Question

Government regulations dictate that for any production process involving a certain toxic chemical, the water in the output of the process must not exceed 7890 parts per million (ppm) of the chemical. For a particular process of concern, the water sample was collected by a manufacturer 25 times randomly and the sample average x was 7914 ppm. It is known from historical data that the standard deviation ơ is 120 ppm. Complete parts (a) and (b) below. Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table.

(a) What is the probability that the sample average in this experiment would exceed the government limit if the population mean is equal to the limit? Use the Central Limit Theorem. The probablity is 0.5000 (Round to four decimal places as needed.)
(b) Is an observed x 7914 in this experiment firm evidence that the population mean for the process exceeds the government limit?

Answer 1

To determine the probability that the sample average in the experiment exceeds the government limit, you can use the Central Limit Theorem. The probability is approximately 0.1587, or 15.87%. However, based solely on the observed sample average of 7914 ppm, we cannot definitively conclude whether the population mean for the process exceeds the government limit.

Step-by-step explanation:

To determine the probability that the sample average in the experiment exceeds the government limit, we can use the Central Limit Theorem. According to the Central Limit Theorem, the distribution of sample means will be approximately normally distributed, regardless of the shape of the population distribution, as long as the sample size is large enough. In this case, since the sample size is 25, we can assume that the distribution of sample means follows a normal distribution.

To calculate the probability, we need to determine the z-score, which measures how many standard deviations the sample average is from the population mean. The formula for calculating the z-score is:

z = (x - μ) / (σ / sqrt(n))

Where:

• z is the z-score
• x is the sample average (7914 ppm)
• μ is the population mean (7890 ppm)
• σ is the standard deviation (120 ppm)
• n is the sample size (25)

Plugging in the values, we have:

z = (7914 - 7890) / (120 / sqrt(25))

z = 24 / (120 / 5)

z = 24 / 24

z = 1

Now, we can use the standard normal distribution table to find the probability associated with a z-score of 1. The table provides the area to the left of the z-score, so we need to subtract this value from 1 to find the probability that the sample average exceeds the government limit. Looking up a z-score of 1 in the table, we find that the area to the left is 0.8413. Subtracting this from 1 gives us the probability of 0.1587. Therefore, the probability that the sample average in this experiment would exceed the government limit is 0.1587 or 15.87%.

For part (b), we cannot make a definitive conclusion about whether the population mean for the process exceeds the government limit based solely on the observed sample average of 7914 ppm. The probability we calculated in part (a) indicates that it is possible for the sample average to exceed the government limit even if the population mean is equal to the limit. To make a more confident inference, additional hypothesis testing or confidence interval calculations would be needed.