Download Questions Question 14 Evolutionary theories often emphasize that

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Question 14
Evolutionary theories often emphasize that humans have adapted to their physical
environment. One such theory hypothesizes that people should spontaneously
follow a 24-hour cycle of sleeping and waking—even if they are not exposed to the
usual pattern of sunlight. To test this notion, eight paid volunteers were placed
(individually) in a room in which there was no light from the outside and no clocks
or other indications of time. They could turn the lights on and off as they wished.
After a month in the room, each individual tended to develop a steady cycle. Their
cycles at the end of the study were as follows: 25, 27, 25, 23, 24, 25, 26, and 25.Using
the 5% level of significance, what should we conclude about the theory that 24
hours is the natural cycle? (That is, does the average cycle length under these
conditions differ significantly from 24 hours?) (a) Use the steps of hypothesis
testing. (b) Sketch the distributions involved. (c) Explain your answer to someone
who has never taken a course in statistics.
Answer:
a)
Null hypothesis: µ = 24
Alternate hypothesis: µ ≠ 24
Sample mean = 25, Sample standard deviation = 1.195229
t-statistic = (25 – 24) / (1.195229/sqrt(8)) = 2.3664
t-critical = 2.3646
Since t-statistic is greater than critical value, we reject null hypothesis.
We conclude that natural cycle isn’t 24 hours.
b)
c) Given the significance level, we find the critical value which defines the
minimum difference between the mean and hypothesized value to make it
statistically significant. We find our t-score is just greater than critical t, so this small
difference is statistically significant, making natural cycle different from 24 hours.
Question 18
Twenty students randomly assigned to an experimental group receive an
instructional program; 30 in a control group do not. After 6 months, both groups are
tested on their knowledge. The experimental group has a mean of 38 on the test
(with an estimated population standard deviation of 3); the control group has a
mean of 35 (with an estimated population standard deviation of 5). Using the .05
level, what should the experimenter conclude? (a) Use the steps of hypothesis
testing, (b) sketch the distributions involved, and (c) explain your answer to
someone who is familiar with the t test for a single sample, but not with the t test for
independent means.
Answer
a)
Null hypothesis: µ1 = µ2
Alternate hypothesis: µ1 ≠ µ2
Pooled standard deviation = sqrt((19*3^2+29*5^2)/(20+30-2)) = 4.32
t-statistic = (38 – 30)/(4.32*sqrt(1/20+1/30)) = 2.4056
degree of freedom = 20+30-2 = 48
t-critical = 2.0106
As t-statistic is greater than critical t, we reject null hypothesis.
Scores for two groups are not same. The experimental group had seemingly higher
scores.
b)
c) T-test for two sample independent means is same as using the t-test for single
sample for the difference of two means. Only the standard error in this case is
computed slightly differently as shown above using pooled standard deviation.
Question 17
Do students at various colleges differ in how sociable they are? Twenty-five
students were randomly selected from each of three colleges in a particular region
and were asked to report on the amount of time they spent socializing each day with
other students. The results for College X was a mean of 5 hours and an estimated
population variance of 2 hours; for College Y, M = 4, S2 = 1.5; and for College Z, M =
6, S2 = 2.5. What should you conclude? Use the .05 level. (a) Use the steps of
hypothesis testing, (b) figure the effect size for the study; and (c) explain your
answers to (a) and (b) to someone who has never had a course in statistics.
Answer
a)
Null hypothesis: Mean time for each college is same
Alternate hypothesis: At least one mean is different
Group mean = (5+4+6)/3 = 5
Between group variance = 25 * ((5-5)^2+(4-5)^2+(6-5)^2)/(3-1) = 25
Within group variance = (24*2+24*1.5+24*2.5)/(25+25+25-3) = 2
F-statistic = 25/2 = 12.5
Numerator degree of freedom = 3-1=2
Denominator degree of freedom = 3*25-3 = 72
F-critical = 3.1239
Since F-statistic is greater than critical value, we reject null hypothesis.
Mean time of at least one college is different from others.
b) Effect size = (25*2) / (25*2+2*72) = 0.2577
c) F-statistic is ratio of between group variance to within group variance. As the
ratio increases, the probability of the means being different increases. Critical value
is found out using significance level and the two degree of freedoms as shown
above. Effect size shows times in one college can explain about 25.77% of the
variations in the times of other colleges.
Question 11
Make up a scatter diagram with 10 dots for each of the following situations: (a)
perfect positive linear correlation, (b) large but not perfect positive linear
correlation, (c) small positive linear correlation, (d) large but not perfect negative
linear correlation, (e) no correlation, (f) clear curvilinear correlation.
For problems 12, do the following: (a) Make a scatter diagram of the
scores; (b) describe in words the general pattern of correlation, if any; (c) figure
the correlation coefficient; (d) figure whether the correlation is statistically
significant (use the .05 significance level, two-tailed); (e) explain the logic of what
you have done, writing as if you are speaking to someone who has never heard
of correlation (but who does understand the mean, deviation scores, and hypothesis
testing); and (f) give three logically possible directions of causality, indicating for
each direction whether it is a reasonable explanation for the
correlation in light of the variables involved (and why).
Question 12
Four research participants take a test of manual dexterity (high scores mean
better dexterity) and an anxiety test (high scores mean more anxiety). The
scores are as follows.
Person-Dexterity-Anxiety
1
2
3
4
1
1
2
4
10
8
4
Negative 2
Answer
a)
b) There seems to be negative correlation.
c) Correlation coefficient r = -0.979958
d) t-statistic= r*sqrt((n-2)/(1-r²))= -0.979958*sqrt(2/(1-(-0.979958)^2))=-6.957
p-value = 0.02
So, the correlation is statistically significant.
e) Significance test as done above shows that the data above could not have
occurred by chance. The higher the value of ‘r’ is, the more the test statistic deviates
from zero. As the p-value is less than significance level, the correlation coefficient is
statistically significant.
f) High dexterity might cause less anxiety.
Or, High anxiety might cause less dexterity.
Or, a third external factor might simultaneously decrease dexterity and increase
anxiety or vice-versa.