Question

In a random sample of 651 computer scientists who subscribed to a web-based daily news update, it was found that the average salary was $46,816 with a population standard deviation of $12,557. Calculate a 91 percent confidence interval for the mean salary of computer scientists. Choose the closest value from your table.

Answer 1

Answer:

$46816-1.695\frac{12557}{\sqrt{651}}=45981.81$    

$46816+ 1.695\frac{12557}{\sqrt{651}}=46650.19$    

And the confidence interval would be given by: (45981.81; 46650.19)

Step-by-step explanation:

Data given

$\bar X=46816$ represent the sample mean

$\mu$ population mean (variable of interest)

$\sigma=12557$ represent the population standard deviation

n=651 represent the sample size  

Confidence interval

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 0.91 or 91%, the significacne is  $\alpha=0.09$ and $\alpha/2 =0.045$, and the critical value is $z_{\alpha/2}=1.695$

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

$46816-1.695\frac{12557}{\sqrt{651}}=45981.81$    

$46816+ 1.695\frac{12557}{\sqrt{651}}=46650.19$    

And the confidence interval would be given by: (45981.81; 46650.19)