Question

Sales personnel for Skillings Distributors submit weekly reports listing the customer contacts made during the week.

Answer 1

Answer:

a) The 90% confidence interval would be given by (18.52;20.49)  

b) The 95% confidence interval would be given by (18.32;20.68)    

Step-by-step explanation:

1) 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".

$\bar X=19.5$ represent the sample mean for the sample  

$\mu$ population mean (variable of interest)

$\sigma=5.2$ represent the population standard deviation

n=75 represent the sample size  

2) Confidence interval 90%

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 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.005$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.005,0,1)".And we see that $z_{\alpha/2}=1.64$

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

$19.5-1.64\frac{5.2}{\sqrt{75}}=18.515$    

$19.5+1.64\frac{5.2}{\sqrt{75}}=20.485$

So on this case the 90% confidence interval would be given by (18.52;20.49)    

3) Confidence interval 95%

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):

$19.5-1.96\frac{5.2}{\sqrt{75}}=18.323$    

$19.5+1.96\frac{5.2}{\sqrt{75}}=20.677$

So on this case the 95% confidence interval would be given by (18.32;20.68)