Stat 301 – 2015 -- HW 4 answers (corrected)
1 point for each part of problems 1 and 5, 2 points each for 2, 3, and 4, and 2 points “for free”.
1. Attitudes to father and mother
a. Null: no difference in mean attitude toward father and toward mother
b. p = 0.30 No evidence of a difference in attitudes
c. (-0.32, 0.93)
Notes: These results are from a paired analysis, i.e. by computing the difference for each
individual, then analyzing those differences. The appropriate data file is fmattitudes.txt,
followed by computing the difference or using the JMP paired analysis dialog.
If you got p = 0.41 and (-0.46, 1.07), you treated the data as two samples (a total of 26 students,
not 13).
2. Yes, causal conclusions are appropriate. The treatment (Spiegel or control) was randomly
assigned to each subject.
Note: how subjects were recruited into the study (volunteers in Polk county) definitely affects
whether these results apply to a broad population (e.g., all US smokers) or something smaller.
But, causal conclusions are still appropriate for the study population.
3. No, causal conclusions are not appropriate. The “neighborhood poverty” characteristic was not
randomly assigned to neighborhoods.
Note: Neighborhood poverty just was or was not. The “poor” neighborhoods may differ in many
other ways, any of which might be the driving variables for reading proficiency.
4. Yes, causal conclusions are appropriate. The treatment (chemical method) is randomly assigned
to sections of meat.
Note: because the sections come from the same piece, the data should be analyzed using a
paired analysis. That is a different issue from whether treatments are randomly assigned.
5. Orange juice sweetness
a. Y = 6.25 – 0.0023 Pectin
b. The mean sweetness index in samples with no pectin is 6.25. An additional 1 ppm of
Pectin is associated with a 0.0023 decrease in sweetness index.
Note: You could express the slope many ways, each 100 ppm increase in Pectin is
associated with a 0.23 decrease in the sweetness index.
c. 5.56
Note: calculated as 6.25 – 0.0023*300
d. (-0.00419, -0.00043)
e. Yes, the p-value for the slope is 0.018 (Or, the 95% confidence interval does not include
0).
f. No, this is an observational study, so a causal conclusion (pectin did it) is not
appropriate
g. The estimated difference is 0.35 units of sweetness.
Note: calculated as -0.002311 * 150.
h. It is not possible to calculate this difference because the study is observational.
Note: g is asking to describe the difference; that can be predicted. Two runs differing by
150 pectin will also differ in everything else that changes with pectin level. h is asking
for a causal type estimate (change pectin and only pectin).
i. (5.22, 6.13)
Note: you want the prediction interval here because you are describing a single run.
Note: The first version of the HW 4 answers gave the answer as (5.10, 6.02), which is the
prediction interval for pectin = 300, not pectin = 250.