Calculate n 1 p ^ 1 = 44 × 0.523 = 23.012 .
Calculate n 1 ( 1 − p ^ 1 ) = 44 × ( 1 − 0.523 ) = 20.988 .
Calculate n 2 p ^ 2 = 45 × 0.689 = 31.005 .
Calculate n 2 ( 1 − p ^ 2 ) = 45 × ( 1 − 0.689 ) = 13.995 . Since all calculated values are greater than or equal to 10, the normal distribution can be used.
Yes, the conditions are met.
Explanation
Analyze the problem and data We are given two sample sizes n 1 = 44 and n 2 = 45 , and their respective sample proportions p ^ 1 = 0.523 and p ^ 2 = 0.689 . We need to check if the normal distribution can be used to compare the population proportions. To do this, we need to verify that n 1 p ^ 1 ≥ 10 , n 1 ( 1 − p ^ 1 ) ≥ 10 , n 2 p ^ 2 ≥ 10 , and n 2 ( 1 − p ^ 2 ) ≥ 10 .
Calculate the required values Let's calculate each of these values:
n 1 p ^ 1 = 44 × 0.523 = 23.012
n 1 ( 1 − p ^ 1 ) = 44 × ( 1 − 0.523 ) = 44 × 0.477 = 20.988
n 2 p ^ 2 = 45 × 0.689 = 31.005
n 2 ( 1 − p ^ 2 ) = 45 × ( 1 − 0.689 ) = 45 × 0.311 = 13.995
Verify the conditions and conclude Now, let's check if the conditions are met:
n 1 p ^ 1 = 23.012 ≥ 10 - This condition is met.
n 1 ( 1 − p ^ 1 ) = 20.988 ≥ 10 - This condition is met.
n 2 p ^ 2 = 31.005 ≥ 10 - This condition is met.
n 2 ( 1 − p ^ 2 ) = 13.995 ≥ 10 - This condition is met.
Since all four conditions are met, the normal distribution can be used to compare the population proportions.
Examples
In quality control, you might want to compare the proportion of defective items produced by two different machines. If the sample sizes are large enough and the proportions satisfy the conditions for using the normal distribution, you can use a normal approximation to perform a hypothesis test and determine if there is a significant difference in the defect rates between the two machines. This helps in making informed decisions about which machine is more reliable.