Chi-Square?

[Terry L. Robinson]

Hi,

I am experimenting with statistics to learn and apply this to my job.

If K=10 df=(K-1)=9 and the alpha margin of error is .05. My hypothetical problem is did patient care increase by 15% from year 2000 to 2004. I still have to develop some numbers for hospitals 1-10 for year 2004 to complete this.

Does anyone know how to do this and is willing to work with me to understand this? I'm using the function of =CHITEST(B3:B12,C3:C12) but the result is 0. Do I have to plug in the Year 2004 data before this will work? Then the follow on question is how to chart this?

Thank you for your help in my learning how to do statistics via excel.

--
Terry L. Robinson


Year 2000
Observed Expected Total
Hospital 1 156,279 174,892 331,171
Hospital 2 196,453 203,618 400,071
Hospital 3 205,784 205,784 411,568
Hospital 4 179,520 183,629 363,149
Hospital 5 192,759 192,563 385,322
Hospital 6 298,627 300,090 598,717
Hospital 7 289,993 289,784 579,777
Hospital 8 294,865 276,592 571,457
Hospital 9 103,582 105,820 209,402
Hospital 10 274,917 314,761 589,678

Chi Square 0

 

[AlfD]

Hi!

Just may be that you data streams are so closely correlated that th
chi-sq test is telling you there is no way the similarity could b
accidental.

You might also try CORREL() on your observed and expected data. 0.97 i
a heck of a correlation, I believe.

Also try graphing the data. Maybe sort by Observed data and then plot
scatter graph. Add trendlines (right click on a graph point and selec
add a trendline: use linear). Do it for both data series. Everythin
points to a nigh dead-cert correlation which is unlikely to be th
result of pure chance.

But what is this column called Total. Can you explain to me the meanin
of total when you add Observed to Expected?

I'm not a statistician by training, so I would not be the person t
take you forward.
Al

 

[Terry L. Robinson]

Sir,

Thank you for your reply. Below is what I worked out today. The Total column
was actually a mistake. I re-labeled the columns to make more sense with the
theory of Chi-Square null hypothesis test.

I will give the CORREL() a try and see what happens. As for graphing the
data, that is my next task if the data makes sense. What I'm trying to prove
is that based on a null hypothesis, that from year 2000 to year 2004
combining hospitals 1-10 that the observed result vs. the expected result is
that their is a 15% increase is satisfied customers without skewing the
results to make the test work. It's OK if the results do not meet the 15%
increase.

Year 2000 Year 2004
fo fe fo-fe (fo-fe)2/fe fo fe fo-fe (fo-fe)2/fe

H1 156,279 174,892 18,613 1980.90 H1 203,861 205,639 1,778 15.37
H2 196,453 203,618 7,165 252.13 H2 187,436 198,734 11,298 642.29
H3 205,784 205,784 0 0.00 H3 305,821 325,857 20,036 1231.96
H4 179,520 183,629 4,109 91.95 H4 267,382 284,391 17,009 1017.28
H5 192,759 192,563 -196 0.20 H5 199,753 203,642 3,889 74.27
H6 298,627 300,090 1,463 7.13 H6 276,085 305,825 29,740 2892.07
H7 289,993 289,784 -209 0.15 H7 304,691 304,421 -270 0.24
H8 294,865 276,592 -18,273 1207.20 H8 305,726 310,957 5,231 88.00
H9 103,582 105,820 2,238 47.33 H9 186,935 187,583 648 2.24
H10 274,917 314,761 39,844 5043.65 H10 352,749 381,866 29,117 2220.15
Totals 2192779.00 2247533.00 8630.64 2590439.00 2708915.00
8183.87

Mean 398687.09 408642.36 1569.21 470988.91 492530.00 1487.98
Median 201118.50 204701.00 69.64 271733.50 294406.00 365.14
Standard Deviation 598422.074 613231.5333 15069 2798.13 705248.5411
737834.3116 11514 2426.78

Chi Square Test 0

k= 10
df= 10-1=9
Critical Value 16.919
Margin of error 0.05
Null-Hypothesis 15%