The level of nitrates was measured and recorded for 10 water samples taken at the bottom of a certain lake...

Understanding the Question

We need to determine whether the standard deviation of nitrate measurements from bottom water samples is greater than that from surface water samples.

Given Information:

• 10 water samples from the bottom of a lake
• 10 water samples from the surface of the same lake
• Nitrate levels were measured for each sample

What We Need to Determine:
This is a yes/no question: Is the bottom samples' standard deviation greater than the surface samples' standard deviation?

For sufficiency, we need to definitively answer either:

• YES: The bottom \(\mathrm{SD} > \mathrm{surface\ SD}\)
• NO: The bottom \(\mathrm{SD} \leq \mathrm{surface\ SD}\)

Key Insight: Standard deviation measures how spread out data points are from their mean. Two datasets can have very different standard deviations even if:

• One set has all values higher than the other
• One set has a larger range than the other

The critical factor is how the values are distributed within each dataset.

Analyzing Statement 1

Statement 1: The least of the measurements from the bottom samples exceeded the greatest of the measurements from the surface samples.

What This Tells Us:
The bottom and surface measurements have completely non-overlapping ranges:

• Every bottom measurement > Every surface measurement
• \(\mathrm{Minimum(bottom)} > \mathrm{Maximum(surface)}\)

Testing Different Scenarios:
However, this tells us nothing about the spread within each group. Let's test with concrete examples:

Scenario A - Surface spread out, bottom clustered:

• Surface: {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
  • \(\mathrm{Mean} = 5.5\), \(\mathrm{SD} \approx 2.87\)
• Bottom: {11, 11, 11, 11, 11, 12, 12, 12, 12, 12}
  • \(\mathrm{Mean} = 11.5\), \(\mathrm{SD} \approx 0.5\)
• Result: Surface SD (2.87) > Bottom SD (0.5) → Answer is NO

Scenario B - Surface clustered, bottom spread out:

• Surface: {9, 9, 9, 9, 9, 10, 10, 10, 10, 10}
  • \(\mathrm{Mean} = 9.5\), \(\mathrm{SD} \approx 0.5\)
• Bottom: {11, 12, 13, 14, 15, 16, 17, 18, 19, 20}
  • \(\mathrm{Mean} = 15.5\), \(\mathrm{SD} \approx 2.87\)
• Result: Bottom SD (2.87) > Surface SD (0.5) → Answer is YES

Conclusion: Since we can get different answers (YES in Scenario B, NO in Scenario A), Statement 1 alone is NOT sufficient.

[STOP - Not Sufficient!] This eliminates choices A and D.

Analyzing Statement 2

Now forget Statement 1 completely and analyze Statement 2 independently.

Statement 2: The range of the measurements from the bottom samples was greater than the range of the measurements from the surface samples.

What This Provides:
\(\mathrm{Range(bottom)} > \mathrm{Range(surface)}\), where \(\mathrm{range} = \mathrm{maximum\ value} - \mathrm{minimum\ value}\).

Key Insight: Range and standard deviation measure different aspects of spread:

• Range only considers the two extreme values
• Standard deviation considers how all values distribute around the mean

Testing Whether Larger Range Guarantees Larger SD:

Scenario A - Larger range but smaller SD:

• Surface: {1, 1, 1, 1, 1, 10, 10, 10, 10, 10}
  • \(\mathrm{Range} = 10 - 1 = 9\)
  • \(\mathrm{Mean} = 5.5\), \(\mathrm{SD} \approx 4.5\) (values clustered at extremes)
• Bottom: {1, 2, 3, 4, 5, 6, 7, 8, 9, 11}
  • \(\mathrm{Range} = 11 - 1 = 10\)
  • \(\mathrm{Mean} = 5.6\), \(\mathrm{SD} \approx 3.03\) (values evenly spread)
• Result: Bottom has larger range but smaller SD → Answer is NO

Scenario B - Larger range and larger SD:

• Surface: {5, 5, 5, 6, 6, 6, 6, 7, 7, 7}
  • \(\mathrm{Range} = 7 - 5 = 2\)
  • \(\mathrm{Mean} = 6\), \(\mathrm{SD} \approx 0.82\)
• Bottom: {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
  • \(\mathrm{Range} = 10 - 1 = 9\)
  • \(\mathrm{Mean} = 5.5\), \(\mathrm{SD} \approx 2.87\)
• Result: Bottom has both larger range and larger SD → Answer is YES

Conclusion: Since we can get different answers (NO in Scenario A, YES in Scenario B), Statement 2 alone is NOT sufficient.

[STOP - Not Sufficient!] This eliminates choice B.

Combining Both Statements

Combined Information:
Using both statements together, we know:

1. All bottom measurements > All surface measurements (Statement 1)
2. \(\mathrm{Range(bottom)} > \mathrm{Range(surface)}\) (Statement 2)

Testing if Together They're Sufficient:
Even with both constraints, we still cannot determine the relative standard deviations. Here's proof:

Example 1 - Surface SD > Bottom SD (despite both conditions):

• Surface: {1, 1, 1, 1, 1, 10, 10, 10, 10, 10}
  • \(\mathrm{Range} = 9\), \(\mathrm{SD} \approx 4.5\) (extreme clustering)
• Bottom: {11, 12, 13, 14, 15, 16, 17, 18, 19, 21}
  • \(\mathrm{Range} = 10\), \(\mathrm{SD} \approx 3.03\) (evenly spread)

Verification:

• ✓ All bottom values (11-21) > All surface values (1-10)
• ✓ Bottom range (10) > Surface range (9)
• Result: Surface SD > Bottom SD → Answer is NO

Example 2 - Bottom SD > Surface SD (with same conditions):

• Surface: {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
  • \(\mathrm{Range} = 9\), \(\mathrm{SD} \approx 2.87\)
• Bottom: {11, 11, 11, 11, 11, 21, 21, 21, 21, 21}
  • \(\mathrm{Range} = 10\), \(\mathrm{SD} \approx 5\) (extreme clustering at two values)

Verification:

• ✓ All bottom values (11 or 21) > All surface values (1-10)
• ✓ Bottom range (10) > Surface range (9)
• Result: Bottom SD > Surface SD → Answer is YES

Key Insight: Both examples satisfy both conditions, yet they give opposite answers to our question. The statements don't constrain the internal distribution patterns that determine standard deviation.

The statements together are NOT sufficient.

[STOP - Not Sufficient!] This eliminates choice C.

The Answer: E

The statements together are not sufficient because we can construct valid examples that satisfy both conditions but yield different answers about whether the bottom standard deviation exceeds the surface standard deviation.

Answer Choice E: "The statements together are not sufficient."