Sales Transactions

  1. α = 0.10.
  2. HA: μ > 120 and H0: μ = 120, where μ is the mean sales transaction for this company following the training.
  • The HA is “greater than” because they are trying to see if the mean sales transaction amount increased.
  • Remember to define the parameter.
  1. A 1-sample Z-test will be used because
    1. one group/population is considered (this company after the training),
    2. a quantitative variable (sales transaction) was recorded, and
    3. σ is known (=15).
  • These are the three characteristics that separate a 1-sample Z-test from other tests we will see this semester.
  • Please include some sort of description that shows that the characteristics are met and that you are not just listing the characteristics – in other words, include the parts in parentheses.
  • The σ will be given in the background information NOT in the table of results from the sample. The standard deviation in the table of results from the sample is s, not σ. If you know σ (as here) then you should use it as it is “better” than s.
  1. This study is observational as the researcher did not create groups or impose a treatment on any individuals. It is not clear or obvious whether the individuals were randomly selected or not.
  • The very vast majority of 1-sample tests will be observational, as experiments usually compare among treatments which requires two or more groups.
  • Do not say that the collection was random if it is not obviously so. Ultimately, this is a problem because the p-value requires randomization. However, we will continue with the analysis in this class, but we will be aware of the potential problems that this poses.
  1. The assumptions are met because (i) σ is known (=15) and (ii) n = 48 ≥ 30.
  • If there had been less than 30 then you would need to know the shape (strongly skewed, normal, etc.) of the population distribution.
  • Make sure to address whether σ is known or not, even though it feels redundant with Step 3.
  1. x̄ = $130.26.
  2. Z = \(\frac{130.26 = 120}{\frac{15}{\sqrt{48}}}\) = \(\frac{10.26}{2.1650635}\) = 4.739.
  3. p-value=0.000001 as computed with distrib(130.26,mean=120,sd=15/sqrt(48),lower.tail=FALSE) or distrib(4.739,lower.tail=FALSE).
  4. Reject H0 because the p-value < α.
  5. It appears that the mean sales transaction is greater than $120. Thus, it appears that the training did result in an increase in the mean sales transaction amount.
  • Note how the conclusion is about the MEAN sales transaction amount and not the sales transaction amount. The test is about the summary (i.e., the MEAN) not about individuals.
  • Also note the use of “appears” in the summary. This language acknowledges the fact that an error (Type I or Type II) could have been made. No conclusion from a statistical hypothesis test will ever be 100% definitive.
  • For the sake of learning, if we had not rejected the H0 then we would have concluded that the “mean transaction amount does not appear to be greater than $120”. Note that we would not have said that the mean transaction amount was $120, because it is possible that it was less than $120.


I performed the following steps or calculations in preparation for the next step.

  • 90% confidence (because α=0.1)
  • Lower confidence bound because HA was “greater than.”
  • Z*=-1.282 as computed with distrib(0.90,type="q",lower.tail=FALSE).
  • 130.26 -1.282*2.165 (where 2.165 is the SE computed for Step 7).
  1. I am 90% confident that the mean sales transaction amount for all sales after the training is greater than $127.49. Thus, the mean transacgtion amount appears to have increased by at least $7.49.
  • This confidence region is a lower bound, so the conclusion is that the mean is greater than the calculated amount.
  • The second sentence is a nice addition as it more directly answers the researcher’s research hypothesis.