It is alleged that the mean marrying age of girls in locality of a certain province is 19 years old. A survey of random sample of 20 respondents has shown a mean marrying age of 16 years old with a standard deviation of 2 years. Has the age of the girls marrying in that locality gone lower? Use 0.05 level of significance.

To determine whether the mean marrying age of girls in the locality has gone lower than the alleged 19 years old, we can conduct a hypothesis test. Since the sample size is small (\( n = 20 \)) and the population standard deviation is unknown, we'll use a one-sample t-test for the mean.

### Step 1: Define the Hypotheses

- **Null Hypothesis (\( H_0 \))**: The mean marrying age is 19 years. (\( \mu = 19 \))
- **Alternative Hypothesis (\( H_a \))**: The mean marrying age is less than 19 years. (\( \mu < 19 \))

### Step 2: Level of Significance

The level of significance (\( \alpha \)) is 0.05.

### Step 3: Test Statistic

We use the t-test statistic given by:

\[
t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}}
\]

where:
- \( \bar{x} = 16 \) (sample mean)
- \( \mu = 19 \) (hypothesized population mean)
- \( s = 2 \) (sample standard deviation)
- \( n = 20 \) (sample size)

### Step 4: Calculate the Test Statistic

Substituting the values:

\[
t = \frac{16 - 19}{\frac{2}{\sqrt{20}}}
\]

Let's compute the t-value:It seems there was a technical issue calculating the test statistic. Let's manually compute it step-by-step:

### Calculating the t-value:

\[
t = \frac{16 - 19}{\frac{2}{\sqrt{20}}}
\]

1. The difference between the sample mean and hypothesized mean: \( 16 - 19 = -3 \).
2. Standard error (SE) is given by: 

   \[
   SE = \frac{2}{\sqrt{20}} = \frac{2}{4.472} \approx 0.4472
   \]

3. Now, calculate the t-value:

   \[
   t = \frac{-3}{0.4472} \approx -6.71
   \]

### Step 5: Determine the Critical Value and Decision

Using a t-distribution table with \( df = n - 1 = 19 \) degrees of freedom and a significance level of 0.05 for a one-tailed test:

- The critical t-value for \( \alpha = 0.05 \) and 19 degrees of freedom is approximately \(-1.729\).

### Conclusion

Since the calculated t-value (\(-6.71\)) is much less than the critical value (\(-1.729\)), we reject the null hypothesis.

**Conclusion**: There is sufficient evidence at the 0.05 level of significance to conclude that the mean marrying age of girls in the locality has gone lower than 19 years.

---

Would you like more details on any step, or do you have further questions? Here are some related questions:

1. How is the t-distribution different from the normal distribution?
2. What are the assumptions for using a t-test?
3. How does the level of significance affect hypothesis testing?
4. What steps should be taken if the sample size is larger?
5. How would the results differ if the standard deviation was higher?

**Tip**: Always check assumptions such as normality and randomness when conducting hypothesis tests to ensure valid results.

Conduct a hypothesis test to determine if the mean marrying age of girls in a certain locality has decreased from the alleged 19 years. A sample mean of 16 years with a standard deviation of 2 was obtained from 20 respondents. Learn step-by-step how to calculate the t-value and compare it to the critical t-value at the 0.05 significance level.

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