We

are presents with the following information in which we are presumed to answer.

We should use chi square test program to come up with numerous hypothetical

conclusions.

The

Northeast region is operating the software sold 165 and the group with no

software sold 100, The Southeast region with software sold 200 and the group

with no software sold 125. The Central region

with software sold 175 and the group with no software sold 125. The West group

with software sold 180 and the group with no software sold 130.

Software

program is categorical variable which can have 2 possibilities either Yes, and

or No. this consider the cases whether the software is assigned or not

assigned. Performance is again a categorical variable which can have 2

possibilities either Improved or not improved.

Contingency Table is shown below.

Further calculations

With

Software

Without

Software

Row

Totals

Northeast

Region

165 (159.00)

0.23

100 (106.00)

0.34

265

Southeast Region

200 (195.00)

0.13

125 (130.00)

0.19

325

Central Region

175 (180.00)

0.14

125 (120.00)

0.21

300

West Region

180 (186.00)

0.19

130 (124.00)

0.29

310

Column

Total

720

480

1200

(Grand Total)

In my Chi-Square

Calculation, the compiled the data set in contingency table. Using the

Chi-Square method, the numbers calculate as expected, differences in all the

regions of the sales group sold. The significance level of operation is: 0.05. The

calculations gives the chi-square statistical figure of is 1.7176, the degree

of freedom =3, the probability, p-value is 0.633018. Accordingly, the final development

is not significant at p < 0 .05. This variables are independent. (G F,
2014).
VP
of Sales at WidgeCorp, would like a trial of whether these two variables are
independent or not. A nonparametric test using Chi square test is used to test
the following hypothesis.
Null
Hypothesis Ho: the two variables which are the Software Program and
Performance are independent.
Alternative
Hypothesis Ha: the two variables which are the Software Program and
Performance are not independent.
When
the conclusion is statistically significant (p-value less than 0.05) it is
shown that the alternative hypothesis is; otherwise (p-value greater than
0.05), the null is chosen.
The
relationship is not statistically significant as a result of the p-value,
0.633, is greater than 0.05. Accordingly, it is apparent that it is in favor of
the null hypothesis - the two variables which are Software Program and
Performance are independent.
The VP of Sales at
WidgeCorp, knows quite a bit about statistics, would like to know feasible null
and alternative hypotheses for a nonparametric test on this data using the
chi-square distribution. A nonparametric test is utilized on information that
are qualitative or categorical. This is predominately is utilized when the
information is does not give the right perception when observed at the mean of
such variables.
When looking problem, the
hypothesis shows that the sales estimates are generated from the four regions
are basically random and this is known as the null hypothesis. This also shows
that the null hypothesis is accurate if the observed data sales for every
region did not change from what was originally expected. To know this is only
expected by chance. What usually correlates the null hypothesis is known as the
alternative hypothesis. The null hypothesis is shown as H0 and the alternative
hypothesis as H1. (C. Z. 2013).
The significance level is
usually the acceptable level of type I error, it is usually given as ?. Most of
the time, a significance level of ? = 0 .05 is used. The significance level
form part one is: less than 0.05. The P-value (that is the probability value)
is the statistic value used to test the null hypothesis. If p < ? then we reject the null
hypothesis. Here it is not. 0.633018 is greater than 0.05. Therefore; H0: p =
<0.5, Ha: p <> 0.5

Since P is not less than

?; hence it is reasonable the state to the VP of Sales at WidgeCorp that the

null hypothesis is accurate and should be accepted. And hence state that we

should take the other correlation of the Alternative Hypothesis.