3. Researcher designs a study English for non-English-speaking people. The resea

Question

3. Researcher designs a study English for non-English-speaking people. The researcher randomly assigns 125 st class of 300 students to instruction instructed using traditional method. At the end of 6-month instruction pe given to compare the effectiveness of using computer instruction to teach udents from a using computer software. The remaining 175 students are riod, students are n an examination. Out of 125 computer-instruction students, 94 pass the examination, while 113 students pass in the traditional-instruction students. Does instruction using computer software appear to increase the proportion of students passing the examination? Use +0.05 to decide. What is the p-value? (10 pts)

Explanation / Answer

Data given:

For the students receiving instructions using computer software:

p1 = 94/125 = 0.752, n1 = 125

p2 = 113/175 = 0.646, n2 = 175

Calculating the standard error:

SE = ( p1*(1-p1)/n1 + p2*(1-p2)/n2 )0.5 = ( 0.752*(1-0.752)/125 + 0.646*(1-0.646)/175 )0.5 = 0.0529

Calculating the test statistic:

t = (p1-p2)/SE = (0.752-0.646)/0.0529 = 2.0037

Degrees of freedom = 175+125-2 = 298

The corresponding p-value for this score is:

p = 0.023

Since p < 0.05, so the result is significant.

Thus, instructions from computer software increase the proportion of students passing in the examination.

Hope this helps !

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