Question

Two professors at a local college developed a now teaching curriculum designed to increase students' grades in math classes. In a typical developmental math course, 45% of the students complete the course with a letter grade of A, B, or C. In the experimental course, of the 20 students enrolled, 18 completed the course with a letter grade of A, B, or C. Is the experimental course effective at the alpha = 0.05 level of significance? Identity the correct null and alternative hypotheses. H_0: p = 0.45 versus H_1: p > 0.45 Determine the P-value. P-value = (Round to three decimal places as needed.)

Answer 1

Answer:

We can conclude that the experimental course effective at the alpha = 0.05 level of significance.

Step-by-step explanation:

Let p be the proportion of students from experimental course who complete the course with a letter grade of A, B, or C. Null and alternative hypotheses are:

H_0: p = 0.45

H_1: p > 0.45

Test statistic can be found using the equation

$t=\frac{p(s)-p}{\sqrt{\frac{p*(1-p)}{N} } }$ where

• p(s) is the sample proportion of students enrolled experimental course and complete with a letter grade of A, B, or C ($\frac{18}{20} =0.9$)

• p is the success proportion assumed under null hypothesis. (0.45)

• N is the sample size (20)

Then $t=\frac{0.90-0.45}{\sqrt{\frac{0.45*0.55}{20} } }$ ≈4.045

Since the one tailed 19 degrees of freedom p-value is ≈0.0003<0.05 we can conclude that the experimental course effective at the alpha = 0.05 level of significance.