Question

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 an iron rod whose electrical conductivity is supposed to be 10.1 at 293 kelvin. (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 rod has the Normal distribution with mean μequal to the true conductivity and standard deviation σ = 0.1. Here are six measurements on the same standard iron rod, which is supposed to have conductivity 10.1.
10.07 9.89 10.04 10.16 10.21 10.11

--> NIST wants to give the buyer of this iron rod a 90% confidence interval for its true conductivity. What is this interval? (Round your answers to three decimal places.)

ANSWER : __________ to ___________ microsiemens per centimeter

Answer 1

Answer:

$10.08-1.64\frac{0.1}{\sqrt{6}}=10.013$    

$10.08+1.64\frac{0.1}{\sqrt{6}}=10.147$    

So on this case the 90% confidence interval would be given by (10.013;10.147)    

Step-by-step explanation:

Previous concepts

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".

$\bar X$ represent the sample mean for the sample  

$\mu$ population mean (variable of interest)

$\sigma =0.1$ represent the population standard deviation

n represent the sample size  

Solution to the problem

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

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

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

$\bar X= \sum_{i=1}^n \frac{x_i}{n}$ (2)  

$s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}$ (3)  

The mean calculated for this case is $\bar X=10.08$

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$, 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 $z_{\alpha/2}=1.64$

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

$10.08-1.64\frac{0.1}{\sqrt{6}}=10.013$    

$10.08+1.64\frac{0.1}{\sqrt{6}}=10.147$    

So on this case the 90% confidence interval would be given by (10.013;10.147)