Question

A factory makes propeller drive shafts for ships. A quality assurance engineer at the factory needs to estimate the true mean length of the shafts. She randomly selects four drive shafts made at the factory, measures their lengths, and finds their sample mean to be 1000 mm. The lengths are known to have a normal distribution with a standard deviation is 2 mm. Calculate a 95% confidence interval for the true mean length of the shafts. Input your answers for the lower bound and the upper bound. Input the lower bound.

Answer 1

Answer:

$1000-1.96\frac{2}{\sqrt{4}}=998.04$    

$1000+1.96\frac{2}{\sqrt{4}}=1001.96$    

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".

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)  

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that $z_{\alpha/2}=1.96$

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

$1000-1.96\frac{2}{\sqrt{4}}=998.04$    

$1000+1.96\frac{2}{\sqrt{4}}=1001.96$