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.
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.
5 answers
b. Use the P-value approach at the alpha=0.10
level of significance to test the hypotheses. Identify the P-value
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.
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.
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.
1 answer