interquartile range would be unchanged.
- a. Using a normal model with μ = 16.1 and σ = 0.12, the z -score for . The z
-score for . P (15.7 < x < 16.3) = P (–3.33 < z < 1.67) = 0.048.
b. This is a binomial situation with n = 20 and p = 0.048 (from part a).
c. This is the sampling distribution of , which will be approximately normal with μ = μ = 16.1,
and .
- a. The mean of the distribution appears to be about 26%. This is the best estimate of the proportion
used in the simulation.
b. A sample proportion of 44% or more appeared only 3 times out of 100. Because that is less than
5% of the trials, this is convincing evidence that the proportion that appears in the article is too
low.
- a. Yes. The residual plot shows no pattern and the points show random scatter. A linear model is
appropriate.
b. The reliability of the estimates decreases as the x variable increases. This is evident from the fan
shape in the residual plot.
c. There is a definite curve in the residual plot, which shows that a linear model is not appropriate
for this relationship.
- This is a matched pairs design.
Hypotheses :
Check conditions :
Treatments are assigned to subjects in a random order.
The plot of the differences is reasonably symmetric with no outliers.
Conditions are met for a matched pairs t -test.
Computations :
