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The concept of sampling distribution, where a random sample is drawn from a population to estimate population statistics such as mean age and proportion of smokers. Why random sampling is used, the statistics to calculate, and the concept of sampling variability. It also discusses the sampling distribution of the sample mean and proportion, and how they can be used to solve problems.
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Math 1530 –Lab- Introducing the idea of Sampling distribution (Chapter 18) Drawing a random sample IS a random experiment Imagine you have a population of individuals and you will select a random sample of size n to ask them a few questions, for example their age and if they are or have been smokers in some point of their life. Before drawing the sample we know n of them are going to be in the sample but we don’t know exactly WHO is going to be in the sample.
answers. For example if the answers to the question ‘Have you smoked more than 100 cigarettes in your life?’ are : YES , NO , YES, NO, NO ; the values of Y would be 1,0,1,0,
n y y n i i ^1
The sample proportion can be understood also as the sample mean of a variable that only takes values 1 and 0 (for success and failure, respectively) Below you see the distribution of age for the population. The population mean 42.92 is marked with an arrow. Mark (in the X axis) the values of the sample means for the two samples you got. How far were the means in the samples from the population mean? 15 25 35 45 55 65 75 85 95 15 10 5 0 Age Fr eq ue nc y Age (in years) of 50 individuals Pop mean We know that a proportion only can take values between 0 and 1. Below, in a line that goes from 0 to 1 we have marked the proportion of smokers in this small population (40% of the 50 individuals are or have been smokers). In the same graph, mark the proportion of smokers in the two samples you obtained. 0 1 0.
6.1) Problem 13 on page 351 of Intro Stats by DeVeaux & Velleman says :“When a truckload of apples arrives at a packing plant, a random sample of 150 is selected and examined for bruises, discoloration and other defects. The whole truckload will be rejected if more than 5% of the sample is unsatisfactory. Suppose that in fact 8% of the apples on the truck do not meet the desired standard. What’s the probability that the shipment will be accepted anyway?” So 8% of the apples in the truckload are not good but we don’t know that because we only examine a random sample of 150 apples and maybe by chance those are in a better condition. First, check if the assumptions necessary to use the normal model are fulfilled. a) Is n no larger than 10% of the population. In this case n=150 , it is reasonable to think that a whole truckload has more than 1500 apples. So first condition is fulfilled b) Is np>10? Is n(1-p)>10In this case n=150 and p=0.08 so np=12 and n(1-p)=150*0.92=138. So second condition is fulfilled The distribution of the sample proportion can be assumed to be approximately normal with mean 0.08 and standard deviation
= 0.0221510 The question is P(accepting the shipment even when 8% of
Sketch a normal distribution and shade in the area you want to find. Use the normal table (or Minitab) to find it Report that probability ________________________ 6.2) Solve problem 21 on page 352 of Intro Stats by DeVeaux & Velleman. (In this case the duration of human pregnancies can be described by a normal model so the distribution of the sample mean can be described by a normal model regardless of the sample size). For other examples in which the variable does not have a normal distribution, you can still use the normal model for the sample mean (provided n is large enough) thanks to the Central Limit Theorem. ==================================Appendix========================================= If you want to generate your own 1000 samples you can use the program below samdismp.mtb (it can be downloaded from the web page, but you need to be careful that the extention of the file remains .mtb, for that use the option ‘all files’ in the moment of downloading it and not the option ‘web page’ or other) The program has the following lines: sample 5 c2 c4 c5-c6; replace. let c7(k1)=mean(c5) let c8(k1)=mean(c6) let k1=k1+ The program takes a random sample (with replacement) of size 5 from columns C2 and C where the values of age and Y (smoke Yes=1, no=0) are. The values of age and Y for the sample are placed in c5-c6 respectively. The program calculates the mean of the sample (for age) and places the mean in C It also calculates the proportion of smokers in the sample and places that proportion in C Then you need to initialize the counter k1 by typing at the MTB prompt: MTB> let k1=1. To execute the program in order to take the 1000 samples, from the menu click on FILE> OTHER FILES> run an executable , the following window will appear: Indicate the number of times you want to execute the program, click on Select File to indicate the name of the program. You can browse to find the program samdismp.mtb .The sample means will appear in C7 and the sample proportions in C8. You can later obtain histograms or tables for those variables. You can also change the sample size and observe what happens.