Question

Suppose you work for Fender Guitar Company and you are responsible for testing the integrity of a new formulation of guitar strings. To perform your analysis, you randomly select 52 'high E' strings and put them into a machine that simulates string plucking thousands of times per minute. You record the number of plucks each string takes before failure and compile a dataset. You find that the average number of plucks is 5,314.4 with a standard deviation of 116.68. A 90% confidence interval for the average number of plucks to failure is (5,287.3, 5,341.5).
From the option listed below, what is the appropriate interpretation of this interval?

1) We are 90% confident that the average number of plucks to failure for all 'high E' strings tested is between 5,287.3 and 5,341.5
2) We cannot determine the proper interpretation of this interval.
3) We are 90% confident that the proportion of all 'high E' guitar strings fail with a rate between 5,287.3 and 5,341.5
4) We are certain that 90% of the average number of plucks to failure for all 'high E' strings will be between 5,287.3 and 5,341.5 5
5) We are 90% confident that the average number of plucks to failure for all 'high E' strings is between 5,287.3 and 5,341.5

Answer 1

Answer:

$5314.4-1.675\frac{116.68}{\sqrt{52}}=5287.30$  

$5314.4+1.675\frac{116.68}{\sqrt{52}}=5341.5$  

So on this case the 90% confidence interval would be given by (5287.3;5341.5)

And the best intrpretation is:

 5) We are 90% confident that the average number of plucks to failure for all 'high E' strings is between 5,287.3 and 5,341.5

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".  

$\bar X=5314.4$ represent the sample mean  

$\mu$ population mean (variable of interest)  

$s=116.68$ represent the sample standard deviation  

n=52 represent the sample size  

90% confidence interval  

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

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

The degrees of freedom are given by:

$df = n-1= 52-1=51$

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 table to find the critical value. The excel command would be: "=-T.INV(0.05,51)".And we see that $t_{\alpha/2}=1.675$  

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

$5314.4-1.675\frac{116.68}{\sqrt{52}}=5287.30$  

$5314.4+1.675\frac{116.68}{\sqrt{52}}=5341.5$  

So on this case the 90% confidence interval would be given by (5287.3;5341.5)

And the best intrpretation is:

 5) We are 90% confident that the average number of plucks to failure for all 'high E' strings is between 5,287.3 and 5,341.5