Question

A college entrance exam company determined that a score of 23 on the mathematics portion of the exam suggests mat a student is ready for college-level mathematics. To achieve this goal, the company recommends that students take a core curriculum of math courses in high school. Suppose a random sample of 200 students who completed this core set of courses results n a mean math score of 23.6 on the college entrance exam with a standard deviation of 3.2. Do these results suggest mat students who complete the core curriculum are ready for college-level mathematics? That is, are they scoring above 23 on the math portion of the exam? Complete parts a through d below. a. State the appropriate null and alternative hypotheses. Choose the correct answer below i. H_0: mu = 23.6 versus H_1 = mu notequalto 23.6 ii.H_0: mu = 23.6 versus H_1 = mu > 23.6 iii.H_0: mu = 23.6 versus H_1 = mu < 23.6 iv. H_0: mu = 23.6 versus H_1 = mu > 23 v. H_0: mu = 23.6 versus H_1 = mu < 23 b. Verify that the requirements to perform the test using the t-distribution are satisfied. Is the sample obtained using simple random sampling or from a randomized experiment? i. Yes, because only high school students were sampled. ii. No, because not all students complete the courses.iii. No, because only high school students were sampled.iv. Yes, because the students were randomly sampled. Is the population from which the sample is drawn normally distributed or is the sample size, n, large (n Greaterthanorequalto 30)? i. No, neither of these conditions are true ii. Yes, the sample size is larger man 30 iii. Yes, the population is normally distributed It is impossible to determine using the given information c. Are the sampled values independent of each other? i. Yes, because each student's test score does not affect other students' test scores ii. No. because students from the same class will affect each other's performance iii. Yes, because the students each take their own tests iv. No, because every student takes the same test d. Use the P-value approach at the alpha = 0 10 level of significance to test the hypotheses. (Round to three decimal places as needed) State the conclusion for the test Choose the correct answer below i. Do not reject the null hypothesis because the P-value is greater than the alpha = 0 10 level of significance ii. Reject the null hypothesis because the P-value a less than the alpha = 0 10 level of significance iii. Do not reject (he null hypothesis because the P value is less than the alpha = 0 10 level of Significance. iv. Reject the nun hypothesis because the P-value greater than the alpha = 010 level of significance e. Write a conclusion based on the results. Choose the correct answer below. i. There is sufficient evidence to conclude that the population mean is greater than 23. ii. There is sufficient evidence to conclude that the population mean is less than 23 iii.There is not sufficient evidence to conclude that the population mean is greater than 23 iv. There is not sufficient evidence to conclude that the population mean is less man 23.

Answer 1

Answer:

a) The null hypothesis is represented as

H₀: μ ≤ 23

The alternative hypothesis is given as

Hₐ: μ > 23

b) Check the Explanation

The conditions for a t-test to be performed are satisfied or not?

- Yes, because the students were randomly sampled.

- Yes, the sample size is larger man 30.

And the central limit theorem allows us to approximate that the random sample obtained from the population is a normal distribution.

c) Are the sampled values independent of each other?

Yes, because each student's test score does not affect other students' test scores.

d) p-value obtained = 0.004

Reject the null hypothesis because the P-value a less than the alpha = 0.10 level of significance

e) There is sufficient evidence to conclude that the population mean is greater than 23.

Step-by-step explanation:

For hypothesis testing, the first thing to define is the null and alternative hypothesis.

The null hypothesis plays the devil's advocate and usually takes the form of the opposite of the theory to be tested. It usually contains the signs =, ≤ and ≥ depending on the directions of the test.

While, the alternative hypothesis usually confirms the the theory being tested by the experimental setup. It usually contains the signs ≠, < and > depending on the directions of the test.

For this question, we want to check if results suggest that students who complete the core curriculum are ready for college-level mathematics.

The only condition to be ready for college is scoring above 23.

So, the null hypothesis would be that the mean of test scores of students that complete core curriculum is less than or equal to 23. That is, there isn't significant evidence to conclude that the results suggest that students who complete the core curriculum are ready for college-level mathematics.

And the alternative hypothesis would be that there is significant evidence to conclude that the results suggest that students who complete the core curriculum are ready for college-level mathematics. That is, the mean score of those that complete the core curriculum is above 23 and are ready for college-level mathematics.

Mathematically

The null hypothesis is represented as

H₀: μ ≤ 23

The alternative hypothesis is given as

Hₐ: μ > 23

b) The conditions required before performing t-test.

- The sample should be a random sample

- The dependent variable should be approximately normally distributed.

- The observations are independent of one another.

- The dependent variable should not contain any outliers

All of these conditions are satisfied for our distribution.

c) Are the sampled values independent of each other?

Yes, because each student's test score does not affect other students' test scores.

d) To do this test, we will use the t-distribution because no information on the population standard deviation is known

So, we compute the t-test statistic

t = (x - μ₀)/σₓ

x = sample mean = 23.6

μ₀ = 23

σₓ = standard error = (σ/√n)

where n = Sample size = 200

σ = Sample standard deviation = 3.2

σₓ = (3.2/√200) = 0.226

t = (23.6 - 23) ÷ 0.226 = 2.65

checking the tables for the p-value of this t-statistic

- Degree of freedom = df = n - 1 = 200 - 1 = 199

- Significance level = 0.10

- The hypothesis test uses a one-tailed condition because we're testing only in one direction.

p-value (for t = 2.65, at 0.10 significance level, df = 199, with a one tailed condition) = 0.004348 = 0.004 to 3 d.p.

The interpretation of p-values is that

When the (p-value > significance level), we fail to reject the null hypothesis and when the (p-value < significance level), we reject the null hypothesis and accept the alternative hypothesis.

So, for this question, significance level = 0.10

p-value = 0.004

0.004 < 0.10

Hence,

p-value < significance level

This means that we reject the null hypothesis, accept the alternative hypothesis and say that there is significant evidence to conclude that the results suggest that students who complete the core curriculum are ready for college-level mathematics. That is, the mean score of those that complete the core curriculum is above 23 and are ready for college-level mathematics.

e) The result of the p-value obtained is that there is significant evidence to conclude that the results suggest that students who complete the core curriculum are ready for college-level mathematics. That is, the mean score of those that complete the core curriculum is above 23 and are ready for college-level mathematics.

Hope this Helps!!!