Agricultural Statistics

Chi-square test for population variance:

• Given $n = 20$, sample variance $s^2 = 25$, and hypothesized population standard deviation $\sigma_0 = 8 \implies \sigma_0^2 = 64$. The chi-square test statistic is $\chi^2 = \frac{(n - 1)s^2}{\sigma_0^2} = \frac{(20 - 1) \times 25}{64} = \frac{19 \times 25}{64} = \frac{475}{64} \approx 7.4219 \approx 7.42$.
• $15.62$ results from using incorrect degrees of freedom.
• $51.20$ is an erroneous computation.
• $55.60$ is an incorrect distractor.

Hypothesis testing for correlation coefficient:

• To test $H_0: \rho = 0$, the t-statistic is $t = \frac{r \sqrt{n - 2}}{\sqrt{1 - r^2}} = \frac{0.2 \times \sqrt{623}}{\sqrt{1 - 0.04}} = \frac{0.2 \times 24.96}{\sqrt{0.96}} = \frac{4.992}{0.9798} \approx 5.095$. Since calculated $t = 5.095 \gt 1.96$ (critical table value at $5\%$ level), the correlation coefficient is highly statistically significant.
• The calculated t vastly exceeds $1.96$, so it cannot be non-significant.
• definite decision is reached.
• Large-sample t-test for correlation is perfectly valid.

Definition and properties of chi-square statistic:

• The formula for Pearson's chi-square is $\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}$. When observed frequencies equal expected frequencies ($O_i = E_i$) for every category, each deviation $(O_i - E_i) = 0$, making $\chi^2 = 0$ identically.
• $1$ occurs when deviations sum to unity.
• $2$ is incorrect.
• $0.5$ is incorrect.