Does delaying oral practice hinder learning a foreign language? Researchers randomly assigned 16 beginning students of Russian to begin speaking practice immediately and another 16 to delay speaking for 4 weeks. At the end of the semester both groups took a standard test of comprehension of spoken Russian. Suppose that in the population of all beginning students, the test scores for early speaking vary according to the N(31, 7) distribution and scores for delayed speaking have the N(27, 8) distribution.
(a) What is the sampling distribution of the mean score x in the early speaking group in many repetitions of the experiment? (Round your answers for s to two decimal places.)
Mean | = 31 |
s | = 0.43 |
What is the sampling distribution of the mean score y in
the delayed speaking group?
Mean | =27 |
s | = 0.43 |
(b) If the experiment were repeated many times, what would be the
sampling distribution of the difference y -
xbetween the mean scores in the two groups? (Round your
answer for s to two decimal places.)
Mean | =-4 |
s | = 0.16 |
(c) What is the probability that the experiment will find
(misleadingly) that the mean score for delayed speaking is at least
as large as that for early speaking? (Round your answer to four
decimal places.)
=
please double check part a and b answers and do part C thank you
Solution:
(a) X:
mean=31
s = s/√n = 7/sqrt(16) = 1.75
Y:
mean=27
s=s/√n = 8/sqrt(16) = 2
-------------------------------------------------------------------------------------------------------------
(b) mean = 27-31 = -4
s = sqrt(7^2/16 +8^2/16) = 2.66
-------------------------------------------------------------------------------------------------------------
(c) So the probability is
P(X>0) = P((X-mean)/s >(0-(-4))/2.66)
=P(Z>1.50)
= 0.0668 (from standard normal table)
Get Answers For Free
Most questions answered within 1 hours.