1. Sampling and Sampling Distributions Chapter 8.4 – 8.7

2. Shape of the Sampling Distribution of the Mean for a Normal Population  Theorem 8.1  The Shape of the Sampling Distribution when Sampling from a Normal Population  If the population being sampled is a normal distribution, then the sampling distribution of the mean is a normal distribution regardless of the sample size, n.

3. Shape of the Sampling Distribution of the Mean for a Non-Normal Population  Theorem 8.2  Central Limit Theorem  For any population, the sampling distribution of the mean approaches a normal distribution as the sample size n becomes large. This is true regardless of the shape of the population being randomly sampled.  In other words, the population can have any shape, it is not even necessary to know the shape of the population provided the sample size is large enough.  How large is large enough?

4. Shape of the Sampling Distribution of the Mean for a Non-Normal Population  Consider 4 different shaped, non-normal populations from which samples are being taken of various sizes.

5. Shape of the Sampling Distribution of the Mean for a Non-Normal Population  Given the progression of distribution shapes as n gets larger, we can see that the look of all the graphs is normal around n=30.  General Rule for Applying the Central Limit Theorem: the n greater than 30 rule.  A sample size n greater than 30 is considered large enough to apply the central limit theorem.  Therefore the shape sampling distribution of the mean is normal whenever the sample size is greater than 30.

6. Mean, Standard Deviation, and Shape of the Sampling Distribution of the Mean Sample Size nMean Standard errorShape of the sampling distribution of the mean 4 16 25 49 100 400 Use µ = 8 and σ = 2 to find the mean and standard error for the values of n. Identify the shape of the distribution if possible.

7. Mean, Standard Deviation, and Shape of the Sampling Distribution of the Mean Sample Size nMean Standard errorShape of the sampling distribution of the mean 481unknown 1680.5unknown 2580.4unknown 4980.2857Approximately normal 10080.2Approximately normal 40080.1Approximately normal Use µ = 8 and σ = 2

8. Calculating Probabilities Using the Sampling Distribution of the Mean  The Central Limit Theorem lets us apply all the applications of the normal distribution to a sampling distribution of the mean.  If a sampling distribution is normally distributed, we can use all tools of the normal distribution and apply them to the sampling distribution: ex: normalcdf and invnorm  Therefore, we can solve probability problems for the sampling distribution of the mean if the sample size is large enough (n > 30) using normalcdf.

9. Calculating Probabilities Using the Sampling Distribution of the Mean  Remember:  To calculate a probability for the normal distribution, we calculate the area under the normal curve for a given event.  Our calculator uses normalcdf and 4 input values to do this:  Normalcdf(lower score, upper score, μ, σ )  For a sampling distribution with n>30, we will calculate probability:  Normalcdf(lower sample mean, upper sample mean,, )

10. Example 8.9  At a large public state college in Virginia, the mean Verbal SAT score of all attending students was µ = 600 with a population standard deviation of σ =65. If a random sample of 100 students is selected from the population of students, determine:  (a) The probability that the mean Verbal SAT score of the selected sample will be less than 615.  (b) The probability that the mean Verbal SAT score of the selected sample will be within 10 points of the population mean.

11. Example 8.9  Since n=100 is greater than 30, the Central Limit Theorem says that the sampling distribution of the mean can be approximated by a normal distribution.  First, we need the mean and standard error of the sampling distribution of the mean. (We can’t use normalcdf without these values).

12. Example 8.9  (a) The probability that the mean Verbal SAT score of the selected sample will be less than 615.  Label a normal curve with  To calculate the probability that the sample of 100 students will be less than 615, we need to find the area to the left of 615  Normalcdf(-E99 615, 600, 6.5)  0.9895

13. Example 8.9  (b) The probability that the mean Verbal SAT score of the selected sample will be within 10 points of the population mean.  We need to find the sample means that are within 10 points of the mean.  Since the mean is 600, the sample means which are 10 points away are:  600-10 = 590  600+10 = 610  To find the probability that the mean Verbal SAT of the selected sample will be within 10 points of the mean, we need to find the area between 590 and 610.

