ith reference to the data below given in minutes, using the chi-square goodness-of-fit test, can you conclude that the time lags are not exponentially distributed with θ = 3.2 minutes? Use α = 0.01. 2,3 0,2 2,9 0,4 2,8 2,4 4,4 5,8 2,8 3,3 3,3 9,7 2,5 5,6 9,5 1,8 4,7 0,7 6,2 1,2 7,8 0,8 0,9 0,4 1,3 3,1 3,7 7,2 1,6 1,9 2 .4 4.6 3.8 1.5 2.7 0.4 1.3 1.1 5.5
With reference to the data below given in minutes, using the chi-square goodness-of-fit test, can you conclude that the time lags are not exponentially distributed with θ = 3.2 minutes? Use α = 0.01.
2,3 0,2 2,9 0,4 2,8 2,4 4,4 5,8 2,8 3,3
3,3 9,7 2,5 5,6 9,5 1,8 4,7 0,7 6,2 1,2
7,8 0,8 0,9 0,4 1,3 3,1 3,7 7,2 1,6 1,9
2 .4 4.6 3.8 1.5 2.7 0.4 1.3 1.1 5.5 3.4
4.2 1.2 0.5 6.8 5.2 6.3 7.6 1.4 0.5 1.4
Please be as specific as possible with your calculations.
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Good morning, I have some doubts about this.
I would appreciate your help to clarify them.
To determine the observed frequencies, I took the data and created a table with some intervals as follows:
1). Determine the maximum value, minimum value,
Maximum value=9.7 ; Minimum value=0.2
Range=maximum value-minimum value=9.7-0.2=9.5
Número de intervalos=1+(3,3222*LOG(n))=1+(3,3222*LOG(50))=6.64
Amplitude=(Maximum value-minimum value)/(Number of intervals)=(9.7-0.2)/6.64= 9.5/6.64=1.43
With this, determine the intervals of the data table and then according to those intervals take out the observed frequencies for those intervals, as follows:
Intervals (Minutes) / Observed frequencies (Oi)
[ 0,20 - 1,63 ] / 18
[ 1,64 - 3,07 ] / 10
[ 3,08 - 4,51 ] / 8
[ 4,52 - 5,95 ] / 6
[ 5,96 - 7,39 ] / 4
[ 7,40 - 8,83 ] / 2
[ 8,84 - 10,27] / 2
Then, with these observed frequencies find the exponential probabilities as follows:
Probabilities (I found the probabilities as follows)
Use this formula for all intervals P(X ≤x)=1-e^((-x)/θ)
P6 (X ≤18)=1-e^((-18)/3.2) =0,9964
P5 (X ≤10)=1-e^((-10)/3.2) =0.9561
P4 (X ≤8) =1-e^((-8)/3.2) =0.918
P3 (X ≤6) =1-e^((-6)/3.2) =0.8466
P2 (X ≤4) =1-e^((-4)/3.2) =0.7135
P1 (X ≤2) =1-e^((-2)/3.2) =0.4647
Expected frequency (I found the expected frequencies as follows)
E1=n*P1 (X ≤2) =50*0.4648 =23.24
E2=n*[P2 (X ≤4)- P1(X ≤2)] =50*[0.7135-0.4648] =12.44
E3=n*[P3 (X ≤6)- P2(X ≤4)] =50*[0.8467-0.7135] =6.66
E4=n*[P4 (X ≤8)- P3(X ≤6)] =50*[0.918-0.8467] =3.56
E5=n*[P5 (X ≤10)- P4(X ≤8)] =50*[0.9561-0.918] =1.91
E6=n*[P6 (X ≤18)- P5(X ≤10)]=50*[0.9964-0.9561] =2.02
Adding E1 to E6 gives me an expected frequency of 49.82 which is the same as you calculated when I add up all the E's in your explanation. So, I think I am doing well solving the exercise up to this point. However, I am not sure.
Test statistic (I found the test statistic like this:
Use this formula:
Xi^2= (Oi- Ei )^2/Ei)
X1^2= (2-23.24)^2/23.24 = (-21.24)^2/23.24= 451.14/23.24=19.41
X2^2= (4-12.44)^2/12.44 = (-8.44)^2/12.44 = 71.23/12.44 =5.72
X3^2= (6-6.66)^2/6.66 = (-0.66)^2/6.66 = 0.44/6.66 =0.06
X4^2= (8-3.56)^2/3.56 = (4.44)^2/3.56 = 19.71/3.56 =5.52
X5^2= (10-1.91)^2/1.91 = (8.09)^2/1.91 = 65.45/1.91 =34.33
X6^2= (18-2.02)^2/2.02 = (15.98)^2/2.02 = 255.36/2.02 =126.69
However, my test statistic gives me a very high 191.75 and you give me 5.079. That is why I think my calculation is wrong and I would like you to tell me what my mistake could be.
Thank you for your time and explanation. Next, I will list some additional doubts.
Doubts.
1). Is it necessary to find the intervals with
2). All the probabilities of all the intervals are found with the following formula: P(X ≤x)=1-e^((-x)/θ)?
3). Why are some expected frequencies found with this formula: P(X ≤x)=1-e^((-x)/θ). For example: E1 and E5. But for E2, E3 and E4 this other formula is used: P(X ≤x)=e^((-x)/θ)?
4). Why when calculating E2 it says that: E2=(total number of observations)*[P(X ≤4)-P(X ≤2)]. However, when it solves it says that: E2=50*[(e^((−2)/3.2)-(e^((−4)/3.2))]? When, P(X ≤4)=(e^((-4)/3.2)) and P(X ≤2)=(e^((-2)/3.2). That's my impression, but I think I remain inverted. Or what, is the reason why it is expressed in one way and solved in another. That generates confusion to me.
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