Math Problem Statement

The mean SAT score in mathematics is 554. The standard deviation of these scores is 39. A special preparation course claims that the mean SAT score, μ, of its graduates is greater than 554. An independent researcher tests this by taking a random sample of 60 students who completed the course; the mean SAT score in mathematics for the sample was 567. At the 0.01 level of significance, can we conclude that the population mean SAT score for graduates of the course is greater than 554? Assume that the population standard deviation of the scores of course graduates is also 39.

Solution

Let’s break down and solve the hypothesis testing problem step by step.


Given Data:

  • Population mean (μ0\mu_0): 554
  • Sample mean (xˉ\bar{x}): 567
  • Population standard deviation (σ\sigma): 39
  • Sample size (nn): 60
  • Significance level (α\alpha): 0.01

We need to perform a one-tailed test to determine if the population mean SAT score for the preparation course graduates is greater than 554.


(a) Null and Alternative Hypotheses:

H0:μ=554(The mean SAT score is not greater than 554)H_0: \mu = 554 \quad (\text{The mean SAT score is not greater than 554}) H1:μ>554(The mean SAT score is greater than 554)H_1: \mu > 554 \quad (\text{The mean SAT score is greater than 554})


(b) Test Statistic Type:

Since the population standard deviation (σ\sigma) is known, we use the z-test.


(c) Calculate the Test Statistic:

The formula for the z-test statistic is:

z=xˉμ0σ/nz = \frac{\bar{x} - \mu_0}{\sigma / \sqrt{n}}

Substitute the given values:

z=56755439/60z = \frac{567 - 554}{39 / \sqrt{60}}

First, calculate the standard error (SE\text{SE}):

SE=σn=39605.032\text{SE} = \frac{\sigma}{\sqrt{n}} = \frac{39}{\sqrt{60}} \approx 5.032

Now, compute zz:

z=5675545.032135.0322.584z = \frac{567 - 554}{5.032} \approx \frac{13}{5.032} \approx 2.584

So, z2.584z \approx 2.584.


(d) Find the p-value:

For a one-tailed test, the p-value corresponds to the area to the right of z=2.584z = 2.584 in the standard normal distribution.

Using a z-table or statistical software, the p-value for z=2.584z = 2.584 is approximately:

p0.0049p \approx 0.0049


(e) Decision:

Compare the p-value to the significance level (α=0.01\alpha = 0.01):

  • p=0.0049<0.01p = 0.0049 < 0.01, so we reject the null hypothesis.

Conclusion:

Yes, we can support the preparation course’s claim that the population mean SAT score of its graduates is greater than 554.


Would you like further explanation of any step or need additional help with hypothesis testing concepts?


Related Questions:

  1. What is the role of a significance level in hypothesis testing?
  2. How does a one-tailed test differ from a two-tailed test?
  3. What assumptions are made when using the z-test?
  4. How would the result differ if the population standard deviation was unknown?
  5. Can you explain how to calculate a confidence interval for the mean SAT score?

Tip:

Always verify whether the sample size is large enough or the population follows a normal distribution to ensure the validity of the z-test.

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Math Problem Analysis

Mathematical Concepts

Hypothesis Testing
Z-Test
Significance Level

Formulas

z = (x̄ - μ₀) / (σ / √n)
p-value calculation for a one-tailed z-test

Theorems

Central Limit Theorem
Properties of the Standard Normal Distribution

Suitable Grade Level

College-Level Statistics