14. Example 8.9  Label a normal curve with  Normalcdf(590, 610, 600, 6.5)  0.8761

15. Example 8.10  The population mean weight of newborn babies for a western suburb is 7.4 lbs. with a standard deviation of 0.8 lbs. What is the probability that a sample of 64 newborns selected at random will have a mean weight greater than 7.5 lbs.?  We need the mean of the sampling distribution:  We need the standard error of the sampling distribution:

16. Example 8.10  Label a normal curve with  To find the probability that the sample mean will be greater than 7.5 lbs. we must find area to the right of 7.5 lbs.  Normalcdf(7.5, E99, 7.4, 0.1)  0.1587

17. Example 8.11  The population of the ages of all U.S. college students is skewed to the right with a mean of 27.4 years and a standard deviation of 5.8 years. Determine the probability that a random sample of 49 college students selected from the population will have a sample mean age within one year of the population mean age.  We need the mean of the sampling distribution:  We need the standard error of the sampling distribution:

18. Example 8.11  We need to find the sample ages which are 1 year less than the population mean of 27.4 and 1 year greater than 27.4  27.4-1 = 26.4  27.4+1 = 28.4  To find the probability that the selected sample will have a mean between 26.4 and 28.4 years, we need to calculate the area between these two sample ages.  Normalcdf(26.4,28.4,27.4,0.8286)  0.7725

19. 8.6 The Effect of Sample Size on the Standard Error of the Mean  Example 8.12  Consider the distribution of 5 feet 9 inch American males whose mean weight is 160 lbs. with a standard deviation of 10 lbs. Determine the mean,, and the standard error,, for the sampling distribution of the mean for samples of size 36, 100, 400, and 1600.

20. 8.6 The Effect of Sample Size on the Standard Error of the Mean  μ =160 and σ =10. Sample Size nMean Standard error 361601.6667 1001601 4001600.5 16001600.25

21. 8.6 The Effect of Sample Size on the Standard Error of the Mean  As the sample size increases, the standard error of the sampling distribution of the mean …  Decreases.  When the sample size is large and standard error is small, more of the data values are clustered about the mean.

22. 8.6 The Effect of Sample Size on the Standard Error of the Mean  You have greater confidence that a data value chosen at random is closest to the true mean if you choose from the sample of largest size: n=1600. The range is smaller and therefore there is less variability in the data.

23. The Sampling Distribution of the Proportion  A proportion is a fraction, ratio or percentage that indicates the part of a population or sample that possesses a particular characteristic.  For example, statisticians might be interested in inferences about the proportion of  1. US adults who wear seat belts.  2. Individuals not catching a cold who take large doses of vitamin C.  3. Children who live with only one parent in the US.  4. Defective computer memory chips produced using a new innovative manufacturing process.  Etc…

24. The Sampling Distribution of the Proportion  To determine the proportion, we need to count the number of data values within the population that possess a particular characteristic.  The population proportion is the ratio of the number of data values in the population that possess a particular characteristic to the population size.  Where p = population proportion  X = number of occurrences in the population possessing the particular characteristic  N = population size  p is often expressed as a percent.

25. Using a sample proportion to estimate the population proportion  For a large population, it can be impractical to calculate the population proportion, and so we use a sample and calculate a sample proportion.  The sample proportion is the ratio of the number of data values in the population that possess a particular characteristic to the sample size.  Where = sample proportion; called “p-hat”  x = number of occurrences in the sample possessing the particular characteristic  n= population size  is often expressed as a percent.

26. Means vs. Proportions MeansProportions Population mean:Population proportion: Sample mean:Sample proportion Sampling Distribution of the MeanSampling Distribution of the Proportion Mean of the sampling distribution of the mean: Mean of the sampling distribution of the proportion: Sampling error of the mean:Sampling error of the proportion: Standard error of the sampling distribution of the mean: Standard error of the sampling distribution of the proportion: Z-score for a data value (sample mean) in the sampling distribution of the mean. Z-score of the data value (sample proportion) in the sampling distribution of the proportion.

