Question

Students M and N sampled body lengths of two populations of a fish species using the same study design. The table below summarizes their data as mean \(\pm\) standard error.

Body length (cm)	Population I	Population II
Student M	\(8.0 \pm 1.2\)	\(6.2 \pm 0.7\)
Student N	\(10.0 \pm 0.6\)	\(5.1 \pm 0.3\)

Given that the data are normally distributed, when comparing the means of the populations using a t-test, which student will find a lower p-value?

• A. Student M
• B. Student N
• C. Insufficient information to comment
• D. p-values will be zero in both cases

Hint:

A smaller p-value comes from a bigger mean difference relative to a smaller combined standard error; compare that ratio for each student.

Answer 1

The Correct Option is B

Step 1: Understanding the Question.
Both students compared the same two fish populations with a t-test and report their body length data as mean \(\pm\) standard error (SE). A smaller p-value in a t-test comes from a larger t-statistic, which happens when the difference between the two means is large relative to the combined standard error of that difference. We need to work out whose data give the bigger t-statistic.

Step 2: Key Formula or Approach.
For comparing two independent means, the t-statistic is
\[ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{SE_1^2 + SE_2^2}} \]
A bigger numerator (bigger mean difference) or a smaller denominator (smaller combined SE) both push \(t\) up and the p-value down.

Step 3: Detailed Explanation.
For Student M: mean difference \(= 8.0 - 6.2 = 1.8\), and the combined SE is
\[ \sqrt{1.2^2 + 0.7^2} = \sqrt{1.44 + 0.49} = \sqrt{1.93} \approx 1.39 \]
So Student M's t-value is roughly
\[ t_M = \frac{1.8}{1.39} \approx 1.30 \]
For Student N: mean difference \(= 10.0 - 5.1 = 4.9\), and the combined SE is
\[ \sqrt{0.6^2 + 0.3^2} = \sqrt{0.36 + 0.09} = \sqrt{0.45} \approx 0.67 \]
So Student N's t-value is roughly
\[ t_N = \frac{4.9}{0.67} \approx 7.31 \]
Student N's data give a much bigger mean difference and a much smaller standard error at the same time, so Student N's t-value is far larger than Student M's. A larger t-value on the same degrees of freedom always gives a smaller p-value. Option (C) is wrong because both datasets give enough information, the sample means and the standard errors, to compute a t-statistic directly. Option (D) is wrong because a p-value is never exactly zero, it only gets very small.

Step 4: Final Answer.
Student N's populations show a bigger separation between means and tighter standard errors, so the t-test on Student N's data returns the lower p-value.
\[ \boxed{\text{Student N}} \]