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.)
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
where
p(s) is the sample proportion of students enrolled experimental course and complete with a letter grade of A, B, or C ()
p is the success proportion assumed under null hypothesis. (0.45)
N is the sample size (20)
Then ≈4.045
Since the <em>one tailed</em> 19 <em>degrees of freedom </em>p-value is ≈0.0003<0.05 we can conclude that the experimental course effective at the alpha = 0.05 level of significance.
A compound interest equation is represented as A = P(1 + r/n)^nt, where n=the number of times per year the interest is compounded. In the given equation, n=12.