I have sampled from two separate populations, and I want to figure out which population is better.

Population 1 has an average score of 84.1 and a standard deviation of 11.8. Population 2 has an average score of 95.8 and a standard deviation of 13.5.

Assume that all samples are independent, and the sample size is 500 for each population.

What is the best way to prove that a random sample from population 2 will result in a greater score (on average) than a random sample from population 1?

What statistic do I need? Confidence interval? Tolerance interval?

  • $\begingroup$ You cannot prove that conclusion, because it is false: these statistics suffice to demonstrate the possibility that a random sample of Population 2 could have a smaller mean score than an independent random sample of Population 1. What you can do is to test the hypothesis that the population mean scores have the same value. $\endgroup$ – whuber Apr 16 '15 at 21:40

As the data is big enough, you can score of populations are taken,I assume scores made independently....

So, Normal Assumption is the Best for this situation



if $H_0$ is true, then $\bar{P_2}-\bar{P_1} \sim N(0,\frac{\sigma_{P_1}^2}{500}+\frac{\sigma_{P_2}^2}{500}) $

You can get $Z=\Bigg(\frac{(\bar{P_2}-\bar{P_1})-0}{\sqrt{\frac{\sigma_{P_1}^2}{500}+\frac{\sigma_{P_2}^2}{500}}}\Bigg)$

Now, you can check it out whether the difference is very high from the "Standard Normal Tables" at any error level,say, 5%(2.5%&97.5%)

Note:- suppose error-level taken as 5%(rejection criterion will be below -1.96 and above 1.96)....

So if Z is above 1.96, $P_2$ will be better. & if Z is below -1.96 $P_1$ will be better. Otherwise $H_0$ holds

  • $\begingroup$ I'm not sure if I fully understand. Is this a one-tailed test? Can you do an example please? $\endgroup$ – Greg Apr 16 '15 at 16:49
  • $\begingroup$ A one-tailed test is inappropriate for this question. When it is not known a priori which population might have the greater mean, a two-tailed test is needed. The best clue to interpretation is the key phrase, "figure out which population is better." $\endgroup$ – whuber Apr 16 '15 at 21:41
  • $\begingroup$ Ohh yes! Pardon, previously I misread! Now cleared :) $\endgroup$ – Hemant Rupani Apr 16 '15 at 22:02

If m1 and m2 are means, s1 and s2 are standard deviations and n1 and n2 are sample sizes, you can use following equations (code in R):

mean_diff = m1-m2
se <- sqrt( (s1^2/n1) + (s2^2/n2) )
df <- ( (s1^2/n1 + s2^2/n2)^2 )/( (s1^2/n1)^2/(n1-1) + (s2^2/n2)^2/(n2-1) )
t <- (m1-m2)/se 
P_value = 2*pt(-abs(t),df)

'pt' is a function in R to get P values. See ?pt for help.

  • $\begingroup$ Please express your formulas in math or English (or both). Relying on a computing language not only can be inaccurate, it also limits your audience unnecessarily. $\endgroup$ – whuber Apr 16 '15 at 21:42

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