AngleWyrm said: So if I'm understanding that right, Standard Error for 15 samples: $=\frac{SD(X_{i})}{\sqrt{n}}=\frac{48.60}{\sqrt{15}} \simeq$ 12.55 Standard Error for 50 samples: $=\frac{SD(X_{i})}{\sqrt{n}}=\frac{48.60}{\sqrt{50}} \simeq$ 6.87 So then a z-score based on Standard Error would be $z_{15} = \frac{\text{observation - mean}}{\text{standard error}} = \frac{400-381.35}{12.55} \simeq$ 1.49 $z_{50}= \frac{\text{observation - mean}}{\text{standard error}} = \frac{400-381.35}{6.87} \simeq$ 2.71 Then this calculator describes the z scores in terms of percentage areas My first guess: z=1.49 confidence: 86.38%, two-tail region: 13.62% z=2.71 confidence: 99.33%, two-tail region: 0.67% So is it reasonable then to say with a sample size of 15 there's a 13.62% assurance the value 400 is drawn from the same population, but with a sample size of 50 that assurance drops to 0.67%? Or is this a one-tail problem? Also, when is it the right thing to do to use Standard Error vs Standard Deviation? Click to expand... Depending on which test you're performing, the test statistic is different. In this particular exercise, we're performing a Z-test, which uses the standard error.