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This presentation is the third of three presentations on calculating and interpreting a z-test.
In this presentation, we will learn how to interpret the results of the z-test. Let's
look over the results of our z-test calculation from our previous presentation.
_ Z-test = X - µ / ∂ = 6.79 -- 8.45 / 2.56
= -1.66 / 2.56 = -.65, | -.65 | = .65
What does this calculation mean? To interpret the result, and thus come to a conclusion
about the null hypothesis, we need to first turn to page 143 in the *** 1 textbook.
This page is headed by "Table 6-3: Proportions of the Normal Curve above the Absolute Value
of Z". This table tells us what percentage of observations lie above the absolute value
of our z-test calculation. To create an absolute value of any number, simply make the number
be a positive number. So, our z-test result of -.65 in absolute terms is .65.
What percentage of observations lie above .65? To use Table 6-3 correctly, look at the
column on the far left of the table, headed by "First digit and first decimal of Z". The
first digit of our z-test result is 6, so proceed down to "0.6". We then need to consider
which column is correct to go to under the heading "Second decimal of Z". For our z-test
result, the correct column is ".05", since our z-test calculation in absolute terms is
.65.
Thus, the percentage of observations that lie above the calculated absolute z-test value
is 25.78%, which is the result when we make a percentage out of a decimal, moving the
decimal point two places to the right.
Let's re-examine the normal curve on page 139 of the *** 1 textbook. You can see
that 25.78% of observations would put the z-test result, and thus the difference between
the sample mean and the population mean (standardized by the population standard deviation) pretty
close to zero. In other words, that is a very common difference, and is very close to zero,
all things considered. We're not seeing anything very special in the difference in mean absentee
rate between the vocational training students and the high school population.
But we need a more specific judge of "not seeing anything very special". We need a standard
for "special". That standard is found on page 139, where we see the arrow pointing to Z
= 2, with the arrow being labeled 2.5%. By agreement amongst social scientists, this
is the point at which we make our decision to accept or reject the null hypothesis. In
other words, any percentage of observations that lie above a z-test result of 2, in absolute
terms, is a rare enough percentage to be called "very special".
To return to the table on page 143, notice that there is one proportion on the table
that is in bold type. It is .0250. In percentage terms, it is 2.5%. Any percentage that is
2.5% or lower indicates a "very special" z-test. We will accept the null hypothesis (notated
as Ho) that there is no statistically significant difference between the mean absentee rate
for the vocational training students (the sample mean) and the mean absentee rate for
all of the students in the local school district (the population mean), and reject the alternative
hypothesis (notated as HA) that there is a statistically significant difference between
the mean absentee rate for the vocational training students (the sample mean) and the
mean absentee rate for all of the students in the local school district (the population
mean).
In mathematical terms, the percent of normal curve above the | Z-test | = .2578 = 25.78%,
and we accept Ho because .2578 > .0250, and reject HA.
In practical terms, the students in the vocational training class are absent less frequently
than are students at the high school on the whole, but they are not statistically significantly
less absent. There's a difference, but it's not large enough for it to be considered statistically
meaningful.
This concludes our three-part presentation of the z-test.