Market research indicates that a new product has the potential to make the company an additional $1.6 million, with a standard deviation of $2.5 million. If these estimates were based on a sample of 12 customers from a normally distributed data set, what would be the 95% confidence interval?a. (0.00, 3.69) b. (-0.13, 3.33) c. (-0.49, 3.69) d.(-0.09, 4.10)

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:

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

In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:

[tex]df=n-1=12-1=11[/tex]

Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,11)".And we see that [tex]t_{\alpha/2}=2.201[/tex]

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

[tex]1.6-2.201\frac{2.5}{\sqrt{12}}=0.011[/tex]

[tex]1.6+2.201\frac{2.5}{\sqrt{12}}=3.188[/tex]

So on this case the 95% confidence interval would be given by (0.011;3.2188)

But on this case the most accurate option seems to be : a. (0.00, 3.69)