Sampling Distributions for MeansWorksheet 1Open the interactive version of the normal tables. (Opens a separate browser window.) Obtain some pictures of normal curves. Print them out and use them to help solve problems. Each of these problems requires you to be familiar with the "Central Limit Theorem." This theorem -- which involves averages computed from random samples of data -- is described below. The basic setting is as follows:
The sample mean x-bar is the focus here. It is a variable (each random sample results in a different sample mean); as such it has a distribution.
Avoid using this result for situations in which the combination of both nonnormal data and small sample size are present. 1. A bottling company uses a filling machine to fill plastic bottles with a popular cola. The bottles are supposed to contain 300 millilters (ml). In fact, the contents vary according to a normal distribution with mean m = 303 ml and standard deviation s = 3 ml.
2. For 1998 as a whole, the mean return of all common stocks listed on the New York Stock Exchange (NYSE) was m = 16% and standard deviation s = 26%. Assume that the distribution of returns is roughly normal.
3. The length of human pregnancies from conception to birth varies according to a distribution that is approximately normal with mean 264 days and standard deviation 16 days. (See a previous worksheet for some questions involving this distribution.) Consider 15 pregnant women from a rural area. Assume they are equivalent to a random sample from all women.
Solutions(Note: sqrt(#) stands for "square root of #".) 1. a) 0.1587, b) mean: 303, stdev: 3/sqrt(10) = 0.94868, c) 0.0008 (1 in 1250 -- very unlikely). 2. a) 0.2692, b) mean: 16, stdev: 26/sqrt(8) = 9.1924, c) 0.0409 (1 in 24); this is the probability that the average of 8 randomly selected stocks loses money; the 0.2692 is the probability a single stock loses money, d) need z = 1.645. then 16 + 16.45*9.1924 = 31.12%. That is, 5% of these portfolios will make more than 31.12%. 3. a) mean: 264, stdev: 16/sqrt(15) = 4.1312, b) need z = 1.645 and z = -1.645; go 1.645 st dev. from mean in either direction. 264 + 1.645(4.1312) = 270.8; 264 - 16.45(4.1312) = 257.2. So, between 257.2 and 270.8, c) 0.0003 (1 in 3333), d) Assume the toxin has no effect on length of pregnancy -- the average length of pregnancy for all people (including people exposed to the toxin) is 264. The chance of an average length of pregnancy at least as low as the observed 250 is very remote -- it should occur in 1 in 3333 trials on average. This leads one to believe that perhaps the result isn't due to chance alone and, instead, that our assumption of 264 days on average is in question. (This result is "beyond a reasonable doubt.") |