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This is a presentation on computing another Z-test. We will present a second problem for
which a Z-test is an appropriate test of a hypothesis, practicing the formulation of
the hypothesis and the calculation of the test statistic, and interpreting the results.
Here is the problem:
Information collected from the population of students at a local university indicates
that students are working, on average, 15.23 hours per week while in school with a standard
deviation of 2.50 hours. Information collected from a random sample of 20 fraternity members,
however, reveals a mean of 17.34 hours per week. What is the null hypothesis? What is
the alternative hypothesis? What is the result of the z-test? Do you accept or reject the
null hypothesis? Do you accept or reject the alternative hypothesis?
We can use a Z-test since the population mean and standard deviation are defined for us.
If these population numbers were not defined for us, then we could not use this test. (Recall
that the population numbers are given to us by an omniscient personality.)
Let's further recall the Z-test formula: Z-test = x-bar minus "mew" divided by "theta"
Where x-bar = mean value of sample variable X
"mew" = population mean value from which sample variable X was drawn
"theta" = population standard deviation value from which sample variable X was drawn
We simply need to write both the null hypothesis and alternative hypothesis, calculate the
Z-test, and correctly interpret the result. What is the null hypothesis? It is: There
is no statistically significant difference between the mean number of hours worked by
the sample of fraternity members per week while in school and the mean number of hours
worked by the population of students at the local university.
In contrast, the alternative hypothesis is: There is a statistically significant difference
between the mean number of hours worked by the sample of fraternity members per week
while in school and the mean number of hours worked by the population of students at the
local university.
Now, for the test calculation. We insert the numbers into the formula in the following
way:
Z-test = 17.34 – 15.23 divided by 2.50
Then, we calculate the result of the Z-test:
Z-test = 17.34 – 15.23 divided by 2.50 = 2.11 divided by 2.50 = .844
We then consult Table 6-3, "Proportions of the Normal Curve above the Absolute Value
of Z", which we find on page 143 of the *** I book. Find the correct proportion of the
normal curve above the absolute value of .844. (Since our calculated Z-test value is positive,
the absolute value is automatically the positive result.) On the far left side of the page,
we find the first digit and first decimal for .844, which is 0.8. We then proceed across
that row to the column headed with our second decimal of z, which is 4, under the .04 column.
The proportion of the normal curve above .844 is .2005, or in percentage terms, 20.05%.
We can now apply our decision rule and correctly accept or reject the null hypothesis. We recall
that any proportion of the normal curve that is either .0250 or below .0250 means that
we have an unusually large difference in mean values between the sample and the population,
and that we consequently reject the null hypothesis. Since .2005 is greater than .0250, we accept
the null hypothesis, that there is no statistically significant difference between the sample
mean value of hours worked by the fraternity brothers and the population mean value of
hours worked by all of the university students.
Furthermore, if we accept the null hypothesis, then we reject the alternative hypothesis
that there is a statistically significant difference between the sample mean value and
the population mean value.