The problem asks for the $\chi^2$ (chi-square) value given the population variance, sample size, and the sum of squares of deviations from the sample mean.
The sample variance ($s^2$) is calculated using the sum of squares and the sample size ($n$). The formula is:
$ s^2 = \frac{\sum (x_i - \bar{x})^2}{n-1} $
Substitute the given values:
$ s^2 = \frac{120}{15-1} = \frac{120}{14} $
The formula for the chi-square statistic for testing population variance is:
$ \chi^2 = \frac{(n-1)s^2}{\sigma^2} $
Substitute the calculated sample variance and the given population variance:
$ \chi^2 = \frac{(15-1) \times \left(\frac{120}{14}\right)}{6} $
Simplify the expression:
$ \chi^2 = \frac{14 \times \left(\frac{120}{14}\right)}{6} $
$ \chi^2 = \frac{120}{6} $
$ \chi^2 = 20 $
The calculated $\chi^2$ value is 20.
| LIST-I | LIST-II | |
| A. One-Tailed Test | I. | Null hypothesis is rejected if the sample value is significantly higher or lower than the hypothesized value of the population parameter |
| B. Paired difference Test | II. | A hypothesis test of the difference between the sample means of two independent samples |
| C. Two-Tailed Test | III. | A sample value significantly above the hypothesized population value will lead to rejection of the null hypothesis |
| D. Upper-Tailed Test | IV. | Concerned only with whether the observed value deviates from the hypothesized value in one direction |