Answer: a) 0.7878, b) 0.8723
Step-by-step explanation:
Since we have given that
Nicotine present Nicotine absent
Self-reported non smoker 82 14
Self-reported smoker 12 52
(a) What is the sensitivity of the self-reported smoking status?
[tex]P(smoker|Nicotine\ present)=\dfrac{52}{52+14}=\dfrac{52}{66}=0.7878[/tex]
(b) What is the specificity of the self-reported smoking status?
[tex]P(Non-smoker|Nicotine\ absent}=\dfrac{82}{82+12}=\dfrac{82}{94}=0.8723[/tex]
Hence, a) 0.7878, b) 0.8723