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Section Q Distribution of the Sample Mean and the Central Limit Theorem Up to this point, the probabilities we have found have been based on individuals in a sample, but suppose we want to find probabilities based on the mean of a sample. In order for us to find these probabilities we need to know determine the sampling distribution of the sample mean. Knowing the sampling distribution of the sample mean will not only allow us to find probabilities, but it is the underlying concept that allows us to estimate the population mean and draw conclusions about the population mean which is what inferential statistics is all about. Sampling Error : The error resulting from using a sample to estimate a population characteristic.
For a variable x and a given sample size n, the distribution of the variable x̅ (all possible sample means of size
n) is called the sampling distribution of the mean. Note: The larger the sample size the smaller the sampling error tends to be in estimating a population mean,
, by a sample mean x̅.
Mean of x̅ : denoted μx̅
For samples of size n, the mean of the variable x̅ equals the mean of the variable under consideration,
i.e. μx̅ = μ , where μx̅ is the mean of variable x̅ and is the population mean.
In other words, the mean of all possible sample means of size n equals the population mean. Example: The following data represent the ages of the winners (age, in years, at time of award given) of the Academy Award for Best Actress for the years 2012 – 2017. 2012: Meryl Streep 62 2013: Jennifer Lawrence 22 2014: Cate Blanchett 44 2015: Julianne Moore 54 2016: Brie Larson 26 2017: Emma Stone 28
a) Calculate the population mean, μ. ___________________
b) The following table consisting of all possible samples with size n = 2 and calculate their corresponding means. Sample Mean Sample Mean Sample Mean Sample Mean Sample Mean Sample Mean 62, 62 22, 62 44, 62 54, 62 26, 62 28, 62 62, 22 22, 22 44, 22 54, 22 26, 22 28, 22 62, 44 22, 44 44, 44 54, 44 26, 44 28, 44 62, 54 22, 54 44, 54 54, 54 26, 54 28, 54 62, 26 22, 26 44, 26 54, 26 26, 26 28, 26 62, 28 22, 28 44, 28 54, 28 26, 28 28, 28
c) Calculate the mean of the sampling distribution of the mean, μx̅. _________________________
(i.e. calculate the mean of the sample means)
d) What do you conclude about μ and μx̅? ________________
Note: The above example is exactly that an example, it is not a proof of μx̅ = μ.
Standard deviation of x̅ denoted σx̅
For samples of size n, the standard deviation of the variable x̅ equals the standard deviation of the
population under consideration divided by the square root of n.
So, σx̅ =
σ √n
σx̅ is smaller than
σx̅ is sometimes called the standard error of the mean.
Example: Using the example above, a) Calculate the population standard deviation, .____________
b) Calculate the standard deviation of the sampling distribution of the mean, σx̅. ______________________
(i.e. calculate the standard deviation of the sample means) n = _______ ∑ x̅ = ___________ ∑ x̅ 2 = ___________
c) Use the formula σx̅ =
σ √n^
to calculate σx̅. ____________________
d) What do you conclude from parts b and c?______________________ Again, the example is an example not a proof.
The Sampling Distribution of the Sample Mean
Suppose that a variable x of a population has mean, and standard deviation, . Then, for samples of size n,
1) The mean of x̅ equals the population mean, , in other words: μx̅ = μ
2) The standard deviation of x̅ equals the population standard deviation divided by the
square root of the sample size, in other words: σx̅ =
σ √n
3 ) If x is normally distributed, so is x̅ , regardless of sample size
4) If the sample size is large (n > 30), x̅ is approximately normally distributed, regardless of the
distribution of x.
Therefore, we can say, x̅ is normally distributed with parameters μx̅ and σx̅ , where μx̅ = μ and σx̅ =
σ √n
Note: Since the sampling distribution of the sample mean is normally under certain conditions you can use
the normal approximation to find probabilities, therefore you need convert x̅ to a z-score.
Converting x̅ to a z-score: z =
x ̅ − μx̅
σx̅
Examples:
- The times that college students spend studying per week have a distribution that is right skewed with a mean of 8.4 hours and a standard deviation of 2.7 hours. Suppose a random sample of 45 students is selected. a) What is the sampling distribution of the mean number of hours these 45 students spend studying per
week? (i.e. What is the sampling distribution of x̅ ?)
n= _______ μ = ______ σ = ______ b) Find the probability that the mean time spent studying per week is between 8 and 9 hours. c) Find the probability that the mean time spent studying per week is greater than 9.5 hours.
2 ) At an urban hospital the weights of newborn babies are normally distributed, with a mean of 7.2 pounds and standard deviation of 1.2 pounds. Suppose a random sample of 30 is selected. a) What is the sampling distribution of the mean weight of these newborn babies?
(i.e. What is the sampling distribution of x̅ ?)
n= _______ μ = ______ σ = ______ b) Find the probability the mean weight is less than 6.9 pounds? c) Find the probability the mean weight is between 6.5 and 7.5 pounds? d) Find the probability the mean weight is greater than 8 pounds?
- A battery manufacturer claims that the lifetime of a certain battery has a mean of 40 hours and a standard deviation of 5 hours. A simple random sample of 100 batteries is selected. a) What is the sampling distribution of the mean life of the batteries?
(i.e. What is the sampling distribution of x̅ ?)
n= _______ μ = ______ σ = ______ b) What is the probability the mean life is less than 38.5? Would this be unusual?
