Prep Courses? Scores for men (nationwide) on the verbal portion of the SAT test are normally distributed with a mean of 509 and a standard deviation of 112. Randomly selected men are given the Columbia review course before taking the SAT test. After the course, a sample of 49 men revealed an average of 535 points and a standard deviation of 90 points. Using a significance level of 0.05, test the claim that the review course students have a mean score greater than or equal to the normal population. (use the 5-step method).
I don't know what your 5-step method is. Here is what I do.
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√(n-1)
If can only calculate one SEm, you can use just that to determine SEdiff.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to that Z score.
To test the claim that the review course students have a mean score greater than or equal to the normal population, we can use the 5-step hypothesis testing method. Here are the steps:
Step 1: State the null hypothesis (H0) and the alternative hypothesis (Ha).
In this case, the null hypothesis is that there is no difference between the mean scores of the review course students and the normal population. The alternative hypothesis is that the review course students have a mean score greater than the normal population.
H0: μ ≤ μ0
Ha: μ > μ0
Step 2: Choose the appropriate test statistic.
Since we are comparing the means of two samples (review course students and the normal population), and the population standard deviation is known, we can use a z-test.
Step 3: Determine the significance level (α).
The significance level (α) is given as 0.05, which means we will reject the null hypothesis if the p-value is less than 0.05.
Step 4: Calculate the test statistic.
The formula for the z-test is:
z = (x̄ - μ0) / (σ / √n)
where x̄ is the sample mean, μ0 is the population mean, σ is the population standard deviation, and n is the sample size.
In this case, x̄ = 535, μ0 = 509, σ = 112, and n = 49.
z = (535 - 509) / (112 / √49)
z = 26 / (112 / 7)
z = 1.857
Step 5: Determine the critical region and make a decision.
Since we are testing the claim that the mean score is greater than or equal to the normal population, we have a right-tailed test.
By looking up the critical value for a right-tailed test with α = 0.05 in the standard normal distribution table, we find that the critical value is 1.645 (approximately).
Since our test statistic (1.857) is greater than the critical value (1.645), we can reject the null hypothesis.
Therefore, we have enough evidence to support the claim that the review course students have a mean score greater than the normal population.
Note: This is a simplified explanation. In practice, it is important to perform the calculations accurately and interpret the results correctly.