Comparing Sample Means to Population Means

Suppose you took 100 unbiased random samples of the heights of U.S. women (recall that height is normally distributed), each sample containing 30 women. What can you say about the means of the samples $$(\bar{x}_{1},\bar{x}_{2},...\bar{x}_{100})$$ compared to the population mean?

Since height is normally distributed, we know that approximately 95% of women will have a height within two standard deviations of the mean (remember the Empirical Rule?). That means that out of 100 samples, we can assume that 95 of them will have a mean within 2 standard deviations of the population mean.

Predicting Population Means 

Suppose the mean of the means of our 100 samples from this example is 5′5″, in other words, $$\bar{x}$$=5′5′′. Within what range of heights can we expect the population mean to be, with 95% confidence? Assume a standard deviation of 1.5″.

Remember that since height is normally distributed, 95% of the values lie within 2 standard deviations of the mean, we need to identify that range of values.

  • First we need to use $$Z_{\frac{a}{2}}\times \frac{\sigma }{\sqrt{n}}$$ to identify the margin of error (since we are looking for a 95% confidence level, this is the range of values within 2 standard deviations of the sample mean). Since $$\sigma$$=1.5′′, in this case we get $$1.96 \times \frac{1.5}{\sqrt{100}} = 1.96 \times \frac{1.5}{10}=1.96\times0.15\approx 0.3"$$ above and below $$\bar{x}$$.
  • The interval then is 5′ 4.7″ to 5′ 5.3″, or .3 inches above and below the mean of 5′5″.

We can say that there is a 95% probability that the mean of our 100 samples would be within 0.3 inches either way of the population mean. Since the mean of our sample is 5′5″, we can say that the population mean is between 5′4.7″ and 5′5.3″ with 95% confidence.

Mathematically: 5′4.7′′< $$\mu$$ <5′5.3′′