Question

JP Two professors at a local college developed a new teaching curriculum designed to increase students' grades in math classes. In a typical developmental math course, 43% of the students complete the course with a letter grade of A, B, or C. In the experimental course of the 17 students enrolled, 11 completed the course with a letter grade of A, B, or C. Is the experimental course effective at the a=0.1 level of significance? Complete parts (a) through (9) (a) State the appropriate null and alternative hypotheses Ho: P = 0.48 versus H P > 0.48 (Type integers or decimals. Do not round) (b) Verify that the normal model may not be used to estimate the P-value Because npo (1-P) = 42<10, the normal model may not be used to approximate the P-value (Round to one decimal place as needed.) (c) Explain why this is a binomial experiment. There is a fixed number of trials with two mutually exclusive outcomes. (Type an integer or a decimal. Do not round.) The trials are independent and the probability of success is fixed at 0.48 for each trial. (d) Determine the P-value using the binomial probability distribution. State your conclusion to the hypothesis test. First determine the P-value. P-value = 0.128 (Round to three decimal places as needed.) Is there sufficient evidence to support the research that the experimental course is effective? OA. Yes, reject the null hypothesis because the P-value is less than a. There is sufficient evidence to conclude that the experimental course is effective. OB. Yes, do not reject the null hypothesis because the P-value is greater than a. There is sufficient evidence to conclude that the experimental course is effective Xc No. reject the null hypothesis because the P-value is less than a. There is insufficient evidence to conclude that the experimental course is effective. * No, do not reject the null hypothesis because the P-value is greater than . There is insufficient evidence to conclude that the experimental course is effective te) Suppose the course is taught with 51 students and 33 complete the course with a letter grade of A, B. or C. Verify whether the normal model may now be used to estimate the P-value. 5% of the population size, and the sample the normal model be used to Because npo (1-P) - 10. the sample size is approximate the p-value (Round to one decimal place as needed Question Viewer Enter your answer in the answer box and then click Check Answer

Answer 1

Objective: To test whether the experimental course was effective in improving students' grades.

For this purpose, students were first separated into two groups - One group of them is made to undergo the typical developmental math program (T) and the other groups was subjected to the experimental course (E).

Let p_T, p_E denote the proportion of students who completed the course with grade A, B or C under the typical developmental math program (T) and the other groups was subjected to the experimental course. We would obtain sufficient evidence in favor of the experimental course if p_E is found to be significantly greater than P_T.

Given that p_T = 0.48; hence, it would suffice if we prove p_E to be greater than 0.48. From the given sample, by definition of proportion,

$\widehat{p}_{E}=$ No. of favorable cases / Total No. of cases

= x / n

= 11 / 17

  $\approx 0.65$

(a) To test: H₀ : p = 0.48 Vs H_a: p > 0.48

(b) We may use a normal approximation if np₀ (1-p₀) > 10. Here,

np₀ (1-p₀) = 17 x 0.48 x (1-0.48)

= 4.24 < 10.

Hence, we need to use the exact test instead of normal approximation to test the above hypothesis.

(c) The features of binomial distribution include - fixed independent repeated no. of trials, constant probability of success and the experiment resulting in two disjoint outcomes. Here, we find that:

There is a fixed no. of trials (n = 17), with two mutually exclusive outcomes (Obtain grade A, B or C (or) Not) and the probability of success is fixed at 0.48 for each trial.

(d) Hence, the normal model may not be used to approximate the p-value. Using the exact method, using Rstudio,

Console > binom.test(11,17,p=0.48, alternative = "greater") Exact binomial test data: 11 and 17 number of successes = 11, number of trials = 17, p-value = 0.1279 alternative hypothesis: true probability of success is greater than 0.48 95 per cent confidence interval: 0.4197054 1.0000000 sample estimates: probability of success 0.6470588

We get p-value = 0.128.

SInce, the p-value 0.128 > 0.10 is not significant, we fail to reject the null hypothesis. The correct option would be:

No. Do not reject the null hypothesis because the p-value is greater than alpha. There is insufficient evidence to conclude that the experimental course is better.

(e) When n = 51 and x = 32, the proportion becomes:

p = 32 / 51 = 0.63

Because np₀(1-p₀) = 51(0.63)(1-0.63) = 11.89 > 10, the sample size is less than 5% of the population and hence, the normal model can be used to approximate the p-value.