hypothesis is that the average household TV viewing per week is 26.4 hours: Ho : μ = 26.4
  2. The Alternative Hypothesis is that the average household TV viewing per week is not equal to 26.4 hours (its either more or less). Ha (or H1) : μ 26.4
  3. The test statistic for a 1-sample z-test is z*.
  4. For an alpha of .01 we need to find the critical value by looking up z* in a table of normal values or use the Excel function =NORMSINV(.99) and we get 2.3264, which is our critical value. So we need a test statistic to exceed this value to reject the Null Hypothesis.
  5. The test statistic z* is made up of a numerator which is the observed difference minus the test mean = 27-26.4 = .6. The denominator is how we account for chance fluctuations. It is the standard error of the mean difference (SE) which is the standard deviation divided by the square root of the sample size = 2.3/SQRT(58) = .30200. So the test statistic z* is .6/.30200 = 1.98 which is less than the critical value of 2.3264.
  6. Since the test statistic z* is less than the critical value we fail to reject the Null Hypothesis. In other words, we can conclude that there is not enough evidence to say that the average time watching TV per week is not different than 27 hours.

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