Question

a college entrance exam company determined that a score of 24 on the mathematics portion of the exam suggests that 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 250 students who completed this core set of courses results in a mean math score of 24.6 on the college entrance exam with a standard deviation of 3.7. do these results suggest that students who complete the core curriculum are ready for college - level mathematics? that is, are they scoring above 24 on the mathematics portion of the exam? complete parts a) through d) below

the appropriate null and alternative hypotheses are ( h_0:mu = 24 ) versus ( h_1:mu>24 )

b) verify that the requirements to perform the test using the t - distribution are satisfied. check all that apply

a. the sample data come from a population that is approximately normal

b. the students were randomly sampled.

c. the sample size is larger than 30.

d. a boxplot of the sample data shows no outliers.

e. the students test scores were independent of one another

f. none of the requirements are satisfied.

c) use the p - value approach at the ( alpha = 0.10 ) level of significance to test the hypotheses in part (a).

identify the test statistic.

t_0=( square ) (round to two decimal places as needed.)

Explanation:

Step1: Recall the formula for the t - statistic

The formula for the t - statistic in a one - sample t - test is \(t_{0}=\frac{\bar{x}-\mu}{s/\sqrt{n}}\), where \(\bar{x}\) is the sample mean, \(\mu\) is the population mean under the null hypothesis, \(s\) is the sample standard deviation, and \(n\) is the sample size.

Step2: Identify the values from the problem

We are given that \(\bar{x} = 24.6\), \(\mu = 24\), \(s=3.7\), and \(n = 250\).

Step3: Calculate the denominator

First, calculate \(s/\sqrt{n}\). \(\sqrt{n}=\sqrt{250}\approx15.81\), and \(s/\sqrt{n}=\frac{3.7}{15.81}\approx0.234\).

Step4: Calculate the t - statistic

Now, use the formula \(t_{0}=\frac{\bar{x}-\mu}{s/\sqrt{n}}\). Substitute the values: \(t_{0}=\frac{24.6 - 24}{0.234}=\frac{0.6}{0.234}\approx2.56\).

Answer:

\(t_{0}\approx2.56\)