Find the sample proportions and test statistic for equal proportions. Is the decision close? Find the p-value.
a. Dissatisfied workers in two companies: x1 = 40, n1 = 100, x2 = 30, n2 = 100, α = .05, two tailed test.
1. Set up the hypotheses:
or, equivalently,
2. Determine the critical region for the test statistic Z:
Since , for rejection of , and since
this is a 2-tailed test, we require that either
or
3. Calculate the three "p-hats":
[Note: I can't put a hat on on here, so I will
write for the point estimate of ,
for the point estimate of , and
for the pooled estimate of . The
stands for "hat".
and
4. Calculate the test-statistic by this
terrible formula:
This is not in the rejection-region so we cannot
reject the hypothesis that the proportions of
dissatisfied workers are equal in the two
companies. So, yes the decision that they are
close is good.
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The p-value is the probability that the test
statistic Z would be either at least as high as
1.48 or at least as low as -1.48, if in fact
the two proportions were equal:
So we use a standard normal table to find the
area to the left of z=-1.48, which is .0694.
That's the probability that it is at least as
low as 1.48. The prbability that it is at
least as high as +1.48 is the same so we
double the value .0694 and get a p-value of
.1388.
Since this p-value is greater than .05, this
is another way to determine that we cannot
reject , so the proportions are
closely the same.
Now if you have a TI-83 or 84 calculator,
you could just clear the screen and do this:
press STAT
use arrow keys to move cursor to hilite TESTS
press 6
beside x1: type 40
beside n1: type 100
beside x2: type 30
beside n2: type 100
beside p1: place cursor on
press ENTER
press cursor on Calculate
press ENTER
You see
2-propZtest
p(hat)1=.4
p(hat)2=.3
p(hat)=.35
Notice that the p-value found by the table was .1388, and
the p-value found by calculator was .1382077316, but that
difference is because with the table we have to round the
z-value from 1.482498633 to 1.48.
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b. Rooms rented at least a week in advance at two hotels: x1 = 24, n1 = 200, x2 = 12, n2 = 50, α = .01, left-tailed test.
1. Set up the hypotheses:
or, equivalently,
2. Determine the critical region for the test statistic Z:
Since , for rejection of , and since
this is a left-tailed test, we require that ,
since -2.33 is the z-value that has the area .01 to the left
of it.
3. Calculate the three "p-hats" same as the other problem:
and
4. Calculate the test-statistic by the same
terrible formula:
This is not in the rejection-region so we cannot
reject the hypothesis that the proportions of
rooms rented at least a week in advance are equal
in the two hotels. So, yes the decision that they
are close is good.
The p-value is the probability that the test
statistic Z would be at least as low as -2.16,
if in fact the two proportions were equal:
So we use a standard normal table to find the
area to the left of z=-2.16, which is .0154.
That's the probability that it is at least as
low as -2.16. So that is the p-value. [We
only double in the case of a two-tailed test.]
Since this p-value is greater than .01, this
is another way to determine that we cannot
reject , so the proportions are
closely the same.
Now if you have a TI-83 or 84 calculator,
you could just clear the screen and do this:
press STAT
use arrow keys to move cursor to hilite TESTS
press 6
beside x1: type 24
beside n1: type 200
beside x2: type 12
beside n2: type 50
beside p1: place cursor on
press ENTER
put cursor on Calculate
press ENTER
You see
2-propZtest
p(hat)1=.12
p(hat)2=.24
p(hat)=.144
Notice that the p-value found by the table was .0154, and
the p-value found by calculator was 0153210422, but that
difference is because with the table we have to round the
z-value from -2.161688506 to -2.16. That's not as big a
round-off error as in the preceding problem, so it wasn't
as far off.
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c. Home equity loan default rates in two banks: x1 = 36, n1 = 480, x2 = 26, n2 = 520, α = .05, right-tailed test.
1. Set up the hypotheses:
or, equivalently,
2. Determine the critical region for the test statistic Z:
Since , for rejection of , and since
this is a left-tailed test, we require that ,
since 1.64 is the z-value that has the area .05 to the right
of it.
3. Calculate the three "p-hats" same as the other 2 problems:
and
4. Calculate the test-statistic by the same
terrible formula:
This is exactly at the boundary of the rejection-region so
it is really a toss-up as to whether we should reject the
hypothesis that the propportions of loan default-rates are
equal at the two banks. So, this test is a failure. We
would need more data.
The p-value is the probability that the test
statistic Z would be at least as high as 1.64,
if in fact the two proportions were equal:
So if we use a standard normal table to find the
area to the right of z=1.64, we should find it to
be which is what we expect to happen
whenever the test-statistic turns out to be the same
as the boundary of the rejection region, as it did in
his case.
Now if you have a TI-83 or 84 calculator,
we will be much more accurate. Just clear the screen
and do this:
press STAT
use arrow keys to move cursor to hilite TESTS
press 6
beside x1: type 36
beside n1: type 480
beside x2: type 26
beside n2: type 520
beside p1: place cursor on
press ENTER
put cursor on Calculate
press ENTER
You see
2-propZtest
p(hat)1=.075
p(hat)2=.05
p(hat)=.062
Notice that the p-value found with the calculator is
very slightly more than .05, so according to this
more accurate calculation, we should not reject .
However it is still so very close to .05 that more data
should be obtained before a decision is made.
Edwin