A college entrance exam company determined that a score of 20 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 20.8 on the college entrance exam with a standard deviation of 3.9 Do these results suggest that students who complete the core curriculum are ready for college-level mathematics? That is, are they scoring above 20 on the mathematics portion of the exam?
a. Identify the test statistic.
To determine if the results suggest that students who complete the core curriculum are ready for college-level mathematics, we need to conduct a hypothesis test comparing the sample mean to the population mean of 20.
The test statistic we will use in this case is the t-score.
b. Use the P-value approach at the alpha=0.10
level of significance to test the hypotheses. Identify the P-value
To determine the p-value, we need to conduct a one-sample t-test.
The null hypothesis (H0) is that the sample mean is equal to the population mean: μ = 20.
The alternative hypothesis (Ha) is that the sample mean is greater than the population mean: μ > 20.
Given that the sample size is large (n = 250), we can assume that the sampling distribution of the sample mean is approximately normal.
Using the formula for the t-score:
t = (sample mean - population mean) / (standard deviation / √sample size)
t = (20.8 - 20) / (3.9 / √250)
t = 0.8 / (3.9 / √250)
t ≈ 0.8 / (0.276639)
t ≈ 2.8945
From a t-distribution table, with a significance level of 0.10 and 249 degrees of freedom, we can find the critical value associated with this test statistic. The critical value for a one-tailed test at 0.10 level of significance is approximately 1.292.
Since the t-score of 2.8945 is greater than the critical value of 1.292, we can reject the null hypothesis.
Now we need to calculate the p-value to further analyze the results.
The p-value is the probability of obtaining a test statistic as extreme as the one observed, assuming the null hypothesis is true. In this case, the p-value represents the probability of obtaining a t-score higher than 2.8945.
To find the p-value, we can use a t-distribution table or a calculator. The p-value for a t-score of 2.8945 with 249 degrees of freedom is approximately 0.0018.
Therefore, the p-value is 0.0018.
I got t=3.243361703
Apologies for the confusion. If the correct t-score is 3.243361703, then we will need to recalculate the p-value.
Using the formula for the t-score:
t = (sample mean - population mean) / (standard deviation / √sample size)
t = (20.8 - 20) / (3.9 / √250)
t ≈ 3.243361703
Using a t-distribution table or a calculator, with 249 degrees of freedom, the p-value for a t-score of 3.243361703 is approximately 0.0007.
Therefore, the p-value is 0.0007.
To determine whether the students who complete the core curriculum are ready for college-level mathematics, we need to conduct a hypothesis test. The test statistic for this hypothesis test would be the z-score.
The formula to calculate the z-score is:
z = (X - μ) / σ
Where:
X = sample mean
μ = population mean
σ = population standard deviation
In this case:
X = 20.8 (sample mean)
μ = 20 (population mean)
σ = 3.9 (population standard deviation)
Using these values, we can calculate the z-score.