The design specifications for a new line of disk drives indicates 
that the mean seek time is 12.5 ms. A quality assurance analyst 
believes that the disk drives are faster and decides to investigate.

1.  Give the appropriate null and one-sided research hypothesis
    that corresponds to this quality assurance analysts belief.

    Answer:     H0: mu(y) = 12.5ms
                Ha: mu(y) < 12.5ms


2.  The analyst collects a random sample of 100 disk drives from
    a production run and determines that the mean seek time is
    12.2 ms with a standard deviation of 0.3 ms.

    a. Calculate the p-value for the hypotheses stated above?

       Answer: Note that in this case we have a large sample.
       We first need to compute the z value for a mean
       seek time of 12.2 ms.

               s(ybar) = 0.3/sqrt(100)
	               = 0.3/10
                       = 0.03

       Hence, since mu(ybar)=mu(y)=12.5 then:

               z = (12.2-12.5)/0.03
	         = -0.3/0.03
	         = -10	

       We need to find the area under the standard normal
       curve for z less than -10. Notice that this z value
       is larger than any value in the z table. Hence, the 
       required area is approximately zero. Hence the
       p-value is approximately zero (i.e. much less than
       1% or, expressed as a probability, 0.01).


    b. Is this significant, highly significant, or non-significant?

       Answer: Highly significant.


    c. Comment on the analyst's point of view?

       answer: Since this is a highly significant result, reject H0. 
       The analyst therefore has strong evidence to support her
       point of view.



3.  Suppose that the sample size is 9 instead of 100, with
    the same mean and standard deviation. how would this affect
    your answers to Q2 above?

    Answer: Since the sample is small we must assume that seek time
    is normally distributed in order to re-compute the p-value.

               s(ybar) = 0.3/sqrt(9)
                       = 0.3/3
                       = 0.1

               t = (12.2-12.5)/0.1
	         = -0.3/0.1
	         = -3

    From the t table, since we have 8 degrees of freedom, this is 
    between 1% and 0.5% (as a probability 0.01 and 0.005) and so 
    the p-value is highly significant and our conclusion remains
    the same.