Twenty students took a management course examination. The sample variance is found to be 80. The teacher claims that based on his past experinece the true variance has been 100. Does the sample result show that variance in the sample is significantly different ?

To determine whether the sample variance (80) is significantly different from the true variance claimed by the teacher (100), we can use a **Chi-Square test** for the population variance. This test is used to determine if there is a significant difference between the sample variance and the population variance.

### Step-by-Step Solution:

1. **Null and Alternative Hypotheses**:
   - Null Hypothesis \(H_0\): The population variance is equal to 100. \(\sigma^2 = 100\).
   - Alternative Hypothesis \(H_1\): The population variance is not equal to 100. \(\sigma^2 \neq 100\).

2. **Test Statistic**:
   The test statistic for the chi-square test is given by:

   \[
   \chi^2 = \frac{(n-1) \cdot s^2}{\sigma^2}
   \]

   where:
   - \(n\) is the sample size (20 students),
   - \(s^2\) is the sample variance (80),
   - \(\sigma^2\) is the claimed population variance (100).

   Substituting the values:

   \[
   \chi^2 = \frac{(20-1) \cdot 80}{100} = \frac{19 \cdot 80}{100} = \frac{1520}{100} = 15.2
   \]

3. **Degrees of Freedom**:
   The degrees of freedom (df) is \(n - 1 = 20 - 1 = 19\).

4. **Significance Level**:
   Typically, a significance level of \(\alpha = 0.05\) is chosen. This means we are testing with a 95% confidence level.

5. **Critical Values**:
   To determine the critical values, we use a chi-square distribution table for \(\alpha = 0.05\) with 19 degrees of freedom. For a two-tailed test:
   - The lower critical value is \(\chi^2_{0.025, 19} = 8.907\),
   - The upper critical value is \(\chi^2_{0.975, 19} = 30.144\).

6. **Decision**:
   - If the test statistic falls between the two critical values (i.e., \(8.907 \leq \chi^2 \leq 30.144\)), we fail to reject the null hypothesis.
   - If the test statistic falls outside this range, we reject the null hypothesis.

   Since the test statistic \(\chi^2 = 15.2\) falls between the critical values \(8.907\) and \(30.144\), **we fail to reject the null hypothesis**.

### Conclusion:
At the 5% significance level, the sample variance is not significantly different from the claimed population variance. Therefore, based on the sample result, there is not enough evidence to suggest that the variance in the sample is significantly different from the teacher's claim of a population variance of 100.

Would you like more details on this or have any further questions?

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Here are 5 related questions you might explore:

1. How would the conclusion change if the significance level was 1%?
2. What happens to the chi-square statistic if the sample size is doubled but the variances remain the same?
3. Can we apply the same test if the sample size was smaller, say 10 students?
4. How is this test related to a confidence interval for the population variance?
5. How does the chi-square distribution change as degrees of freedom increase?

**Tip**: Always check if your data is normally distributed when applying chi-square tests for variance comparison.

Twenty students took a management course examination. The sample variance is found to be 80. The teacher claims that based on his past experience the true variance has been 100. Does the sample result show that variance in the sample is significantly different?

Learn how to perform a Chi-Square test to determine if the sample variance of 80 is significantly different from the population variance of 100. This example walks through hypothesis testing, calculating the chi-square test statistic, and interpreting the results with a 5% significance level. Suitable for undergraduate statistics students.

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