expert on excel using megastat

Output

Binomial distribution
		6	n
		0.5	p
			cumulative
	X	P(X)	probability
	0	0.01563	0.01563
	1	0.09375	0.10938
	2	0.23438	0.34375
	3	0.31250	0.65625
	4	0.23438	0.89063
	5	0.09375	0.98438
	6	0.01563	1.00000
		1.00000
		3.000	expected value
		1.500	variance
		1.225	standard deviation
				3.0
Normal distribution
		P(lower)	P(upper)	z	X	mean	std.dev
		.9257	.0743	1.44	60,000	49,400	7,340
				2.0
Normal distribution
		P(lower)	P(upper)	z	X	mean	std.dev
		.2744	.7256	-0.60	45,000	49,400	7,340
Uniformly Distributed Random Numbers
Minimum	1
Maximum	100
	Uniform
	70.8492036462
Uniformly Distributed Random Numbers
Minimum	1
Maximum	100
	Uniform
	53.8089779615
	58.3723430037
	29.6666839123
	30.8928530812
	77.6992698908
	2.3877465129
	76.3116354942
	81.6345120072
	71.1947520971

Binomial distribution (n = 6, p = 0.5)

0.0 1.0 2.0 3.0 4.0 5.0 6.0 0.015625 0.09375 0.234375 0.3125 0.234375 0.09375 0.0156 25

X

P(X)

3

2

1

0

-1

-2

-3

f(z)

z

1.44

-1.28

3

2

1

0

-1

-2

-3

f(z)

