Note: When performing the 11 steps, make sure to …
  • Define the parameter μ in Step 2.
  • Provide some evidence for the characteristics of the test in Step 3 (note the items in parentheses).
  • The decision is about H0 in Step 9. However, in Step 10 you will likely be describing what you decided about HA. For example, in the first example below, Step 10 is saying that HA is not supported. However, the opposite is true in the second example. This happens because the researcher’s hypothesis (what they are trying to see if there is evidence for) is almost always in HA. Therefore, answering the researcher’s question requires you to say whether HA was supported or not.
  • In situations where the p-value is VERY small, I will often show the p-value as “<0.00005” (i.e., I round p-values to four decimals so if it is not 0.0001 then it must be less than 0.00005) rather than showing the scientific notation or the many decimal places.
  • You values may differ slightly from mine as I calculated everything in R that holds many decimal places on intermediate values. Note that you do not need to use R as a calculator as I did for the hand calculations, but you do need it for distrib().

Diet and Weight Gain

  1. α=0.05.
  2. HA: μ>5 and H0: μ=5, where μ is the mean weight gain for all women that followed the diet.
  3. 1-sample Z-test because (i) one group or population (women that followed this diet) was sampled, (ii) a quantitative variable (weight gain) was recorded, and (iii) σ is assumed to be known (=7.1).
  4. An observational study as the women chose to be on the diet. There is clear randomization in this study.
  5. The assumptions are met because (i) σ is known (=7.1) and (ii) n=35≥30.
  6. x̄=6.7 pounds gained.
  7. Z=\(\frac{6.7-5}{\frac{7.1}{\sqrt{35}}}\)=1.417.
  8. p-value=0.0783.
  9. Do not reject H0 because the p-value>α.
  10. There is not enough evidence to conclude that the mean weight gain for ALL women on this diet is greater than 5 pounds.
  11. I am 95% confident that the mean weight gain for ALL women is greater than \(6.7-1.645\frac{7.1}{\sqrt{35}}\)=4.73, which further indicates that the mean weight gain is not greater than 5 pounds.

R Appendix.

xbar <- 6.7; sigma <- 7.1; n <- 35
se <- sigma/sqrt(n)
( Z <- (xbar-5)/se )
( pv <- distrib(xbar,mean=5,sd=sigma/sqrt(n),lower.tail=FALSE) )
( pv <- distrib(Z,lower.tail=FALSE) )  # same as above (but uses test stat)
( Zstar <- distrib(0.95,lower.tail=FALSE,type="q") )
( lcb <- xbar+Zstar*se )

Cactus Growth

  1. α=0.05.
  2. H0: μ = 4, HA: μ < 4 where μ is the mean growth of all cacti.
  3. A one-sample Z-test is required because (i) a quantitative variable (growth) was measured, (ii) individuals from only one group/population were sampled (only one set of conditions considered), and (iii) σ is known (given in the background).
  4. The data are part of an experiment in which cacti were randomly selected.
  5. σ is known. In addition, because the population distribution is normal (stated in the background), we do not have to worry about the size of the sample.
  6. x¯ = 3.26.
  7. Z = \(\frac{3.26-4}{\frac{1.40}{\sqrt{10}}}\) = \(\frac{-0.74}{0.44}\) = -1.671.
  8. The p-value is 0.0473.
  9. H0 is rejected because the p-value<α.
  10. The mean growth rate of all cacti under the experimental conditions does appear to be less than 4 cm.
  11. One is 95% confident that the mean growth of all cacti under the experimental conditions is less than \(3.26+1.645\frac{1.40}{\sqrt{10}}\)=3.988 cm, which further indicates that mean growth is less than 4 cm.

R Appendix

> xbar <- 3.26; sigma <- 1.40; n <- 10
> se <- sigma/sqrt(n)
> ( Z <- (xbar-4)/se )
> ( pv <- distrib(xbar,mean=4,sd=sigma/sqrt(n)) )
> ( pv <- distrib(Z) )  # same as above (but uses test stat)
> ( Zstar <- distrib(0.95,type="q") )
> ( ucb <- xbar+Zstar*se )