In a random sample of 95 sections of pipe in a chemical plant, 16 showed signs of serious corrosion. Construct a 98% confidence interval for the true proportion of pipe sections showing signs of serious corrosion, using the large sample confidence interval formula.
To construct a confidence interval for the true proportion of pipe sections showing signs of serious corrosion, we will use the large sample confidence interval formula.
The formula for a confidence interval is:
CI = p̂ ± Z * √( p̂ * (1 - p̂) / n )
Where:
- CI represents the confidence interval
- p̂ is the sample proportion (number of successes divided by the sample size)
- Z is the Z-score corresponding to the desired level of confidence
- n is the sample size
In this case, we have a random sample of 95 sections of pipe, and 16 of them showed signs of serious corrosion. So, the sample proportion (p̂) is 16/95.
Now, we need to find the Z-score for a 98% confidence level. Since the confidence level is 98%, we need to find the Z-score that corresponds to an area of 0.98 in the standard normal distribution table. The Z-score for a 98% confidence level is approximately 2.33.
Now, we can plug in the values into the formula:
CI = 16/95 ± 2.33 * √((16/95) * (1 - 16/95) / 95)
Calculating this, we find:
CI = 0.168 ± 2.33 * √(0.168 * 0.832 / 95)
Simplifying further, we get:
CI = 0.168 ± 2.33 * sqrt(0.00174904)
CI = 0.168 ± 2.33 * 0.0418
Finally, we have:
CI ≈ 0.168 ± 0.0976
Therefore, the 98% confidence interval for the true proportion of pipe sections showing signs of serious corrosion is approximately 0.07 to 0.27.