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Assume a researcher recruits 150 African American and Caucasian individuals taking warfarin to determine if there is a difference in the mean dosage of the medication needed to cause a decrease in their INR blood test. If the mean dosage for 75 Caucasian individuals required to get their INR blood test in range is 6.1 mg with a standard deviation of 1.1 mg and the mean dosage for 75 African American individuals required to get their INR blood test in range is 4.3 mg with a standard deviation of 0.9 mg, the z value obtained while calculating the test statistic is approximately:

4.4
1.8
11.0
5.16

Respuesta :

Answer:

Z=11.0

Step-by-step explanation:

Let Caucasian individuals be the population 1 and African American be the population 2. So, we are given that

n=150

n1=75

mean1=xbar1=6.1.

standard deviation1=S.D1=σ1=1.1.

Variance1=V(x1)=σ1²=1.1²=1.21.

n2=75.

mean2=xbar2=4.3.

standard deviation2=S.D2=σ2=0.9.

Variance2=V(x2)=σ2²=0.9²=0.81.

The z-statistic is

[tex]Z=\frac{xbar1-xbar2}{\sqrt{\frac{V(x1)}{n1}+ \frac{V(x2)}{n2} } }[/tex]

[tex]Z=\frac{6.1-4.3}{\sqrt{\frac{1.21}{75}+ \frac{0.81}{75} } }[/tex]

[tex]Z=\frac{1.8}{\sqrt{0.0161+ 0.0108 } }[/tex]

[tex]Z=\frac{1.8}{\sqrt{0.0269 } }[/tex]

[tex]Z=\frac{1.8}{0.164 } }[/tex]

Z=10.97

Z=11.0

So, the z value obtained while calculating the test statistic is approximately 11.0.