Let's break down and solve the problem step by step:

Given Data:

• Sample Scores (X): 110, 100, 110, 120, 110
• US National Average Reading Score: 100
• Test Type: One-tailed t-test
• Alpha (significance level): 5% or 0.05

a. Directional Hypothesis

We need to establish a hypothesis based on the scenario. Since we're comparing the reading scores of 3rd graders at Spring Run Elementary to the national average, the hypothesis will address whether their scores are higher than the national average.

• Null Hypothesis (H₀): The mean reading score of 3rd graders at Spring Run Elementary is equal to the national average.
  $H₀: \mu = 100$

• Alternative Hypothesis (H₁): The mean reading score of 3rd graders at Spring Run Elementary is greater than the national average.
  $H₁: \mu > 100$

This is a one-tailed test since we're only interested in whether the reading scores are higher than the national average.

b. Critical Value for a One-Tailed t-Test (alpha = 0.05)

We need the critical value for a one-tailed t-test at $\alpha = 0.05$.

• Degrees of Freedom (df): $n - 1 = 5 - 1 = 4$

Using a t-distribution table for a one-tailed test with 4 degrees of freedom at $\alpha = 0.05$, the critical value is approximately 2.132.

c. Perform a t-Test

To perform the t-test, we first calculate the test statistic using the following formula:

$t = \frac{\bar{X} - \mu_0}{\frac{s}{\sqrt{n}}}$

Where:

• $\bar{X}$ = Sample mean
• $\mu_0$ = Population mean (100)
• $s$ = Sample standard deviation
• $n$ = Sample size

Step 1: Calculate the Sample Mean ($\bar{X}$)

$\bar{X} = \frac{110 + 100 + 110 + 120 + 110}{5} = \frac{550}{5} = 110$

Step 2: Calculate the Sample Standard Deviation (s)

$s = \sqrt{\frac{\sum (X_i - \bar{X})^2}{n - 1}}$

$s = \sqrt{\frac{(110 - 110)^2 + (100 - 110)^2 + (110 - 110)^2 + (120 - 110)^2 + (110 - 110)^2}{4}}$ $s = \sqrt{\frac{0 + (-10)^2 + 0 + 10^2 + 0}{4}} = \sqrt{\frac{0 + 100 + 0 + 100 + 0}{4}} = \sqrt{\frac{200}{4}} = \sqrt{50} \approx 7.071$

Step 3: Calculate the t-Statistic

$t = \frac{110 - 100}{\frac{7.071}{\sqrt{5}}} = \frac{10}{\frac{7.071}{2.236}} = \frac{10}{3.162} \approx 3.162$

d. Was the Directional Hypothesis Supported?

• The calculated t-statistic is 3.162.
• The critical value at $\alpha = 0.05$ and 4 degrees of freedom is 2.132.

Since the calculated t-statistic (3.162) is greater than the critical value (2.132), we reject the null hypothesis. This means the directional hypothesis is supported — the mean reading score of 3rd graders at Spring Run Elementary is significantly higher than the national average.

Summary:

• The hypothesis that the mean reading score is greater than the national average is supported.

---

Would you like more details on any part of this solution?

Here are 5 related questions you might find helpful:

1. How does changing the significance level affect the critical value?
2. What is the effect of sample size on the t-test results?
3. How do you calculate the p-value from the t-statistic?
4. When would you use a two-tailed test instead of a one-tailed test?
5. How can you interpret a t-test when the sample standard deviation is large?

Tip: Always check if your hypothesis should be one-tailed or two-tailed before performing a t-test.