A manufacturer wants to assess the proportion of defective items in a large batch produce by a particular machine. He tests a random sample of 300 items and finds that 45 items are defective. If 200 such tests are performed and a 95% confidence interval calculated for each, what is the probability that more than 194 of the confidence intervals cover the true proportions?
To solve this problem, we need to understand the concept of confidence intervals and how they are calculated. A confidence interval is a range of values that is used to estimate an unknown population parameter, in this case, the proportion of defective items in the large batch produced by the machine.
In this scenario, the manufacturer performs 200 tests and calculates a 95% confidence interval for each test. This means that if we were to repeat the experiment many times, we would expect the true proportion of defective items to be within the calculated confidence interval in approximately 95% of the tests.
To calculate a confidence interval for a proportion, we can use the formula:
CI = p̂ ± z * sqrt((p̂ * (1 - p̂)) / n)
Where:
- CI is the confidence interval
- p̂ is the sample proportion of defective items
- z is the z-score associated with the desired level of confidence (in this case, 95%)
- sqrt is the square root function
- n is the sample size
Using this formula, we can calculate the upper and lower bounds for each confidence interval. If the true proportion falls within these bounds, we consider the confidence interval to cover the true proportion.
Now, let's calculate the confidence intervals for the 200 tests and determine how many of them cover the true proportion.
Given that the sample size is n = 300 and the number of defective items is x = 45, the estimated proportion of defective items is p̂ = x / n = 45 / 300 = 0.15.
For a 95% confidence level, the corresponding z-score is approximately 1.96.
Now, we can calculate the upper and lower bounds for the confidence interval:
CI = p̂ ± z * sqrt((p̂ * (1 - p̂)) / n)
CI = 0.15 ± 1.96 * sqrt((0.15 * (1 - 0.15)) / 300)
Calculating the upper and lower bounds, we get:
Lower bound = 0.15 - 1.96 * sqrt((0.15 * (1 - 0.15)) / 300)
Upper bound = 0.15 + 1.96 * sqrt((0.15 * (1 - 0.15)) / 300)
Now, for each of the 200 tests, we can determine if the true proportion falls within the confidence interval. If it does, we count it as covering the true proportion. Finally, we calculate the probability that more than 194 of the confidence intervals cover the true proportions.
To determine this probability, we can use the binomial distribution formula:
P(X > 194) = 1 - P(X ≤ 194) = 1 - ∑ (200, 0.95, x) [nCx * p^x * (1-p)^(n-x)]
Where:
- P(X > 194) is the probability of having more than 194 confidence intervals covering the true proportions
- P(X ≤ 194) is the probability of having 194 or fewer confidence intervals covering the true proportions
- ∑ represents the sum from x = 0 to x = 194
- n is the number of trials (200 tests)
- p is the probability of a single test covering the true proportion (approximately 0.95)
Using the formula, we can calculate the probability that more than 194 of the confidence intervals cover the true proportions. The exact calculation would involve a sum of many terms, but it can be approximated using statistical software or online calculators.
Please note that this calculation assumes that the sample is truly random, the sample size is large enough for approximations, and the assumptions of the binomial distribution are met.