A one-way ANOVA test or analysis of variance

is used to compare three or more population means and variances.

We draw simple random samples from each population and use

the data to test the null hypothesis that the population means are all equal.

The manual method for calculating is too complicated to show here,

so we use statistical tools such as Microsoft Excel or Minitab to assist us.

The assumptions for a one-way ANOVA test is,

one-way means that there is one factor

that describes the cause of the variation in the data.

ANOVA consists of two parts,

the variation between treatment means and the variation within the treatments.

The populations must be normally distributed.

The samples must be independent from each other.

Each population must have the same variance.

The procedure using Excel is as follows: set up your hypothesis of

H-nought equals Mu sub 1 equals Mu sub 2 equals Mu sub 3 so forth to Mu sub k,

and the alternative is that all are not equal.

This will always be a right-tailed test.

Number two, pick a confidence level or Alpha.

Number three, enter the data into Excel in each column of a blank spreadsheet.

Number four, select data analysis from the data tab or install using Excel Add-Ins.

Number five, select ANOVA: Single Factor from the Data Analysis dialog box.

Number six, set up the ANOVA dialog box with the data range,

the Alpha level, et cetera.

Number seven, interpret the results.

Let's take an example.

We have a casting process that can be run at three different temperatures,

200 degrees, 220 degrees,

or 240 degrees Fahrenheit.

We ask, does the temperature significantly affect the moisture content?

Let's use Alpha equal 0.05.

We enter the data into the table.

This data is the moisture values or

the dependent y-values for the oven temperatures or the independent x-values.

Then we choose "Anova: Single Factor" from the Data Analysis box.

In the Anova: Single Factor dialog box,

we enter ranges, include the check mark, and enter Okay.

The resultant values include the variances within

the sample groups and between the sample groups.

The p-value of 0.004813 is then compared to the Alpha level of 0.05.

Since the p-value is less than or equal to Alpha or 0.004813 is less than 0.05,

we reject H_0 and conclude that at the Alpha equal 0.05 confidence level,

the data indicate that temperature does not have an impact on the moisture content.

The Chi-Square statistic is used in instances where comparing

the population variance becomes more important than comparing the population means.

The two most popular cases to apply Chi-Square is, number one,

comparing variances when the variance of the population is known and, two,

comparing observed and expected frequencies of

test outcomes when there was no defined population variance.

The hypothesis tests for the Chi-Square is as follows: H-nought is that Sigma

squared of x is equal to Sigma

squared of naught and its alternative that they are not equal,

or Sigmas square sub x is less than or equal to

Sigma square naught and

its alternative that Sigma squared x is greater than Sigma squared naught,

or that Sigma squared x is greater than or equal to Sigma squared

naught and the alternative is that Sigma squared x is less than Sigma squared naught.

These hypothesis test compare a population variance,

x sub squared x with a fixed value Sigma sub squared naught.

The test statistic is given by: Chi-Square equal n minus one times x

squared over Sigma squared x and its degree of freedom is n minus one,

where the number of samples is n and the sample variance is s squared.

If the problem calls for a population variance to equal

the fixed value or that Sigma squared sub x equals Sigma squared naught,

and its alternative that they don't equal,

then the two-tailed test is being requested.

Likewise, if the problem calls for anything less than or greater than the fixed value,

then we use the other two choices.

One is the right-tailed test and one is a left-tailed test. Let's take an example.

An R&D department of a steel plant has tried to

develop a new steel alloy with less tensile variability.

This department claims that the new material will show

a 4-Sigma tensile variation less than or equal to 60 psi,

95 percent of the time.

An eight sample test yielded a standard deviation of 8 psi.

Can a reduction in tensile strength variation be

validated with the 95 percent confidence?

The best range of variation given in the problem is 60.

At the 4-Sigma range,

this is 60 divided by 4 equals 15 for one Sigma.

So, Sigma square sub naught or 15 squared is our fixed value to compare our sample to.

So, the null hypothesis will test the claim as follows:

H sub naught is Sigma square sub 1 is greater than or equal to 15 squared,

and its alternative is that Sigma squared sub 1 is less than 15 squared.

From the Chi-Square table,

with degrees of freedom of seven and Chi-Square of 0.95,

the left-tailed test, we find that Chi-Square sub c is 2.17.

This is the value corresponding to

the five percent of the variation that will not be met.

Now let's compare that with the test statistic.

We plug in the numbers,

and we get a Chi-Square equal 1.99 with a degree of freedom of seven.

For our hypothesis, we're testing that if

Sigma square sub 1 is greater than 15 squared or its alternative,

Sigma square sub 1 is less than 15 squared.

Since Chi-Square is equal to 1.99 is less than Chi-Square sub c of 2.17,

we must reject the null hypothesis.

Our conclusion is that there is significant evidence to indicate a reduced variation.

Thus, an improvement to the manufacturing process is statistically valid.