Exercise 10-21

This is a radomized experiment.
The participants are not randomly selected out of all students from the company. The results can be applied to all students ‘like these’ in the experiment group.

pA: proportion of students like these with A as instructor that pass the exam
pB: proportion of students like these with B as instructor that pass the exam

The assumption is that A is more effective as B, so the alternative hypothesis is pA is greater than pB

H0: pA - pB = 0
HA: pA - pB > 0
\(\alpha\) = 0.05

Sample data:
nA = 50, xA = 30 and so \(\hat{p}\)A = 0.60
nB = 50, xB = 22 and so \(\hat{p}\)B = 0.44

Conditions check

  1. Independent random samples? Yes, because student’s are randomly assigned to instructor A or instructor B
  2. 10% condition; doesn’t have to be checked, because no sampling took place (see book, p. 625)
  3. Large Count condition
  • A group: number of successes (30) and number of failures (20) in the sample are both more than 10
  • B group: number of successes (22) and number of failures (28) in the sample are both more than 10

Test is allowed, test results:
z = 1.601; p-value = 0.055
p-value > \(\alpha\) so do not reject H0

The sample data do not provide convincing evidence for the hypothesis that the proportion of students from instructor A that pass the exam is higher than that of instructor B.
No support found for the believe that instructor A is more effective than instructor B.

Note
The believe of the owner that instructor A is more effective than instructor B is operationalized (made measurable) by comparing the proportions of students from each of the instructors that pass the exam.
We can think of other ways to compare the effectiveness of both instructors. For instance: how long does it on average take for a student from the first lesson to pass the exam.

Exercise 10-23

p1: actual rate for women like those in the study that get pregnant with intercession prayer
p2: actual rate for women like those in the study that get pregnant without intercession prayer
H0: p1 = p2
H1: p1 > p2

  1. The test is a 2-proportion Z-test
    Conditions are met:
  • random samples: the group is randomly divided intot two groups
  • 10% condition, not applicable, it is a randomized experiment; in case of a randomized experiment the conclusion hlds for a theoretical infinite group of elements like these in the experiment
  • LC condition, group 1: nr of successes 44 > 10, number of failures 44 > 10;
    group 2: nr of successes 21 > 10, number of failures 60 > 10
  1. P-value = 0.0007; the probability to find of a difference between the sample proportions equal or more than what is found - that is 44/80 - 21/81 = 0.291 by chance alone, is 0.0007.
  2. if \(\alpha\) = 0.05, than P-value < \(\alpha\) and H0 would be rejected; the difference between the proportions in the samples, provides convincing evidence that p1 > p2
  3. That the women didn’t know whether they were prayed for is important, because of a possible psychological effect that could have influence on becoming pregnant, or an effect on the way women behave that influences pregnancy probability

Exercise 10-24

p1: actual rate for women like those in the study that get pregnant with acupuncture
p2: actual rate for women like those in the study that get pregnant without acupuncture
H0: p1 = p2
H1: p1 > p2

  1. The test is a 2-proportion Z-test
    Conditions are met:
  • random samples: the group is randomly divided intot two groups
  • 10% condition, not applicable, it is a randomized experiment; in case of a randomized experiment the conclusion hlds for a theoretical infinite group of elements like these in the experiment
  • LC condition, group 1: nr of successes 34 > 10, number of failures 46 > 10;
    group 2: nr of successes 21 > 10, number of failures 59 > 10
  1. P-value = 0.0152; the probability to find of a difference between the sample proportions equal or more than what is found - that is 34/80 - 21/80 = 0.162 by chance alone, is 0.0152.
  2. if \(\alpha\) = 0.05, than P-value < \(\alpha\) and H0 would be rejected; the difference between the proportions in the samples, provides convincing evidence that p1 > p2
  3. That the women did know whether they received acupuncture is important, because of a possible psychological effect that could have influence on becoming pregnant, or an effect on the way women behave that influences pregnancy probability; the statistical relationship in that case doesn’t mean that there is a cause and effect (causal) relationship

Exercise 10-25

  1. Correct
  2. Wrong; the research question is whether there is a difference, no direction is mentioned
  3. Wrong; see (b)
  4. Wrong; H0 hypothesis must have an equal (‘=’) sign
  5. Wrong; see (d)

Exercise 10-26

  1. Wrong; if the P-value is less than the significance level, H0 is rejected
  2. Wrong; see (a)
  3. Wrong; H0 is rejected, this means there is convincig evidence for the HA hypothese, which states that the two proportions differ
  4. Wrong; the conclusion holds for the populations from which the samples are taken
  5. Correct

Exercise 10-27

The pooled or combined sample proportion for the two sample is: \(\hat{p}_C\) = \(\frac{410+484}{500+550}\) = 0.851.
The critical z-value for a 99% CI = 2.576

  1. Correct
  2. Wrong; this is the margin of error for a one-sample proportion case
  3. Wrong; the samples are not pooled
  4. Wrong, wrong z-value
  5. Wrong, wrong z-value

Exercise 10-28

  1. Wrong; z = 2.25 and P = 0.0123 is found if the test is performed; but the conditons to allow the use of this test are not met (LC is not met)
  2. Wrong
  3. Wrong
  4. Wrong; the rats are randomly divided into two groups
  5. Correct; for both groups the LC condition is not met

Note. This doesn’t mean that significance testing can not be used. What not may used is the Two Samples Proportion Z-test. A test that can be used is the so called Fisher Exact test. Another option is to make use of simulation.

1.0.0.1 Exercise T10.1

Correct answer: (e)

1.0.0.2 Exercise T10.2

sample A: nA = 125, xA = 68, \(\hat{p}_A\) = 35/125 = 0.28
sample B: nB = 125, xB = 68, \(\hat{p}_B\) = 17/68 = 0.25

Correct answer: (b)
(a) misses the critical z-value
(c) this is the formula for a 90% CI
(d) uses the pooled proportion which is used in hypothesis testing when it is is assumed that the two proportions are the same
(e) see (c) and (d)

1.0.0.3 Exercise T10.3

Correct answer: (a)

1.0.0.4 Exercise T10.4

Correct answer: (a)
Janelle’s method is theoretically correct. But because the formula for calculating df by hand is complicated, as a rule of thumb it’s allowed to use df = min(n1 - 1, n2 - 1). A lower number of df gives a higher t-value and so also a wider CI.

1.0.0.5 Exercise T10.5

Correct answer: (e)

1.0.0.6 Exercise T10.6

Correct answer: (e)
You can use your calculator to check the P-value.
P-value < \(\alpha\), so rejct H0; there is convincing evidence for the HA hypothesis, that is for the hypothesis that there is a difference in average time spent on extracurricular activities by students in the suburban and city school districts.

1.0.0.7 Exercise T10.7

Correct answer: (c)
Answer (a) is incorrect, because it suggests that support is found for the H0 hypothesis, but that is njot the case. The data do not contradict hte H0 hypothesis, that is not the same as proving it.

1.0.0.8 Exercise T10.8

Correct answer: (c)
mean \(\mu_\hat{p}\) = pA - pB = 0.8 - 0.5 = 0.3
sd \(\sigma_\hat{p}\) = \(\sqrt{\frac{p_A(1-p_A)}{n_A}+\frac{p_B(1-p_B)}{n_B})}\) = \(\sqrt{\frac{0.8\times 0.2}{100}+\frac{0.5\times 0.5}{400})}\) = 0.0471699

1.0.0.9 Exercise T10.9

Correct answer: (b)

1.0.0.10 Exercise T10.10

Correct answer: (a)