Question

You are contacted by the top cookie-selling Girl Scout troop in the city. As part of their sales plan for next year’s cookie selling season, they need to know (with 99% confidence, or α = 0.01) if the types of cookies purchased by their customers depends on where the scouts sell them. Here is their confidential sales data from last year:

Answer 1

The types of cookies purchased by the customers depends on where the scouts sell them.

Explanation:

The data provided is:

Sales Type of Cookie Sold

Method

Thin Mints Tagalongs Do-si-dos Trefoils Samoas

At

parents’ 472 641 350 223 401

workplace

Door

-to- 840 1092 673 511 542

Door

Community 683 742 491 420 374

Events

A Chi-square test for goodness of fit will be used to determine whether the types of cookies purchased by the customers depends on where the scouts sell them.

The hypothesis can be defined as:

H₀: The observed frequencies are same as the expected frequencies.

Hₐ: The observed frequencies are not same as the expected frequencies.

The test statistic is given as follows:

`\chi^(2)=\sum\limits^(n)_(i=1)((O_(i)-E_(i))^(2))/(E_(i))`

The formula to compute the expected frequency is:

`E_(i)=\frac{i^(th)\ \text{Row Total}* i^(th)\ \text{Column Total}}{N}`

Consider the Excel output for the expected frequency values.

The Chi-square test statistic value is:

`\chi^(2)=\sum\limits^(n)_(i=1)((O_(i)-E_(i))^(2))/(E_(i))=56.788`

The degrees of freedom is:

df = (r - 1)(c - 1)

= (3 - 1)(5 - 1)

= 2 × 4

= 8

The p-value is:

`P(\chi^(2)_(8)>56.788)<0.00001`

*Use a Chi-square table.

The p-value of the test is very small.

The null hypothesis will be rejected at 1% level of significance.

Conclusion:

There is enough evidence to support the claim that the types of cookies purchased by the customers depends on where the scouts sell them.