Practice Test 3 – Answer Key
1.
(a) A strong tendency automatically rules out choices (b), (c), and (d). (The closer r is to 1 in
absolute value, the stronger the relationship.) The relationship here is positive since as the
height of one child increases, we would expect the height of the second child to also increase.
2.
(b) Correlation is independent of units (which rules out choices (a) & (c)), but is affected by
outliers.
3.
(d) The P-value is big in this case; therefore, we will fail to reject the null. Hence, choice (b)
cannot be correct. Choice (c) is incorrect because we cannot show that the null is true by using
a hypothesis test. (The P-value is not related to the margin of error.)
4.
(b) Statistically significant at the 1% level means that we would reject at this level. Therefore, we
need a P-value less than .01.
5.
(b) The hypotheses would be H0: μ ≥ 130; Ha: μ < 130. If we fail to reject H0, there is
insufficient evidence to reject H0 (μ ≥ 130) and conclude that μ < 130.
6.
(c) We want to show our city’s rate is different which automatically eliminates choices (b) and (d).
(Choice (b) would be correct if we wanted to show our city’s rate was less than the national
rate. Choice (d) would be correct if we wanted to show our city’s rate was greater than the
national rate.) Choice (a) represents 43%.
7.
(a) Using the t-distribution for a left-tailed test with α = 0.05 and d.f. = 9, we get tc = -1.833.
8.
(c) The “equal” value always goes in the null hypothesis.
9.
(b) Since this is a two-sided test and n ≥ 30, we want the critical values ±zc, where -zc corresponds
to a cumulative area of α/2 = 0.05. zc = -1.645 is the most accurate such z-score.
10. (e) A standard score of 2.5 would give us a P-value of .0062. This is smaller than 1% and 5%.
Therefore, we would reject the null hypothesis. Rejecting the null hypothesis also implies that
our results are statistically significant.
11. (b) By definition.
12. (d) Height is a quantitative variable so we need to use µ. “Over 6 feet” would mean that we want µ
to be greater than 6.
1090  1120
 2.16  0.0154 . Since this is a
13.86
35
2-sided test, we have P = 2(0.0154) = 0.0308.
13. (a)  x 
82
 13.86. Since n ≥ 30, z 
14. (c) The P-value gives the probability of making a Type I error. This means that the P-value is the
chance we would be wrong if we reject. (i.e. The possibility the results occurred by chance
alone.)
15. (a) The phrase “at least 12 ounces” implies that the hypotheses should be H0: μ ≥ 12; Ha: μ < 12.
The alternative hypothesis implies that this is a two-tailed test.
16. (c) By definition.
17. (a) Our hypotheses would be H0: μ = 47.4; Ha: μ ≠ 47.4. If we reject H0, then we reject the claim
that μ = 47.4 and conclude that μ ≠ 47.4.
18. μ = the mean waiting time (in minutes) for all customers at the fast food outlet.
H0: µ ≥ 4.9
n = 60 > 30  we can use a normal distribution
Ha : µ < 4.9
α = 0.05, left-tailed
Assume H0 is true (µ = 4.9).
4.8  4.9
.6
z
 1.29  0.0985 = P-value
x 
 .0775
.0775
60
Since P > 0.05, we fail to reject H0 and fail to conclude that the mean waiting time for all customers
is less than 4.9 minutes. (P = 0.0985)
19. μ = the average number of days worked by all
laborers using the employment agency
H0: µ = 120
α = 0.05
HA : µ ≠ 120
(Two-tailed)
Assume H0 is true (µ = 120)
45
x 107 n 100  x 
 4.5
100
z
107 120
  2.89  0.0019
4.5
P-value = 2(0.0019) = 0.0038 < 0.05
Reject H0 and conclude that µ ≠ 120.
The average number of day worked by
all laborers was not 120 (P = 0.0038).
20. This means that there is only a probability of .38% chance that this sample or one that was more
extreme would occur by chance and that the null hypothesis is correct. In other words, given this
sample result, there is only a .38% chance we have rejected the null hypothesis when we should not
have.
21. μ = the mean waiting time (in minutes) for all passengers catching a bus during rush hour.
H0: µ ≥ 7
n = 20 < 30, population normal  use a t-distribution
Ha : µ < 7
α = 0.01, left-tailed  Critical value: tc = -2.539
Assume H0 is true (µ = 7).
Rejection region: t < -2.539
5.2  7
2.1
t
 3.83
x 
 0.4696
0.4696
20
Since t = -3.83 lies in the rejection region, we reject H0 in favor of Ha and conclude that the
mean waiting time for all passengers is less than 7 minutes. (P < 0.005)
22. p = the proportion of all registered voters in the district that plan to vote for a Republican.
H0: p ≥ 0.5
Ha : p < 0.5
Assume H0 is true (p = 0.5).
α = 0.05, left-tailed
pˆ  579 / 1200  .4825
.4825  .5
(0.5)(0.5)
z
 1.22  0.1112 = P-value.
 .0144
.0144
1200
Since P > 0.05, we fail to reject H0 and fail to conclude that the Republican candidate will lose.
(P = 0.1112)
 pˆ 
23. a)
b)
c)
d)
Null: The new anesthetic does not result in lower death rates than the other anesthetics.
Alternative: The new anesthetic does result in lower death rates than the other anesthetics.
Type I: We conclude that the new anesthetic does result in lower death rates than the other
anesthetics when it actually does not.
Type II: We fail to conclude that the new anesthetic does result in lower death rates than the
other anesthetics when it actually does.