Question

In a chemical process, the amount of a certain type of impurity in the output is difficult to control and is thus a random variable. It is known that the standard deviation of the impurity is 0.0015 following an approximately normal distribution. If a random sample of 75 outputs has an average impurity of 0.310, find a 95% confidence interval for the mean amount of the impurity.

Answer 1

Answer:

+/- 0.00033

Explanation:

For a 95 % confidence interval the range is given by

+/- Z * s/sqrt(n)

where Z value is 1.916 for a 95% confidence interval

Therefore the interval is

s is the standard deviation in this case 0.0015

n is 75 the number of outposts

calculating,

+/- 1.916 * 0.0015/sqrt(75) = +/- 0.00033

Answer 2

Answer:

$\\ (Lower\;value =0.3097 ; Upper\;value = 0.3103)$.

We can also represent it as $\\ P(0.3097 \leq \mu \leq 0.3103) = 0.95$.

Explanation:

This is a case of finding two values that define an interval in which is included the population mean for the impurity in the chemical process with a probability of 95%.

We have full information to this respect to solve the question:

1. The population standard deviation, which is $\\ \sigma = 0.0015$.
2. The sample mean, which is $\\ \bar{x} = 0.310$.
3. The sample size $\\ n = 75$.

That is, we can find the 95% confidence interval for a given population standard deviation.

The formula for finding these two values, in these conditions, is as follows:

$\\ Lower\;value = \bar{x} - 1.96\frac{\sigma}{\sqrt{n}}$.

$\\ Upper\;value = \bar{x} + 1.96\frac{\sigma}{\sqrt{n}}$.

We have already confirmed that we have all these values: the population standard deviation, the sample mean and the sample size.

The number 1.96 corresponds to a z-score that is 1.96 times the standard deviation from the mean, and represents the confidence coefficient and depends on the confidence level, which is in this case of 5% (or 0.05). A confidence level of 5% determines this 95 % confidence interval.

Notice that we are going to use a mean obtained from a sample of 75 elements $\\ \bar{x} = 0.310$, which represents a sample mean.

As a result, in order to find the two values that define the 95% confidence interval, we can proceed as follows:

Lower value of the 95% confidence interval

$\\ Lower\;value = \bar{x} - 1.96\frac{\sigma}{\sqrt{n}}$.

$\\ Lower\;value = 0.310 - 1.96\frac{0.0015}{\sqrt{75}}$.

$\\ Lower\;value =0.3097$.

Upper value of the 95% confidence interval

$\\ Upper\;value = \bar{x} + 1.96\frac{\sigma}{\sqrt{n}}$.

$\\ Upper\;value = 0.310 + 1.96\frac{0.0015}{\sqrt{75}}$.

$\\ Upper\;value = 0.3103$.

Thus, the 95% confidence interval, or the interval for which there is a probability of 95% to find the population mean for this certain type of impurity:

$\\ Lower\;value =0.3097$.

$\\ Upper\;value = 0.3103$.

That is:

$\\ P(0.3097 \leq \mu \leq 0.3103) = 0.95$.