Question

In: Statistics and Probability

Hand disinfection is frequently recommended for prevention of transmission of the rhinovirus that causes the common...

Hand disinfection is frequently recommended for prevention of transmission of the rhinovirus that causes the common cold. In particular, hand lotion containing 2% citric acid and 2% malic acid in 70% ethanol (HL+) has been found to have both immediate and persistent ability to inactivate rhinovirus (RV) on the hands in an experimental setting. Is hand disinfection effective in reducing the risk of infection in a natural setting? A total of 209 volunteers were assigned at random to either the HL+ group, which used the hand lotion every three hours or after hand washing, and a control group, which was asked to use routine hand washing but to avoid the use of alcohol-based hand sanitizers. Here are the data on the numbers of subjects with and without RV infection in the two groups over the 10-week study period: RV INFECTION Yes No HL+ 50 66 Control group 48 45 Let p1 p 1 is the proportion for HL+ group. Let p2 p 2 is the proportion for control group. Do we have evidence at α=5% α = 5 % that hand sanitizers reduce the chance of an RV infection? (Use pˆ1−pˆ2 p ^ 1 − p ^ 2 for z z calculations.) z(±0.0001)= z ( ± 0.0001 ) = P P -value (± ± 0.0001) = We have evidence that the proportions are different. We don't have evidence that the proportions are different.

Solutions

Expert Solution

Answer:

Given that,

Hand disinfection is frequently recommended for prevention of transmission of the rhinovirus that causes the common cold.

In particular, hand lotion containing 2% citric acid and 2% malic acid in 70% ethanol (HL+) has been found to have both immediate and persistent ability to inactivate rhinovirus (RV) on the hands in an experimental setting.

Given table:

RV Infection
Yes No
HL+ 50 66
Control Group 48 45

Calculations:

HL+ Control Group
x1=50 x2=48
n1=116 n2=94
p1=x1/n1=50/116=0.431 p2=x2/n2=48/94=0.511

Standard error:

=0.048477

=0.0485

Therefore, Standard error is 0.0485.

Test Statistic (z):

=-1.649484

Therefore, z=-1.6495

P-value:

Evidence at α=5%=0.05

From the P-value calculator at 0.05 level of significance.

The P-Value is 0.0496.

We do not have evidence that the proportion are different. As p value is greater then 5%.



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