Question

Consider a computer test to indicate if you like Math 123 or dislike Math 123. If positive, you like the class and if negative you dislike the class. In a sample of 25 thousand people, 80% of the population indicated they liked the class. If the test is 85.5% accurate, what are the chances that the test showed the person disliked Math 123, but in reality they actually liked the course? Round your percentage to 2 decimals.

Answer 1

To find the chances that the test showed a person disliked Math 123, but in reality they actually liked the course, we need to calculate the probability of a false negative. The chances that the test showed the person disliked Math 123, but in reality they actually liked the course is 29%.

Step-by-step explanation:

To find the chances that the test showed a person disliked Math 123, but in reality, they actually liked the course, we need to calculate the probability of a false negative. The test accuracy is given as 85.5%, which means it correctly identifies a positive (liking the class) 85.5% of the time. Therefore, the test incorrectly identifies a negative (disliking the class) 14.5% of the time. We are given that 80% of the population actually likes the class. So, the probability of a false negative can be calculated as follows:

Probability of false negative = Probability of disliking the class (negative) × Probability of test showing negative given that the person likes the class × 100

Probability of false negative = (100 - 80) × (14.5/100) × 100

Probability of false negative = 20 × 0.145 × 100 = 29%

Therefore, the chances that the test showed the person disliked Math 123, but in reality they actually liked the course is 29%.

Answer 2

14.5%

Step-by-step explanation:

80% of the population indicated they liked the class.

Therefore: (80% X 25000)=20000 liked the class

Out of the 20,000, the test indicates that (85.5% of 20,000)=17100 are positive.

The table summarizes the computation.

`\left|\begin{array}c&Positive&Negative&Total\\---&----&----&----\\Liked&17100&2900&20000\\Disliked&725&4275&5000\\\---&----&----&----\\Total&17825&7175&25000\end{array}\right|`

The table below show the various probabilities.

`\left|\begin{array}c&Positive&Negative&Total\\---&----&----&----\\Liked&0.855&0.145&0.8\\Disliked&0.145&0.855&0.2\\\---&----&----&----\\Total&0.71&0.29&1\end{array}\right|`

Therefore:

P(Negative|Liked)=0.145

= 0.145X100=14.5% (to two decimal places)