Stat6 chi square test
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The chi-square test is used to determine if differences in frequencies observed in qualitative variable categories are statistically significant or likely due to chance. An example compares influenza rates in a vaccine vs placebo group in a clinical trial. The expected and observed frequencies are calculated. The chi-square test statistic is greater than the critical value, so the null hypothesis that the vaccine and placebo have the same influenza proportion is rejected. Therefore, the difference is likely due to the vaccine's effectiveness rather than chance.
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2) Compare the means of two independent samples.
3) Compare the values of one sample at two different time points.
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Hyperemia is an active process resulting from increased blood flow due to arteriolar dilation, causing engorged tissue that appears red. Congestion is a passive process resulting from impaired outflow, causing tissue to appear bluish-red due to accumulation of deoxygenated blood.
Pulmonary congestion microscopically shows engorged alveolar capillaries and edema, while chronic pulmonary congestion shows thickened fibrotic septa and hemosiderin
Hemodynamic disorders
This document summarizes key concepts related to hemodynamic disorders, thrombosis, and shock. It discusses edema, including the mechanisms and clinical significance of edema. It also covers hyperemia and congestion, hemorrhage, and thrombosis. For edema, it describes how fluid moves between vascular and interstitial spaces and the causes of increased interstitial fluid. It discusses the pathologic features and clinical significance of pulmonary, subcutaneous, and brain edema. For thrombosis, hemorrhage, hyperemia and congestion, it outlines the mechanisms, morphological changes, and clinical implications.
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Hemodynamic disorders, thrombosis and shock (practical pathology)
This document discusses hemodynamic disorders and thrombosis. It covers several topics including edema, congestion, hemorrhage, thrombosis, embolism, and infarction. Edema is an accumulation of fluid in tissues and organs, and can occur in the lungs (pulmonary edema), abdomen (ascites), and brain. Congestion and hyperemia involve increased blood volume in organs and tissues, seen in conditions like heart failure and liver disease. Thrombosis is the formation of a clot (thrombus) in a blood vessel. Key factors in thrombosis are described by Virchow's triad. Thrombi can embolize and block vessels in other organs, potentially leading to infarction or tissue death.
Hemodynamic Disorders
The document discusses various hemodynamic disorders including hyperemia, congestion, thrombosis, embolism, and infarction. Hyperemia is an increased blood volume in tissue from vasodilation. Congestion is increased blood volume from impaired venous return. Thrombosis is the formation of a blood clot within vessels. An embolism occurs when a piece of thrombus or other material blocks a vessel. Infarction is tissue death from blocked arteries or veins.
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Dementia Drugs
Placebo
Number of patients
500
450
Total deaths
23
12
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Individual
1
2
3
4
5
6
7
8
9
10
VA Lines Read Before
3
4
2
4
5
3
4
5
3
3
Laser Procedure
VA Lines Read After
4
4
5
5
6
3
7
6
3
5
Laser Procedure
Construct a 95% confidence interval estimate of the mean difference between the before and after number of visual acuity lines read. Give your answer with two decimals, e.g., (12.34,56.78) (Points : 0.5)
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Conclusion: r results in a value close to 1
Conclusion: an increase in floating insects is related to an increase in insect-eating fish
Conclusion: r results in a value close to -1
Conclusion: an decrease in floating insects causes a decrease in insect-eating fish
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Mon
Tues
Wed
Thurs
Fri
Sat
Sun
10
9
5
8
15
12
11
Determine the value of the χ2 test statistic. Give your answer to two decimals, e.g., 12.34 . (Points : 0.5)
____________________________________________________________________________________
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Let
X
= (
X
1
,...,Xn
) be the blood pressure (measured in
mmHg
) and let
Y
= (
Y
1
,...,Yn
) be the cortisol level (measured in
mcg/dL
) recorded for
n
= 79 patients recruited for a study in a hospital (
Xi
and
Yi
are measurements for the same patient). What test is most appropriate to gather evidence towards the alternative hypothesis that blood pressure is associated with cortisol level? Please provide the reasoning in detail for your answer.
A) The two-sample paired
t
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X
and
Y
differ.
B) The test with the null hypothesis that the Pearson correlation coefficient between
X
and
Y
is zero.
