Question # or Section
Answer
Explanation
Part I.
z*
(Fill in the blank on your study sheet.)
Part II.
1-4.
False, False, True, False
5.
False
P-value in one-sided test exceeds 0.05.
6.
False
P-value in two-sided test exceeds 0.10.
7.
False
This could be a case of matched pairs (in which case we would use one-sample z or t procedures).
8.
True
This is nearly the only way you can correctly use the word "probability" in a sentence about confidence intervals.
9.
False
This is a common misconception.
10.
False
Since m.o.e. = (critical value)(s.e.), s.e. must equal m.o.e./(critical value) = 0.428/(z* or t*). If we knew this was a z confidence interval, our answer would be 0.428/1.645 = 0.260. Of course, we don’t even know for certain that this is a z confidence interval. We might need to be using t* instead of z*, but we wouldn’t get very far without knowing the df. There is not nearly enough information to answer this question.
11.
False
See explanation for #10. This value is a bit more plausible, since 0.778 divided by z* gives 0.473. However, the answer is still fallacious for two reasons: (1) the numerator should be m.o.e., not the upper endpoint of the C.I., and (2) as in #10, we might need to be using t* instead of z*. Again, there is not nearly enough information to answer this question.
12.
False
Although a 1-prop. z test is tempting (and reports a P-value of 0.0416, implying statistical significance), the assumptions are not satisfied. Specifically, np0 and nq0 are not at least 10. We must use the binomial distribution instead, which gives a P-value of 0.0730.
Part III.
0, 1/4, 2/3, 1.5, 4, B
We have discussed these issues before. The expected value is positive if you can eliminate one or more choices, but you are likely to fall into a trap if you make an educated guess. Either solve the problem completely, or make a random guess.
C
We know m.o.e. = (z*)(s.e.) < 2.57583Ö(0.5 · 0.5/n) by using a very conservative upper bound for the s.e.; this simplifies to m.o.e. < 1.2879/Ön, which we want to be < 0.02. Solving the inequality 1.2879/Ön < 0.02 gives n > 4146.8. In a situation like this, where we have an estimate for the probability of success (heads), it is also valid to use values of 0.4 for P(heads) and 0.6 for P(tails). The results are similar. (There would be a large difference only if P(heads) differed greatly from 0.5. The rule of thumb is that any value between 0.3 and 0.7 gives a pq product that is close enough that we may safely use 0.5 to estimate p and q.)
14.
B
Use a one-sample t confidence interval since we have matched pairs (plots paired off for control vs. experimental) and do not know the true s.d. of the difference between yields.
15.
E
This question tests your ability to read a table and apply common sense. Note that you must ignore most of the text and focus merely on what is asked. Soil and weather conditions differ between central Illinois and northern Indiana, which is something you can deduce by seeing that the Pioneer website is divided into regions by portions of states.
16.
B
At first, this question may appear to involve proportions since percentages are involved. However, the moisture content is simply another data value, like crop yield. Tests for proportions are valid only when you are considering the proportion of successes in a population, and moisture content does not fit that description. The moisture figure is by lot, not by individual. This is not at all like the classic example of conducting a political poll where each subject who supports the candidate of interest counts as a success.
To compute the P-value (namely, 0.00000008165), use a two-sided one-sample t test with n = 29, since once again, the test plots are matched side-by-side as matched pairs.
Part IV.
Respect the hypothesized (H0) values when conducting a test.
Use the best information available from the data when constructing a C.I.