You wish to examine whether hot air hand blowers or paper in a medical facility have an effect on the rate of respiratory incidence. you follow 98 health care workers in primary care clinic during flu season. you observe the following incidence of respiratory disease and preferred method of hand drying. 46 workers preferred paper for hand drying, 52 preferred hot air blowers. during the study period 29 of the 98 workers experienced a respiratory illness.
observed values sick not sick
paper 19 27 49
hotair 10 42 52
29 68 98
your null and alternative hypothesis for the chi square test comparing the exposure (air vs paper) and outcome (ill or not ill is hnull there is no association between the drying choice and ilness halt there is an association between drying choice and illness
a. true
b. false

There is an association between drying choice and illness

a.True

Step-by-step explanation:

H_0: There is no association between the drying choice and ilness

H_a: There is an association between drying choice and illness

(Two tailed chi square test)

Observations are given as per the following table

Observed    

SickNot sick  

Paper192746

Hot air104252

296998

Expectedrow total*col total/grand total  

Paper13.612244932.387755146

Hot air15.387755136.612244952

296998

Chi square =(observed-expected)^2/Expected    

Paper2.1324847780.8962617183.028746496

Hot air1.8864288420.7928469052.679275747

4.018913621.6891086235.708022243

Thus chi square = 5.708

degrees of freedom = (r-1)(c-1) = 1

p value = 0.0167

Since p <0.05 our alpha we reject null hypothesis

At 5% significance level, we can say

There is an association between drying choice and illness

a.True

1.true 2.false? ? i think. .

step-by-step explanation:

answer: 48

just remember to use pemdas:

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