1. state the null and alternative hypotheses. 2. select a random sample and record observed...
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1/4.25 Hypotheses Goodness of Fit Test H 0 : p C = p L = p S = p A = H a : customers prefer a particular style e i = ( n )( p i ) i.e., there is at least one proportion much greater than.25 e 1 = (0.25)(100) = 25 Expected frequencies e 2 = (0.25)(100) = 25 e 3 = (0.25)(100) = 25 e 4 = (0.25)(100) = 25TRANSCRIPT
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1. State the null and alternative hypotheses.2. Select a random sample and record observed frequency fi for the ith category (k categories).3. Compute expected frequency ei for the ith category:
i ie n p
4. Compute the value of the test statistic.
if ei > 5, this has a chi-square distribution
22
1
( )-stat i
k
i
i
i
efe
5. Reject H0 if 2 2-stat df = k – 1
Goodness of Fit Test
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Goodness of Fit Test
Example: Finger Lakes Homes (A)
Finger Lakes Homes manufactures four models of prefabricated homes, a two-story colonial, a log cabin, a split-level, and an A-frame. To help in production planning, management would like to determine if previous customer purchases indicate that there is a preference in the style selected.
Split- A-Model Colonial Log Level Frame# Sold 30 20 35 15
The number of homes sold of each model for 100 sales over the past two years is shown below.
k = 4
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1/4.25Hypotheses
Goodness of Fit Test
H0: pC = pL = pS = pA = Ha: customers prefer a particular style
ei = (n)(pi)
i.e., there is at least one proportion much greater than .25
e1 = (0.25)(100) = 25Expected
frequencies
2 2 2 225 25 25 2525 25
( ) ( ) ( )25 25
(30 20 35 1 )5 102 -stat
e2 = (0.25)(100) = 25e3 = (0.25)(100) = 25e4 = (0.25)(100) = 25
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Goodness of Fit Test
= .05 (column)
7.815
Do Not Reject H0 Reject H0
2.05 7.815
.05 2
At 5% significance, the assumption that there is no home style preference is rejected.
2 -stat
df = 4 – 1 = 3 (row)
m = 3 10
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1. State the null and alternative hypotheses.2. Select a random sample and record observed frequency fi for each cell of the contingency table.3. Compute expected frequency eij for each cell
(Row Total)(Column Total) ij
i jen
4. Compute the test statistic.2
2 ( )-stat ij ij
i j ij
efe
5. Reject H0 if 2 2-stat df = (m - 1)(k - 1)
Independence Test
if ei > 5, this has a chi-square distribution
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The number of homes sold for each model and price for the past two years is shown below. For convenience, the price of the home is listed as either $99,000 or less or more than $99,000.
Each home sold by Finger Lakes Homes can be classified according to price and to style. Finger Lakes’ manager would like to determine if the price of the home and the style of the home are independent variables.
Example: Finger Lakes Homes (B)
Price Colonial Log Split-Level A-Frame
> $99,000< $99,000
k = 4
Independence Test
18 6 19 1212 14 16 3
m = 2
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45
Price Colonial Log Split-Level A-Frame Total< $99K> $99K Total
3012 14 16 3
18 6 19 1255
Price Colonial Log Split-Level A-Frame Total< $99K> $99K Total
16.5 11 19.25
8.2513.5 9 15.7
56.75
55
30
Expected Frequencies (ei)
Observed Frequencies (fi)
Independence Test
20 35 15 100
45 20 35 15 100
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2 2 2 22
2 2 2 2
16.5 11 19.25 8.2516.5 11 19.25 8.25
13.5 9 15.75 6.7513.5 9 15.75 6.7
18 6 19 12
12 14 16
( ) ( ) ( ) ( )-stat
( ) ( ) ( ) 3(5
)
Compute test statistic
9.145
Hypotheses
H0: Price of the home is independent of the style of the home that is purchasedHa: Price of the home is not independent of the style of the home that is purchased
Independence Test
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= .05 (column)
7.815
Do Not Reject H0 Reject H0
2.05 7.815
.05 2
At 5% significance, we reject the assumption that the price of the home is independent of the style of home that is purchased.
2 -stat
df = (4 – 1)(2 – 1) = 3 (row)
9.145m = 3
Independence Test
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1. State the null and alternative hypotheses.
3. Compute ei for each interval.
2. Select a random sample anda. Compute the mean and standard deviation (p = 2).b. Define intervals so that ei > 5 is in the ith intervalc. For each interval, record observed frequencies fi
22
1
( )-stat i
k
i
i
i
efe
4. Compute the value of the test statistic.
2 2-stat5. Reject H0 if df = k – p – 1
if ei > 5, this has a chi-square distribution
Goodness of Fit Test: Normal Distribution
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Example: IQ Computers
IQ Computers (one better than HP?) manufactures and sells a general purpose microcomputer. As part of a study to evaluate sales personnel, management wants to determine, at a 5% significance level, if the annual sales volume (number of units sold by a salesperson) follows a normal probability distribution.
A simple random sample of 33 of the salespeople was taken and their numbers of units sold are below.
33 43 44 45 52 52 56 58 63 63 6464 65 66 68 70 72 73 73 74 74 7583 84 85 86 91 92 94 98 101 102 105
n = 33, x = 71.76, s = 18.47
Goodness of Fit Test: Normal Distribution
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z.
k = 33/5 = 6.6 6 equal intervals.
To ensure the test statistic has a chi-square distribution, the normal distribution is divided into k intervals.
Expected frequency: ei = 33/6 = 5.5
1/6 = .1667
The probability of being in each
interval is equal to
Goodness of Fit Test: Normal Distribution
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= (1)(.1667) = .1667
z.– .97 .1667
Find the z that corresponds to the red tail probability
Goodness of Fit Test: Normal Distribution
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= .3333
–.43
.3333
z.
Goodness of Fit Test: Normal Distribution
Find the z that corresponds to the red tail probability
= (2)(.1667)
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= .5000
0
.5000
z.
Goodness of Fit Test: Normal Distribution
= (3)(.1667)
Find the z that corresponds to the red tail probability
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0
.97 .43 z.–.97 –.43
Goodness of Fit Test: Normal Distribution
Find the remaining z values using symmetry
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.471.76 ( )(18.473 )x 071.76 ( )(18.47)x .4371.76 ( )(18.47)x .9771.76 ( )(18.47)x
( )( )sz x x ( )( )zx x s
.971.76 ( )(18.477 )x
z x xs
53.84 63.81 71.76 79.70 89.68
Convert the z’s to x’s
xz. 0 .97 .43–.97 –.43
Goodness of Fit Test: Normal Distribution
Find the z that corresponds to the red tail probability
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33 43 44 45 52 52 56 58 63 63 6464 65 66 68 70 72 73 73 74 74 7583 84 85 86 91 92 94 98 101 102 105
Observed and Expected Frequencies
0.5-1.5 0.5 0.5-1.5 1.5
5.55.55.55.55.55.533
6
fi ei fi – ei (fi – ei)2/ei
Total
LL UL53.8463.8171.7679.7089.68 ∞
46647
33
Data Table
∞53.8463.8171.7679.7089.68
0.050.410.050.050.410.411.362 -stat
Goodness of Fit Test: Normal Distribution
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= .05 (column)
7.815
Do Not Reject H0 Reject H0
.05 2
At 5% significance, there is no reason to doubt the assumption that the population is normally distributed.
2 -stat
df = 3 (row) 2.05 7.815
1.36 m = 3
Goodness of Fit Test: Normal Distribution