Download Does too much sleep impair intellectual performance? Researchers

Does too much sleep impair intellectual performance?
Researchers examined this commonly held belief by
comparing the performance of subjects on the mornings
following (a) two normal night’s sleep and (b) two nights
of “extended sleep.” In the morning they were given a
number of tests of ability to think quickly and clearly.
One test was for vigilance where the lower the score, the
more vigilant the subject, (vigilance = alertness). The
following data were collected:
Subject 1 2 3 4 5 6 7 8 9 10
Normal
8 9 14 4 12 11 3 26 3 11 – L1
Extended 8 9 15 2 21 16 9 38 10 11 – L2
(L3 -> Difference of L2-L1)
1. If this experiment was properly designed, explain
how the assignment of subjects to treatments could have
been carried out.
2. Carry out an appropriate test to help answer the
researchers’ question.
Sleep and Performance
1. The subjects should all be tested under both
conditions, but randomly assigned to one of 2
groups:
Group 1: 2 nights of “normal sleep” followed by
“extended sleep”
Group 2: 2 nights of “extended sleep” followed
by “normal sleep”
and each should get a break in between to avoid
any effects from the previous sleep impairment.
Sleep and Performance
2. P: μ = mean difference in vigilance score (extended –
normal)
H: Ho: μ = 0, Ha: μ>0
A: SRS, Normality, Indep
T: x-bar = 3.8, s = 4.60, n = 10, df = 9
Table: .01 <p<.02
Calc: p = .015
C: Reject the Ho; there is significance evidence that the
scores for vigilance test were higher, on average, after
extended sleep.
ie: Too much sleep impairs intellectual performance.
12.2 Inference for a Population Proportion
• We are interested in the unknown
proportion p of a population that has some
outcome – call the outcome we are looking
for a “success.”
• The statistic that estimate the parameter p
is the sample proportion
p-hat = count of successes in sample
count of observations sample
Recall: The sampling distribution of p-hat…
• The mean of the sampling distribution is p.
• The sample proportion p-hat is an unbiased
estimator of the population proportion p.
• The standard deviation of p-hat is p(1  p)
n
Only
problem: we
don’t know
st. dev. of
p-hat since
don’t know p
Solution to not knowing p
To test the null hypothesis H 0 : p  p0 that
the unknown p has a specific value p0 ,
replace p by p0 in the test statistic:
z
pˆ  p
p (1  p )
n
z
pˆ  p0
p0 (1  p0 )
n
p: p̂ will
to get
In a confidence interval for
be
close to p if n is large, so replace the std.
dev. by the standard error of p̂ .
pˆ  z
*
pˆ (1  pˆ )
n
Assumptions for inference about a
proportion
• The data are an SRS from the population of
interest
• The population is at least 10 times as large as the
sample (Independence).
• For a test of H 0 : p  p0, the sample size n is
so large that: 10  np0 and 10  n(1  p0 )
(Normality)
• For a confidence interval, n is so large that
and 10  n(1  pˆ )
10  npˆ
(successes)
(failures)
Example
• A coin that is balanced should come up heads
half the time in the long run. The population for
coin tossing contains the results of tossing the
coin forever. The parameter p is the probability
of a head, which is the proportion of all tosses
that give a head. The tosses we actually make are
an SRS from this population. The French
naturalist Count Buffon (1707-1788) tossed a
coin 4040 times. He got 2048 heads. The sample
proportion of heads is 0.5069. That is more than
one-half.
• Is this evidence that Buffon’s coin was not
balanced?
• Find the 95% confidence interval.
Buffon’s Coin
• P: p = probability of a head
p-hat = 2048/4040 = .5069
• H: Ho: p = .5, Ha: p ≠ .5
• A: SRS, Indep, Normality
• T: Stat, Tests, 1-PropZTest (po:.5, x:2048,
n:4040, prop≠po) Calculate =>
z = .88, p-val = .3783
Stat, Tests, 1-PropZInt(x:2048, n:4040, C-level:
.95) Calculate => (.49151, .52235)
• C: Fail to reject Ho; we are 95% confident that
the probability of a heads is between above CI.
Recall: n for a desired margin of error
• To determine the sample size n that will
yield a level C confidence interval for a
population proportion p, solve the
following for n: * p* (1  p* )
z
n
m
• z * is the standard normal critical value for
level of confidence we want
• p* is a guess (or is from a study).
Many colleges that once enrolled only male or only female
students have become coeducational. Some administrators
and alumni were concerned that the academic standards of
the institutions would decrease with the change. One
formerly all-male college undertook a study of the first class
to contain women. The class consisted of 851 students, 214
of whom were women. An examination of first-semester
grades revealed that 15 of the top 30 students were female.
1) What is the proportion of women in the class? Call this
value p-nought.
2) Assume that the number of females in the top 30 is
approximately a binomial random variable with n=30 and
unknown probability p of success. In this case success
corresponds to the student being female. What is the value
of p-hat?
3) Are women more likely to be top students than their
proportion in the class would suggest? State hypotheses that
ask this question, carry out a significance test, and report
your conclusion in non- technical language.
College Study:
1. po = 214/851 = .2515
2. p-hat = 15/30 = .5
P: p = proportion of women who are top
students
3. H: Ho: p = .2515, Ha: p>.2515
A: SRS, Normality, Indep
z = (p-hat – po) / √[(po(1-po))/n] = 3.14
p-val=.00085
C: There is sufficient evidence to reject Ho and conclude that a
higher % of the women are among the top 30 students than would
be expected.
Methods of Poll Explained
• The Times-Dispatch/12 News poll was conducted
by the research department of Medial General,
Inc., parent company of the Times-Dispatch.
Based on telephone interviews October 23
through Wednesday with 502 respondents who
identified themselves as registered voters, the
survey had a sampling error of plus or minus 4.5
percentage points. In other words, one could say
with 95 percent certainty that the results of the
poll would vary 4.5 percent in either direction if
the entire adult population of Virginia had been
polled.
• Verify that the newspaper’s margin of error is
correct. What is the exact margin of error?
Methods Example:
Margin of error = z* √[(p*(1-p*))/n]
No p* given so use .5
m = 1.96 √[(.5)(.5)/502] = .044
=> The newspapers margin of error is
essentially correct.