Final answer:
The question involves performing a one-sample t-test on the I.Q. of brown trout to determine if their mean I.Q. differs from 4. The calculated t-value is 1.95 with a p-value of 0.076, and with a significance level of 0.05, we fail to reject the null hypothesis.
Step-by-step explanation:
The student's question pertains to conducting a one-sample t-test to test the hypothesis about the mean I.Q. of brown trout. To address the specific question regarding whether the mean I.Q. of the brown trout is different from 4, we assume a null hypothesis that μ = 4 (the mean I.Q. is 4) and an alternative hypothesis that μ ≠ 4 (the mean I.Q. is not 4). Given the sample I.Q. scores of 12 brown trout, we calculate the sample mean (X) and perform a t-test to compare it against the hypothesized population mean. With a t-value of 1.95 and a p-value of 0.076, and considering the significance level (α) of 0.05, we would fail to reject the null hypothesis since the p-value is greater than the significance level, indicating that there is not enough evidence to conclude that the average I.Q. of brown trout is significantly different from 4.
As for the information provided without specific context, the SS within and SS between are concepts related to ANOVA (Analysis of Variance), which are used to assess differences among group means in a study. The k represents the number of groups or treatments, and i represents the number of subjects within each group.