Section 5.2 The Sampling Distribution of the Sample Mean

Let be the mean of a SRS of size n from a population with mean µ and standard deviation σ. The mean and standard deviation of are:

Notice that = µ, which is the mean of the population. This makes an unbiased estimator of the population mean.

The standard deviation depends on two things – the standard deviation of the population, and also on the sample size. In order to make the variability smaller (and so guarantee that our estimate is more likely to be close to the truth) we should use a large sample.

Central Limit Theorem: Draw a SRS of size n from any population with mean µ and finite standard deviation σ. If the population is exactly normal, then so is the distribution of. If the population is not normal, then when n is large, the sampling distribution of the sample mean is approximately normal: is ≈

“Large” will depend on the shape of the population, but we can use the rule of thumb that if the population is not normal, then n should be at least 30 in order to say the distribution of is approximately normal.

Suppose we have a normal population with mean µ = 95 and σ = 27. If we take a SRS of size n = 9, what, if anything, can we say about the distribution of ? Because the pop’n is normal, the distribution of is also normal. Therefore, we can say that is N(µ, σ/√n) = N(95, 27/√9) = N(95, 9).

is N(µ, σ/√n) = N(95, 27/√9) = N(95, 9). This allows us to find P(90 ≤ ≤ 100), i.e., the probability that the sample mean (when n is 9) will be within 5 units of the population mean. z = – 0.56 and P(90 ≤ ≤ 100) = when n = 9.

If we take a SRS of size n = 49, what, if anything, can we say about the distribution of ? is N(µ, σ/√n) = N(95, 27/√49) = N(95, 3.857). Find P(90 ≤ ≤ 100).

is N(95, 3.857). Find P(90 ≤ ≤ 100). z = – 1.30 and P(90 ≤ ≤ 100) = when n = 49.

Yet P(90 ≤ ≤ 100) is about 81% when n = 49. P(90 ≤ ≤ 100) is about 42% when n = 9. Why would you expect these results? n = 9 n = 49

We increase our chances of getting a sample mean that is “close” to the truth (in this case, within 5 units of the population mean of 95) when we use a larger sample! n = 9 n = 49

According to the International Mass Retail Association, girls aged 13 to 17 spend an average of $31.20 on shopping trips in a month, with a standard deviation of $8.27. If 85 girls in that age category are randomly selected, what is the probability that their mean monthly shopping expense is between $30 and $33?

µ = and σ = 8.27 n = 85 We want P(30 < < 33) We do not know if the population is normally distributed or not, but with a sample size of 85, we know that will be approximately normal, with = µ = = σ ÷√n = 8.27 /√85

z left = (30 – 31.2) / (8.27 /√85) = – 1.34 z right = (33 – 31.2) / (8.27 /√85) = 2.01 Area = – =