QMS 202 Binomial Distributions, π, Symmetry and the Distribution of Sample Proportions
1)
Collect a sample of n items from a dichotomous
population.
Yes/No or defective/not defective
2)
Only interested in the number (not order) of occurrences
of a result. This number is often denoted by
x = "# of occurrences of one of the results".
Therefore 0 < x < n eg: x = 2 “Yes” in sample of 10.
3)
π is the proportion of of the population that is the result
being counted.
eg: P(Yes)= π = .1
Characteristics:
Under the conditions above, the process has a binomial distribution is uniquely identified by the two parameters n and
π. The sample space is always the integers 0< x<n . The probabilities of the n+1 outcomes is given by the binomial
formula, see QMS102 notes.
Also
Binomial distributions can have many shapes. If n is
small and π is <.5 the distribution would be skewed
right, if π >.5 it would be skewed left.
In general however binomial distributions are
symmetrical if π = .5 or n is large, the binomial
distribution is symmetric.
The test for normality is that nπ >5 and n(1-π )>5.
When a Binomial Distribution is symmetric, the probability of a value falling in an interval can be approximated using
the normal curve with the same mean and standard deviation since the height of the rectangles above is equal to the
height of the normal).
Distribution of Sample Proportions
Create the sampling distribution from the Binomial by dividing each an every x by n yielding p’s (sample proportions).
If nπ>5and n(1-π)>5then the p’s are normally distributed with:
Therefore if [-z , z] are the values that bounds the middle k% of standard normal values then k% is the probability that
p will fall in the interval
“The mean plus/minus z standard deviations”
Using the same switching technique as we did in the case of sample means, we can change the centre from π to p
The probability that this interval contains π is k%, unfortunately we can’t use
it for interval estimates of π since π is still in the formula. However we can
replace the p’s in the standard deviation of p’s part of this formula too, with
little loss of accuracy because and an error in estimating π is compensated
by the opposite error in estimating (1-π).
ie: π(1-π) ~ p(1-p)
If we need to construct a Confidence Interval Estimate of a Proportion,
Level of k% using a random sample, we can obtain the limits using the formulae:
where
z bounds the middle k%
x is the “number” in the sample
p is the sample proportion
n is the sample size
Condition:
This calculations can be done with
•
SPSS with statistics using SPSS as a calculator ie:Transform/Compute
•
Casio 9850 with raw data or statistics using [Stat][INTR][z][1-p]
π , using a Confidence
Question!
In a random sample of 240 Ryerson students, 19 were late for their first class. Construct a 95 % confidence interval
estimate of the proportion of all Ryerson students that are late for their first class.
Define:
π - proportion of students who were late for their first class
Work: [Menu][Stat][INTR][z][1-p]
C-Level: “.95” x: “19” n: “240” [Exe]
Left=.0450 Right=.113
{p^ =.0792, sample proportion}
Therefore;
95%:
4.54 < π < 11.3 % . {Note: The answer was reported to 3 significant figures}
Condition Check:
* note: np always equals x
Your Question!
In a random sample of 60 Ryerson Business students, 21 indicated they were going to the Forum Presentation on Wed.
January 21st. If it is known that there are 4100 Ryerson Business Students, estimate the number of all Ryerson
Business students that are expected to attend {a “practical” range would be useful}.
(Hint: consider using a confidence interval of π).
Define:
π - proportion of students who were were going to the Forum Presentation on Wed. January 21st
Assumption:
Assume a confidence level of 95% is reasonable
Work: [Menu][Stat][INTR][z][1-p]
C-Level: “.95” x: “21” n: “60” [Exe]
Left=.22931 Right=.47068 p^ =.35, sample proportion
Condition Check
Lower Limit of total=.22932*4100=940.212
Upper Limit of total=.47068*4100=1929.788
Therefore:
At the 95% confidence level, the total number of students who were were going to
the Forum Presentation on Wed. January 21 st is between 940 and 1930