Welcome to a lesson
on a test of a single variance,
using the Chi-Square distribution.
A test of a single variance assumes
that the underlying distribution is normal.
The null and alternative hypotheses are stated
in terms of a population variance,
or the population standard deviation.
The test statistic is chi-square,
which is equal to the product of n-1,
and the sample variance, divided by the population variance.
And you may think of s, the sample standard deviation,
as the random variable in this test.
The number of degrees of freedom is equal to n-1,
where n is the sample size, and a test of a single variance
may be right-tailed, left-tailed, or two-tailed.
Let's take a look at an example.
With individual lines at its various windows,
a post office finds that the standard deviation
for normally distributed waiting times for customers
on Friday afternoon is 7.2 minutes.
The post office experiments with a single main waiting line
and finds that for a random sample of 25 customers,
the waiting times for customers have a standard deviation
of 3.5 minutes.
With a significance level of five percent,
test the claim that a single line causes lower variation
among waiting times, or shorter waiting times for customers.
Let's first state the given information.
The population standard deviation is 7.2 minutes.
The sample size n is equal to 25.
The sample standard deviation is 3.5 minutes.
And alpha is equal to 0.05.
And now let's work on stating the null
and alternative hypotheses.
Since the claim is that a single line causes less variation,
this is a test of a single variance.
And we will use the population variance
for the null and alternative hypothesis.
Again, because the claim is a single line
causes less variation,
the alternative hypothesis is
the population variance is less than the square of 7.2.
Notice how we are squaring 7.2 here,
because we are given sigma equals 7.2.
So if we square sigma, we must also square 7.2.
And then it follows a null hypothesis,
is the population variance equals the square of 7.2.
The next step is to work up,
the distribution for this test is a chi-square distribution
with 24 degrees of freedom.
Because the degrees of freedom are equal to n-1,
which in this case is 25-1, which is equal to 24.
Also notice how because the alternative hypothesis
is the population variance is less than
the square root of 7.2, this is a left-tailed test.
Next we need to calculate chi-square, and then the p value.
Using the formula for chi-square,
we substitute 25 for n,
the square root of 3.5 for the sample variance,
and the square of 7.2 for the population variance.
Let's go ahead and verify this quotient on the calculator.
Notice how we do get approximately 5.67 for chi-square.
Also notice how I did put parentheses around the numerator.
And since chi-square is approximately 5.67,
the p value is equal to the probability
that chi-square is less than or equal to 5.67.
And now we will determine the p value
using the TI-84, and the chi-square CDF function.
To do this, we press second vars
for the distribution menu, option eight.
Now we enter the lower bound comma, the upper bound comma,
the degrees of freedom.
Again, because we have a left-tailed test,
the lower bound is zero comma.
The upper bound, 5.67 comma.
The degrees of freedom are 24 comma 24.
Close parentheses and enter.
Notice how we have e-5 here on the right,
which indicates scientific notation.
The p value's approximately 4.2 times 10 to the power of -5.
Let's go ahead and write that down.
If we want to convert this to decimal notation,
we need to move the decimal five places to the left,
which would give us 0.000042,
which we can see matches the p value shown below.
The next step is to compare alpha and the p value
to determine whether we reject the null hypothesis
or do not reject the null hypothesis.
Notice how the p value is low, compared to alpha.
When the p value is low, the null must go.
Or more formally, because the p value is less than
or equal to alpha, or because alpha is greater than
or equal to the p value, we reject the null hypothesis.
So if we reject the null hypothesis,
we do not think that the variation
in waiting times is 7.2 minutes.
We think the variation in waiting time is less.
So our final conclusion, because alpha is equal to 0.05,
at a five percent level of significance from the data,
there is sufficient evidence to conclude
that a single line causes a lower variation
among the waiting times.
Or, with a single line, the customer waiting times
vary less than 7.2 minutes.
I hope you found this helpful.