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Unformatted text preview: s a randomization test. He wrote the students’ grades
on 50 index cards, shuﬄed the cards and dealt them into two groups: cheat sheet and open book.
He then recorded the diﬀerence between the two group averages (¯cheat sheet − xopen book ) and
repeated 100 times. The dot plot below shows the distribution of these diﬀerences (denoted as
diffMean). 5 Calculate the p-value and determine if the data provide evidence to support the professor’s theory.
p-value = P(observed or more extreme outcome | H0 true)
= P(xcheat sheet − xopen book ≥ 6 | µcheat sheet = µopen book )
= 3 / 100 = 0.03
Yes, with p-value < 0.05, we reject H0 . The data provide convincing evidence to support the
professor’s theory that students who take the exam with just a cheat sheet score better on average.
9. We are interested in estimating the proportion of graduates at a mid-sized university who found a job
within one year of completing their undergraduate degree. We conduct a survey and ﬁnd out that
out of the 400 randomly sampled graduates 348 found jobs within within one year of completing their
(a) Deﬁne the sample statistic and the population parameter of interest. What is the value of the
The sample statistic is the proportion of graduates in the sample who found a job within one
year of graduating, p = 348 = 0.87. The population parameter of interest is the proportion of all
graduates from this university who found a job within one year of graduating, p.
(b) Are assumptions and conditions for inference satisﬁed?
1. Independence Assumption:
• Random Sampling Condition: We are told that the sample is random.
• 10% Condition: We can safely assume that 400 < 10% of all students at a mid-sized
Since we have a random sample and the 10% condition is satisﬁed, we can assume that whether
or not one student in the sample has found a job is independent of another.
2. Nearly Normal Condition: First we must check if the success failure condition is met.
np ≥ 10 → 400 ∗ 0.87 = 348 > 10
nq ≥ 10 → 400 ∗ 0.13 = 52 > 10
ˆ Since the observations are independent and the success-failure condition is met, we can assume
that p is nearly normal.
ˆ 6 (c) Construct a 95% conﬁdence interval for the proportion of graduates who who found a job within
one year of completing their undergraduate degree at this university.
A 95% conﬁdence interval can be calculated as follows:
p(1 − p)
0.87 × 0.13
= 0.87 ± 1.96 ×
= 0.87 ± 0.033
= (0.837, 0.903) (d) Explain what this interval means in the context of this question.
We are 95% conﬁdent that the true proportion of graduates from this university who found a job
within one year of completing their undergraduate degree is between 83.7% and 90.3%.
(e) What does “95% conﬁdence” mean?
95% of random samples of 400 would produce a conﬁdence interval that includes the true proportion of students at this university who...
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This document was uploaded on 12/04/2013.
- Winter '13