z

-0.60

Sheet1

Week 3 - Probabilities
	First we cover counting probabilities
	Counting Probability - combinations order does not count- in permutations order counts. This following shows how many combinations one permutations there are of 2 given 4 things.
		Things		Combinations	Permutations
		A		A-B	A-B	B-A
		B		A-C	A-C	C-A
		C		A-D	A-D	D-A
		D		B-C	B-C	C-B
				B-D	B-D	D-B
				C-D	C-D	D-C
			totals	6	12
1.0	If a new shopping center has seven locations, how many arrangements of new businesses locate there?
	Use Factorial to do.					use factorial
2.0	If seven business want to locate in a new shopping center but there are only five locations, how many
	arrangements could there be?					use permutations
3.0	If seven bossiness want to locate in a new shopping center but there are only five locations, how many
	different combinations could be selected?					use combinations
	Discrete Probabilities: Concept and Binomial
	Probability must between 0 and 1; total of all probabilities is 1
	Concept is simply "The Specific Question" divided by the total possibilities (Excel for these)
		Example	Married	Not Married
			B	-B
		Sport car	14	27	41
		Not sport car	43	36	79
			57	63	120
		Question: What is probability of being married = 57/120
		Question: What is the probability of having a sports car and being married = 14/120
		Question: What is the probability of having a spots car or being married =(41+57-14)/120
	Another old example: Refer to the contingency table shown below. (a) Calculate each probability (i–vi) and explain in
	words what it means. (b) Do you see evidence that smoking and race are not independent?
	Explain. (c) Do the smoking rates shown here correspond to your experience? (d) Why might
	public health officials be interested in this type of data? (Data are from Statistical Abstract of the
	United States, 2001, pp. 12 and 16. Note: Actual statistics are applied to a hypothetical sample of
	1,000.) Smoking2
	i. P(S) ii. P(W) iii. P(S | W)
	iv. P(S | B) v. P(S and W) vi. P(N and B)
		Smoking by Race for Males Age 18-24
			Smoker	Nonsmoker	Row Total
			(S)	(N)
		White (W)	290	560	850
		Black (B)	30	120	150
		Column Total	320	680	1000
		a.-i	320/1000 =			probability of a smoker
		a.-ii	850/1000=			probability of a White
		a.iii	290/850=			probability of a While being a smoker
		a.iv	30/150=			probability of a Black being a smoker
		a.v	290/1000=			probability of selecting a White and being a smoker
		a.vi	120/1000=			probability of selecting a Black and being a nonsmokeroer
		b.	The P of smoking and white is .29 and the P(S)*P(W) =.32*.85=
			The P of smoking and Black is .03 and the P(B)*P(S)=.32*.15 =
		d.	If smoking is dependent on race, then health officials might target special
			programs based on race.
4.0	Your exercise to do
			Size of company
			Small	Medium	Large
	Plan MBA?	Yes	24	50	105	179
		No	49	58	100	207
			73	108	205	386
		P(Yes) =
		P(Yes | Medium) =
		P(Yes and Large) =
		P(Small and Yes) =
		P(Medium or Yes) =
		P(Small) =
		P(Medium | Yes) =
	Now Binomial (MegaStat for these)
	Example
	What is the probability of observing two heads is a coin is tossed six times?
		n=6
		P/head = .5
		Use MegaStat to compute
		Results
	Binomial distribution
		6	n
		0.5	p
			cumulative
	X	P(X)	probability
	0	0.01563	0.01563
	1	0.09375	0.10938
	2		0.34375	Note the	Probability of throwing 2 heads is .23438
	3	0.31250	0.65625
	4	0.23438	0.89063
	5	0.09375	0.98438
	6	0.01563	1.00000
		1.00000
		3.000	expected value
		1.500	variance
		1.225	standard deviation
5.0	Your turn:	A basketball team that shoots free throws at 80.5% shoots 14 free throws in their next game.
	What is the Probability they will be successful in 11 of them?
	and what is the probability they will be successful in at least 9 of them?
	Now we do what is the very basis of much of statistics - six sigma -lean. The Normal Distribution
	Example: The mean annual income of workers in a particular industry is $49,400 with a standard deviation of $7340.
	Assume a normal distribution.
	A. What is the probability a worker would have an income below $60,000?
	Use MegaStat for this
	Normal distribution
		P(lower)	P(upper)	z	X	mean	std.dev
		.9257	.0743	1.44	60,000	49,400	7,340
	B. What is the probability a worker would have an income above $60,000?
		.0743
	C. What is the probability a worker would make between $45,000 and $60,000
		.9257	.0743	1.44	60,000	49,400	7,340
		.2744	.7256	-0.60	45,000	49,400	7,340
		0.6512
6.0	Your turn to do the exercise
	Assume a college entrance exam has a mean of 1000 and a standard deviation of 100. Assume a normal distribution.
	A. What is the probability the student's score would be below 1000?
	B. What is the probability the student's score would be below 850?
	C. What is the probability the student's score would b above 1250"
	D. What it the probability the student's score would be between 850 and 1250?
	E. What would be the student's score to have only 5% of the students taking the test have higher >
	Now sampling. Sampling is the basis of nearly all medical improvements and used in all industries
	Sampling requires that one Identifies the Population one is research exactly. Then one must determine the appropriate sample size for the research. The sampling must be designed so everyone in the population as an equal chance of being selected (the procedures are Random, Cluster, Stratified, and Systematic - we will do an exercise using random selection)
	Example - what is the required sample size for a customer with a .035 error and 95% confidence.
	Answer:	784
7.0	Your turn:
	A. Customer states he wants a 3% error with 95% confidence. What is the require sample size
	B. What is the require sample size for a study have an error of 1,000 whose standard deviation is 7,800
	and a 90% confidence level.
	Now we randomly select
	Example - random selection			Random select 5 from the following "population".
			Population
			dog		N count of population =		9
			cat
			bird
			lion
			elephant
			grizzly
			polar bear
			panda
			turtle
					Sort on Uniform
		Uniform	Population		Uniform	Population
		53.8089779615	dog		2.3877465129	grizzly	1
		58.3723430037	cat		29.6666839123	bird	2
		29.6666839123	bird		30.8928530812	lion	3
		30.8928530812	lion		53.8089779615	dog	4
		77.6992698908	elephant		58.3723430037	cat	5
		2.3877465129	grizzly		71.1947520971	turtle
		76.3116354942	polar bear		76.3116354942	polar bear
		81.6345120072	panda		77.6992698908	elephant
		71.1947520971	turtle		81.6345120072	panda
		select top 5
			grizzly
			bird
			lion
			dog
			cat
8.0	Your turn.	Randomly select 7 from the following - show all your steps as in the example
			dog
			cat
			bird
			lion
			elephant
			grizzly
			polar bear
			panda
			turtle
			mouse
			rat
			squirrel
			rabbit
			skunk
	And we end the week with confidence interval
	Example:
	A local retailer wants to know the 95% confidence interval of the proportion of customers who use a credit card. A
	random sample of 142 receipts showed that 83 used a credit card.
	95% confidence interval .503 to .666
9.0	Your Turn
	A. Compute the 95% confidence interval of those people choosing chocolate ice cream. 12 of 27 customer
	chose a favor other than chocolate.
	B. Compute the 90% confidence interval when the mean is 12,000, the standard deviation is 500 and the N count is 86

&P of &N

Binomial distribution (n = 6, p = 0.5)

0.0 1.0 2.0 3.0 4.0 5.0 6.0 0.015625 0.09375 0.234375 0.3125 0.234375 0.09375 0.0156 25

X

P(X)

Sheet2

Sheet3