Question

I need "e" option solution expert please solve option"e".

1) State College is evaluating a new English composition course for freshmen. A random sample of n = 25 freshmen is obtained and the students are placed in the course during their first semester. One year later, a writing sample is obtained for each student and the writing samples are graded using a standardized evaluation technique. The average score for the sample is M = 76. For the general population of college students, writing scores form a normal distribution with a mean of µ = 70.

a) Define the dependent and the independent variables

b) If the writing scores for the population have a standard deviation of σ = 20, does the obtained sample provide enough evidence to conclude that the new composition course has a significant effect on writing ability? Assume a two-tailed test with α = .05. State the null hypothesis, critical z-score(s) for rejecting/retaining H0. Evaluate any significant treatment effect you may obtain by calculating Cohen’s d. Conclude with a statement appropriate for a scientific journal

c) If the population standard deviation is σ = 10, does the sample provide enough evidence to conclude that the new composition course has a significant effect on writing ability? Again, assume a two-tailed test with σ = .05. Evaluate any significant treatment effect you may obtain by calculating Cohen’s d. That is, are you dealing with a small, medium, or large treatment effect?

d) Compare your answers to (b) and (c) above and explain how the magnitude of the population standard deviation influenced the outcome of the hypothesis test.

e) Assume that you expected a 6-point increase as a result of the the English composition course at the outset. What is the power of the test conducted in (c)?

Answer 1

E.

Critical z-value at α = .05 is ±1.96

Critical values of sample mean for the hypothesis test in part C can be calculated in the following way:

(M-μ)/σ /sqroot(n) = ±1.96

Lower critical M value:

(M-70)/10 /sqroot(25) = -1.96

M = 70 - 1.96*2

= 66.08

Upper critical M value:

(M-70)/10 /sqroot(25) = 1.96

M = 70 + 1.96*2

= 73.92

Null hypothesis would be failed to reject between these two limits.

Type 2 error = P(Fail to reject the null hypothesis|When the null hypothesis is false)

= P(66.08<M<73.92) (When μ = 76)

= P((66.08-76)/10 /sqroot(25) < Z < (73.92-76)/10 /sqroot(25))

= P(-4.96<Z<-1.04)

= 0.1492

Power of the test = 1-Type 2 error

= 0.8508

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