C) The test with the null hypothesis that the regression coefficient is zero in a linear regression with response variable
X
(blood pressure) and explanatory variable
Y
(cortisol level).
(5 points)
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the duration
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X
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X
1
, X
2
,...
), and
the durations
for the control sample be
Y
= (
Y
1
, Y
2
,...
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the duration
? Please provide the reasoning for your answer. Please provide the reasoning in detail for your answer.
A) The
one-sided
two-sample
unpaired t
-test with
H
0: The mean of
X
is
greater than or equal to the mean of Y.
B) The
one-sided
two-sample
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X
is
less than or equal to the mean of Y.
C) The test against the null hypothesis that the Spearman’s ranked correlation coefficient between
X
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Y
is zero.
D) The
one-sided
two-sample
paired t
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H
0: The mean difference between
Xi
and
Yi
is
less than or equal to zero
.
E) The
two-sided
two-sample
paired t
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Xi
and
Yi
is
zero
.
(5 points)
iii) Road vehicle accidents involving ambulances have more detrimental outcomes than accidents involving other similarly sized vehicles (Ray and Kupas, 2005). Measures to avoid such accidents are continually being refined by organizations involved in emergency medical services. Suppose that a city council is interested in knowing if adoption of such measures has lead to an improvement over the last decade. Suppose that the ratio between the number of accidents
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Stat6 chi square test
- 1. Chi square testChi square test
- 2. Uses and ApplicationsUses and Applications Used when you have frequencyUsed when you have frequency distribution of qualitative type of variables.distribution of qualitative type of variables. Applications:Applications: To test goodness of fit.To test goodness of fit. In the 2X2 table, it is used to test whetherIn the 2X2 table, it is used to test whether there is anthere is an associationassociation between the rowbetween the row and the column variables; ie whether theand the column variables; ie whether the distribution of individuals among thedistribution of individuals among the categories of one variable iscategories of one variable is independentindependent of their distribution amongof their distribution among the categories of the other.the categories of the other.
- 3. Example, Influenza vaccination trialExample, Influenza vaccination trial InfluenzaInfluenza NoNo influenzainfluenza TotalTotal VaccineVaccine 2020 (8.3%)(8.3%) 220220 240240 PlaceboPlacebo 8080 (36.4%)(36.4%) 140140 220220 TotalTotal 100100 (21.7%)(21.7%) 360360 460460
- 4. The question is:The question is: Is the difference (in the percentages ofIs the difference (in the percentages of influenza) due to vaccination or occurredinfluenza) due to vaccination or occurred by chance?by chance? To answer, a Chi square test is doneTo answer, a Chi square test is done which compares the observed numbers inwhich compares the observed numbers in each of the four categories in theeach of the four categories in the contingency table with the numbers to becontingency table with the numbers to be expected if there was no difference in theexpected if there was no difference in the effectiveness between the vaccine andeffectiveness between the vaccine and placebo.placebo.
- 5. The expected frequencies:The expected frequencies: InfluenzaInfluenza NoNo influenzainfluenza TotalTotal VaccineVaccine 52.252.2 187.8187.8 240240 PlaceboPlacebo 47.847.8 172.2172.2 220220 TotalTotal 100100 360360 460460
- 6. Solution, continuedSolution, continued Ho: The proportion of influenza among the vaccineHo: The proportion of influenza among the vaccine group = The proportion of influenza among thegroup = The proportion of influenza among the placebo group.placebo group. Level of significance (alpha) = 0.05Level of significance (alpha) = 0.05 D.f. = (No. of rows-1) (No. of columns-1).D.f. = (No. of rows-1) (No. of columns-1). Test statistics: Chi square test.Test statistics: Chi square test. E EO 2 2 )( −∑ =χ
- 7. Solution, continuedSolution, continued 09.53 2.172 )2.172140( 8.187 )8.187220( 8.47 )8.4780( 2.52 )2.5220( 2222 2 = − + − + − + − =χ The tabulated value for X2 for 1 d.f. is 3.841 The calculated value (53.09) > tabulated value Therefore reject Ho and conclude that there is statistically significant difference between the 2 proportions. This difference is unlikely to be due to chance. Therefore the vaccine is effective. P < 0.05