Has anyone got mini-tab and if so could they help me with this question?

Question

A company uses 120-min video tapes in its training course. It is concerned that the mean length μ of the tapes that it is using may be less than the 120-minutes. It tests a random sample of 9 tapes and finds that they have the following lengths (in mins):

121,117, 116, 117, 120, 118, 118, 115, 120

It can be assumed these lengths of these tapes are Normally distributed.

i) Perform suitable hypothesis test at the 5% significance levelto check whether there is evidence that the data supports the companys concern. Report conclusion.

ii) Find 95% confidence interval for the mean length of the tapes in the population from which the sample was drawn

Answer 1

Now I'm having to go back 20 years to my college statistics class so I might have a few things wrong in this.  The second link below has an example of exactly this kind of problem and you may want to look at that.

You will need to do a t test and compute the average and standard deviation (see hyperlinks below) of your sample as well as the degrees of freedom.

X=(121+117+116+117+120+118+118+115+120...

X=118

S=2 (long and involved calculation to get this one, see hyperlink below or use a statistical calculator)

Df = n-1 = 9-1 = 8

Compute your T value:

t = (X-μ) / (S/√(n))

t = (118-120) / (2/√(9))

t = -2 / (2/3)

t = -2(3) / 2

t = -3

Now go to your t table (see last hyperlink below, and look up the row for your degree of freedon (e.g. 8), and the column for a 95% confidence (i.e. (1-.95)/2=.025) then find the t value from the table.  The t value is 2.306.  The t value you calculated has an absolute value greater than the tables t value so you can conclude that based on this sample, the population of all tapes does not have a mean length of 120 minutes, and that the actual mean is less than 120 minutes .

Answer 2

One-Sample T: C1

Variable  N     Mean  StDev  SE Mean        95% CI

C1        9  118.000  2.000    0.667  (116.463, 119.537)