- (B) They are comparing means without a known population standard deviation.
- (A) They are comparing means, so a t -test would be the appropriate test. The students were
randomly selected. With such large sample sizes, population shapes are not a concern. Conditions are
met.
- (D) That is the definition of a P -value.
- (B) The critical value of t for n – 1 = 17 degrees of freedom is ±2.11. His test statistic was not
outside those values.
- (C) Reducing measurement variability would give a lower standard error of the mean and, therefore,
a greater test statistic (in absolute value). Thus, the probability of rejecting the null hypothesis
increases.
- (C) This is a confidence interval for a mean with unknown s, so t for 27 degrees of freedom is used
rather than z . The standard error of the mean is .
- (D) As the number of degrees of freedom increases, the entire distribution is moved to the right, so
the critical value for the rejection region increases.
- (B) The official should use the hypothesized value of p , which is 0.5, in the calculation. Also, n =
200, not 108.
Solutions to Practice Test 1, Section II, Part A
- a. The 1.5IQR guideline can be used. IQR = 46.46 – 40.52 = 5.94. 1.5IQR = 8.91 bushels. The
largest possible nonoutlier would be 46.46 + 8.91 = 55.37 bushels. The maximum of 51.5 bushels
is below this value, so it is not an unusually high soybean yield.
(It would be acceptable to use the mean + 2 standard deviations, which is 43.555+2.3.937 =
51.429. Using this guideline, the maximum is more than 2 standard deviations from the mean, so
one might consider it unusual.)
b. Even if the treatment has no effect on soybean yield, the means of the groups will probably be
different due to random variability. If the treatment group’s yield was statistically significantly
higher than that of the control group, that means the difference is too large to be reasonably
attributed to chance.
c. The mean and median would both decrease by 0.5 bushels. The standard deviation and