27. Example 8.16  According to a survey in Men’s Health Magazine (about) 37% of adults aged 18 to 54 have taken a herbal diet supplement. If represents the proportion of adults aged 18 to 54 within a sample of 200 such adults, then:  (a) State the mean of the sampling distribution of the proportion using the appropriate notation.  (b) State the standard error of the proportion using the appropriate notation.

28. Example 8.16  (a) State the mean of the sampling distribution of the proportion using the appropriate notation.  Let p = proportion of all adults aged 18 to 54 who have taken an herbal diet supplement.  p=.37  The mean of the sampling distribution of the proportion is:  (b) Using the formula for the standard error of the proportion:

29. Shape of the Sampling Distribution of the Proportion  The shape of the sampling distribution of the proportion is determined from the Central Limit Theorem, just like the shape of the sampling distribution of the mean.  For the sampling distribution of the mean, we conclude that the shape of the distribution is approximately normal if the sample size, n, is greater than 30.  For the sampling distribution of the proportion, we conclude the shape of the distribution is approximately normal if  np > 5 and n(1-p) >5  Where n is the sample size and p is the population proportion.

30. Example 8.18  According to the US Census Bureau, 20% of men aged from 25 to 29 are still living at home. Assume this proportion is true for the population of all men aged 25-29. If a random sample of size 100 is selected from this population, then determine whether the sampling distribution of the proportion can be approximated by a normal distribution.  Since p is defined to be the proportion of all men aged 25 to 29 who still live at home, then:  p=0.20  If np and n(1-p) are both greater than 5, then the shape of the sampling distribution of the proportion is approximately normal.  np = (100)(0.20) = 20  n(1-p) = (100)(1-0.20) = (100)(0.80) = 80

31. Example 8.18  Since np and n(1-p) are both greater than 5, the shape of the sampling distribution of the proportion is approximately normal and  We can use all the properties of the normal distribution (like probability) to answer questions about the sampling distribution of the proportion.

32. Example 8.20  According to Princeton Survey Research, 47% of drivers in the Midwest say they drive at or below the speed limit. Assume this percentage represents the true population proportion of all Midwest drivers who drive at or below the speed limit.  Determine the probability that more than 53% in a random sample of 200 Midwest drivers will say they drive at or below the speed limit.  We can’t calculate probability without a mean and standard deviation of the sampling distribution of the proportion.  First, note that p =.47 because we are told to assume 47% is the true population proportion.  Therefore, the mean of the sampling distribution of the proportion is

33. Example 8.20  The mean of the sampling distribution of the proportion is  The standard error of the sampling distribution of the proportion is  Since np = 94 and n(1-p) = 106 are both greater than 5, we can concluded the sampling distribution of the proportion is approximately normal.

34. Example 8.20

35.  To calculate the probability that more than 53% in a random sample of 200 will say they drive at or below the speed limit,  we must determine the shaded area under the normal curve to the right of

36. Example 8.20  To find the area to the right the sample proportion  Use normalcdf (lower sample proportion, upper sample proportion, mean of the sampling distribution of the proportion, standard error of the proportion)  Normalcdf(0.53, 100000, 0.47, 0.0353)

37. Example 8.21  According to a survey by the Better Sleep Council, (nearly) one-third of people admit to dozing off at their workplace. If a random sample of 100 workers is selected from this population, then determine the probability that the proportion of workers who admit to dozing off at work falls between 0.24 and 0.42.  The mean of the sampling distribution of the proportion is  The standard error of the sampling distribution of the proportion is  Since np = 33 and n(1-p) = 67 are both greater than 5, we can concluded the sampling distribution of the proportion is approximately normal.

38. Example 8.21  To calculate the probability that the proportion of workers who admit to dozing off at work falls between 0.24 and 0.42  we must determine the shaded area under the normal curve to the between  Normalcdf(0.24, 0.42, 0.33, 0.0470)  = 0.9445  Therefore, the probability is 0.9445 that between 0.24 and 0.42 of the workers in a random sample of 100 workers admit to dozing off at work.