Question

A random sample of n measurements was selected from a population with unknown mean μ and known standard deviation σ. Calculate a 95% confidence interval for μ for the given situation. Round to the nearest hundredth when necessary. n = 125, x-bar = 91, σ = 20 Group of answer choices

Answer 1

Answer:

The 95% confidence interval for μ for the given situation is between 87.49 and 94.51.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$.

So it is z with a pvalue of $1-0.025 = 0.975$, so $z = 1.96$

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 1.96\frac{20}{\sqrt{125}} = 3.51$

The lower end of the interval is the sample mean subtracted by M. So it is 91 - 3.51 = 87.49

The upper end of the interval is the sample mean added to M. So it is 91 + 3.51 = 94.51

The 95% confidence interval for μ for the given situation is between 87.49 and 94.51.

Answer 2

Answer:

$91-1.96\frac{20}{\sqrt{125}}=87.494$    

$91+1.96\frac{20}{\sqrt{125}}=94.506$    

So on this case the 95% confidence interval would be given by (87.494; 94.506). We have 95% of confidence that the true mean is between the limits founded.  

Step-by-step explanation:

Notation

$\bar X=91$ represent the sample mean

$\mu$ population mean (variable of interest)

$\sigma=20$ represent the population standard deviation

n=125 represent the sample size  

Confidence interval

The confidence interval for the mean if we know the deviation 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 significance is $\alpha=0.05$ 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):

$91-1.96\frac{20}{\sqrt{125}}=87.494$    

$91+1.96\frac{20}{\sqrt{125}}=94.506$    

So on this case the 95% confidence interval would be given by (87.494; 94.506). We have 95% of confidence that the true mean is between the limits founded.