Hypothesis Testing for Average Age of CSIT Students **Step 1: Set up Hypotheses** - **Null Hypothesis (H₀):** mu = 22 (The average age of enrolling students is 22 years) - **Alternative Hypothesis (H₁):** mu < 22 (The average age is less than 22 years) This is a **left-tailed test** at 5% level of s

Hypothesis Testing for Average Age of CSIT Students

Step 1: Set up Hypotheses

• Null Hypothesis (H₀): $\mu = 22$ (The average age of enrolling students is 22 years)
• Alternative Hypothesis (H₁): $\mu < 22$ (The average age is less than 22 years)

This is a left-tailed test at 5% level of significance.

Step 2: Sample Data

| 20 | 19 | 22 | 23 | 19 | 20 | 20 | 21 | 22 | 20 | 19 | 20 |

Sample size: $n = 12$

Step 3: Calculate Sample Statistics

Sample mean: $$\bar{x} = \frac{20 + 19 + 22 + 23 + 19 + 20 + 20 + 21 + 22 + 20 + 19 + 20}{12} = \frac{245}{12} = 20.417$$

Sample standard deviation: $$s = \sqrt{\frac{\sum(x_i - \bar{x})^2}{n-1}} = \sqrt{\frac{26.917}{11}} = \sqrt{2.447} = 1.564$$

Step 4: Test Statistic

Since the population is normally distributed and population standard deviation is unknown, we use the t-test:

$$t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} = \frac{20.417 - 22}{1.564/\sqrt{12}} = \frac{-1.583}{0.451} = -3.511$$

Step 5: Critical Value

For a left-tailed test with $\alpha = 0.05$ and $df = n - 1 = 11$: $$t_{0.05, 11} = -1.796$$

Step 6: Decision

Since $t = -3.511 < -1.796$, we reject the null hypothesis.

Conclusion: At the 5% level of significance, there is sufficient evidence to support the researcher's doubt that the average age of CSIT enrolling students is less than 22 years. [5]