A college chemistry instructor thinks the use of embedded tutors will improve the success rate in introductory chemistry courses. The passing rate for introductory chemistry is 6565​%. During one​ semester, 200200 students were enrolled in introductory chemistry courses with an embedded tutor. Of these 200200 ​students, 147147 passed the course. The instructor carried out a hypothesis test and found that the observed value of the test statistic was 2.522.52. The​ p-value associated with this test statistic is 0.00590.0059. Explain the meaning of the​ p-value in this context. Based on this​ result, should the instructor believe the success rate has​ improved?

Answer:Step-by-step explanation:Hello!The chemistry instructor tested the hypothesis that the proportion of students that passed the introductory chemistry class is better with an embedded. If the known proportion for this population is 65%, the tested hypothesis is:H₀: p=0.65H₁: p>0.65The calculated statistic is Z=2.52 and the associated p-value: 0.0059Remember: The p-value is defined as the probability corresponding to the calculated statistic if possible under the null hypothesis (i.e. the probability of obtaining a value as extreme as the value of the statistic under the null hypothesis). In this case:P(Z≥2.52)=0.0059There is no significance level, the most common one is α: 0.05 so I'll use it as an example.To make a decision using the p-value you have to compare it to the α.If p- value>α then you support the null hypothesis (In this case, you can say that there is no change in the proportion of students that passed the introductory chemistry class with an embedded tutor.)If p-value≤α your decision will be to reject the null hypothesis (In this case, there is significant evidence to say that there is an improvement in the success rate of the introductory chemistry class with an embedded tutor?Since the p-value:0.0059 is less than the significance level 0.05, you will decide to reject the null hypothesis.I hope you have a SUPER day!