Acollege 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?

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=0.65

H₁: p>0.65

The calculated statistic is Z=2.52 and the associated p-value: 0.0059

Remember:

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)=0.0059

There 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!

d. 24+-0.114

step-by-step explanation:

the question requires us to calculate the approximate 95% confidence interval for the mean number of ounces of ketchup per bottle in the sample given the following statistics;

sample mean = 24

standard deviation, σ = 0.4

sample size , n = 49

the confidence interval for a population mean is calculated using the formula;

sample mean ± z-score*(σ/sqrt(n))

the z-score associated with a 95% confidence interval, from the standard normal tables, is;

1.96

substituting these values into the formula;

24 ± 1.96*(0.4/7)

= 24 ± 0.112

from the alternatives given, choice d is closest to the above expression;

just sum

10 + 10 = 20