The National Institute of Standards and Technology (NIST) supplies "standard materials" whose physical properties are supposed to be known. For example, you can buy from NIST a liquid whose electrical conductivity is supposed to be 5. (The units for conductivity are microsiemens per centimeter. Distilled water has conductivity 0.5.) Of course, no measurement is exactly correct. NIST knows the variability of its measurements very well, so it is quite realistic to assume that the population of all measurements of the same liquid has the Normal distribution with mean ? equal to the true conductivity and standard deviation ? = 0.2. Here are 6 measurements on the same standard liquid, which is supposed to have conductivity 5:

"5.32 4.88 5.10 4.73 5.15 4.75"

NIST wants to give the buyer of this liquid a 90% confidence interval for its true conductivity. What is this interval? Give a 90% confidence interval for the true conductivity.

A.4.8143 to 5.1623

B.4.8283 to 5.1483

C.4.6593 to 5.3173

D.4.8540 to 5.1226

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

[tex]\bar X[/tex] represent the sample mean for the sample

[tex]\mu[/tex] population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size

Solution to the problem

The confidence interval for the mean is given by the following formula:

[tex]\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex] (1)

In order to calculate the mean and the sample deviation we can use the following formulas:

[tex]\bar X= \sum_{i=1}^n \frac{x_i}{n}[/tex] (2)

(3)

The mean calculated for this case is

The sample deviation calculated [tex]s=0.238[/tex]

We assume that the population deviation is [tex] \sigma =0.2[/tex]

Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-NORM.INV(0.05,0,1)".And we see that [tex]z_{\alpha/2}=1.64[/tex]

Now we have everything in order to replace into formula (1):

[tex]4.988-1.64\frac{0.2}{\sqrt{6}}=4.8540[/tex]

[tex]4.988+1.64\frac{0.2}{\sqrt{6}}=5.1226[/tex]

So on this case the 95% confidence interval would be given by (4.8540;5.1226)

So the correct option would be:

D.4.8540 to 5.1226