A group of Psychology students was interested in writing a paper
for their research class
concerning the difference between girls and boys and their
friendships. They determined
to study primary school aged children on the playgrounds of
schools. They hypothesized
that girls would pursue narrower, more intense friendship groups
than boys, that is, they
would interact with fewer other children than would the boys on
average. They selected
30 boys and 30 girls to watch at their schools. On average the
girls interacted with 3
other children during their recess (with a standard deviation of
1). The boys interacted
with 6 other children on average (with a standard deviation of 3).
Use this data to
answer the following questions.
. The appropriate null hypothesis would be
A. ☐ = 3
B. ☐ = 0
C. ☐ b- g= 3
D. ☐ b - g = 0
E. ☐d = 0
The appropriate alternative hypothesis would be
A. ☐ > 3
B. ☐ 0
C. ☐ b- g 0
D. ☐ b -g > 0
E. ☐g - b > 0
F. ☐d = 0
G. ☐b - g 3
The students (bless their hearts for thinking of this) determined
that they could
assume equal variances for boys and girls and so determined that
the appropriate df
for their problem was
A. ☐29
B. ☐28
C. ☐59
D. ☐58
E. ☐none of the above
Having never had statistics, they hired a consultant to calculate
the appropriate
standard error for their problem. Still assuming equal variances,
the consultant
determined that x-x was
A. ☐10
B. ☐.577
C. ☐.167
D. ☐3
E. ☐4
In order to do this test, the students had to look up a critical
value on the t table for
alpha=.01. They hire you to help them because you know how to use a
t table. You
determine that if df were 40 (it is not, but we will pretend it
is), the 1 tail critical t for
deciding between not rejection and rejection of the null is
A. ☐2.390
B. ☐2.660
C. ☐1.303
D. ☐1.684
E. ☐2.576
Let’s suppose that the students calculate their t score to be 1.833
(I made this up. It is
not the real value.) And suppose they think that their df=12 (It is
not). What should
they say about the one tail p value of their t statistic?
A. ☐p<.01
B. ☐p<.05
C. ☐p<.10
D. ☐p>.10
a).the null hypothesis is:-
[D]
b).the alternative hypothesis is:-
[ D]
[ as they hypothesized that girls would interact with fewer other children than that of the boys[
c).the df be: 58 (D)
[ here, we have,
for an equal variance case,
]
d)
Pooled variance sp2 = ((n1 - 1) s1^2 + (n2-1) s2^2)/(n1 + n2 - 2)
= ((30 - 1) * 1^2 + (30-1) * 3^2)/(30 + 30 - 2)
= 5
Standard error, =
= 0.577
B. 0.577 [answer]
e)
Critical value of t at alpha = 0.01 and df = 58 is 2.39
A. 2.390
f)
Test statistic = (6 - 3) / 0.577 = 5.2
Since the test statistic is greater than critical value.
A. p < 0.01
any doubts please ask thank you !:)
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