Question

Does delaying oral practice hinder learning a foreign language? Researchers randomly assigned 23 beginning students of Russian to begin speaking practice immediately and another 23 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(29, 6) distribution and scores for delayed speaking have the N(28, 3) 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   =
1.3

Incorrect: Your answer is incorrect.
s   =

What is the sampling distribution of the mean score y in the delayed speaking group?
Mean   =
s   =

(b) If the experiment were repeated many times, what would be the sampling distribution of the difference y - x between the mean scores in the two groups? (Round your answer for s to two decimal places.)
Mean   =
s   =

(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.)

Answer 1

Consider the variables,

X₁ : Score for early speaking.

X₂ : Score for delayed speaking.

From the information

a) n₁ : number of students in early speaking group.

n₂: number of students in delayed speaking group.

Let X: Mean score for early speaking.

Y : mean score for delayed speaking.

By using result

If then sampling distribution of mean is

Hence sampling distribution of mean score X for early speaking group is

Mean = 29 and Variance = 6/23 and standard deviation S₁ = 0.5108

Mean = M₁ = 29

Standard Deviation = S₁=0.5108

and

sampling distribution of mean score Y for delayed speaking group is

Mean = 28 and Variance = 3/23 and standard deviation S₁ = 0.3612

Mean = M₂ = 28

Standard Deviation = S₂=0.3612

b) Y- X : Difference between mean score for two groups.

The sampling distribution of Y- X is

Mean( Y- X) = M₂- M₁ = 28 -29 = -1

Var(Y-X) = S₁² + S₂² = 6/23 + 3/23 = 0.3913

Standard deviation (Y-X) = sqrt(0.3913) = 0.6255

c) Required Probability = P ( Y - X > 0)

= P ( Z > 1.5987) since Z = (( Y-X) - E(Y-X)) / SD(Y-X) ~ N(0,1).

From normal probability table

P( Z > 1.5987) = 0.0549

P ( Mean score for delayed speaking is at least as large for that of early speaking) = 0.0549.