Parts a-g!! Test: Chapter 9-10 Time Limit: 02:00:00 Submit Te his Question: 1 pt
ID: 3324755 • Letter: P
Question
Parts a-g!! Test: Chapter 9-10 Time Limit: 02:00:00 Submit Te his Question: 1 pt 11of25 (8 complete) This Test: 25 pts poss Question Help Two professors at a local college developed a new teaching arriculum designed to increase students' gades n math classes. In a typical developmental math course 55% of he sto ens complete the course with a letter grade of A, B, or C. In the experimental course, of the 17 students enrolled, 13 completed the course with a letter grade of A, B, or C. Is the experimental course effective at the 0.025 level of significance? Complete parts (a) through (g). (a) State the appropriate null and altenative hypotheses Type integers or decimals. Do not round.) b) Verity that the normal model may not be used to estimate the P-value Because npo(1-po) =L1 110. the normal model 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 Type an integer or a decimal. Do not round.) d) Determine the P-value using the binomial probability distribution. State your conclusion to the hypothesis test irst determine the P-value. independent and the probability of success is fed atfor each trial. | number of trials with The trials lick to select your answerfs).Explanation / Answer
(a) Ho : p 0.55
Ha : p > 0.55
(b)n Because np0(1-p0) = 17 * 0.55 * 0.45 = 4.2075 < 10 , the normal model cannot be used to approximate the p - value.
(c) There is fixed 17 number of trials with a give probability of 0.55. The trials are independent and the probability of success is fixed 0.55 for each trial./
(d) P - value = BIN (X > = 13; 17; 0.55) = 1 - BIN(X < 13; 17; 0.55) = 1 - 0.9404 = 0.060
(e) No, do not reject the null hypothesis because p - value is greater than alpha = 0.025. There is insufficient evidence to conclude. Option B is correct.
(f) Becuase, np0(1-p0) = 51* 0.55 * 0.45 = 12.6225> 10 , the sample size is less than 5% of population size, and the samples are independently drawn, The normal model can be used to approximate the p - value.
(g) Here
standard error of proportion = sqrt [po * ( 1- po)/N] = sqrt [0.55 * 0.45/ 51] = 0.0697
sample proportion p^ = 39/51 = 0.765
Test statistic
Z = (0.765 -0.55)/0.0697 = 3.085
p - vvalue = Pr(Z >3.085) = 1 -Pr(Z < 3.085) = 1 - 0.999 = 0.001
Yes, there is sufficient evidence to reject the null hypothesis because p - value is less than alpha. Option D is correct.