>>Dr. Kash Barker: Let’s take a look at our first example. Say you’re the quality
controller engineer for the manufacturer, for a manufacturer of electronic subassemblies.
And those subassemblies each require a computer chip. You’re considering two different suppliers,
appropriately named Supplier One and Supplier Two. A random sample of two hundred chips
from Supplier One is examined and eight are found to be defective. From Supplier Two we
have a random sample of two hundred and fifty chips and thirteen of those are found to be
defective. Okay. So here we’re calling our success as a defective. Okay. It’s kind of counterintuitive
but that’s the notation we’ve been using all semester. So we have from population one,
Supplier One, we have a sample of two hundred chips. We found the number of successes to
be eight, therefore p one hat, the proportion of successes is eight over two hundred which
is 0.04. Okay. So what is our estimater, what is our estimate for the proportion of defective
units from Supplier One in general? Our estimate is 0.04. From Supplier Two we collect two
hundred and fifty chips. We see that of those thirteen are defective therefore our point
estimate for the proportion of defective units, yeah defective chips from Supplier Two is
0.052. Okay, and let’s say we want to know if supplier two produces a larger proportion
of defective chips. We want to know if proportion two produces a larger proportion of defective
chips. That would mean p one minus p two would be negative. Okay. If p two is larger than
p one minus p two would be a negative number. Okay. Larger implies something in the alternative,
okay. P one, we want to know if p one minus, if if we want to know if Supplier Two produces
a larger number of defective chips. That means p one minus p two must be less than zero,
strictly less than zero. If p two is larger than p one, p one minus p two would be strictly
less than zero. Okay. Strictly less than zero does not imply equality, equality must be in the
null, so we have this statement in the null hypothesis. Okay, so now we’ve just got
to calculate a test statistic. That’s a z test statistic. That would be p one hat
minus p two hat, from that we’re going to subtract off p one minus p two, and we’re
going to divide that by the standard deviation of the sampling distributions for the differences
in the sample proportions. That is p one hat, one minus p one hat, over n one, plus p two
hat times one minus p two hat, over n two. Okay. We’ve got all of these parameters we
just got to plug them in. P one hat, 0.04. P two hat, 0.052. The difference that we want
to detect? Just zero. Down here we’ve got 0.04 times one minus 0.04 divided
by two hundred plus 0.052 times one minus 0.052 divided by two fifty, and we’re going to
take the square root of all of that. Okay. If we work this out we should get negative
0.608. Okay so our test statistic negative 0.608. And we want to draw conclusions let’s
say with ninety-five percent confidence. So let’s find the p value associated with this
test statistic. Okay. So the p value here would be the area in the lower tail, in the lower tail found from this
test statistic negative 0.608. Okay. This is a z distribution we want to find this area down
here, that is the p value. Okay? If we go use Excel, or actually we could look this up in the z
table. We could go down to the 0.6 row, the 0.01 column. Okay and what is that going to give
us? Remember about our table, okay. The table gives us this area, right, between zero and 0.61.
Okay. We want this area, this area corresponds to that right there. Okay. So we’re going to have
to take what we find in the lookup table which is 0.2291 and we’re going to have to subtract
that from 0.5. Okay. So this area right here the p value is equal to 0.5 minus 0.2291 which
is equal to 0.2709. So this area right up here 0.2709. Okay. That’s our p value.
Okay. That’s a very big p value. Bigger than any alpha that we would ever choose okay.
Much bigger than any alpha we would ever choose. We do not reject the null. P value greater
than alpha, we do not reject the null. And we conclude that the proportion for population
two is not greater than the proportion for population one. The proportion of defectives
from Supplier Two is not greater than the proportion of defectives, defective chips from population
one. And we draw that, well we could say with ninety-five percent confidence, but our p
value is just very large. It would be larger than any amount of confidence we would ever
choose.