Q.50 - The Practice of Statistics for AP

Question

Market research Before bringing a new product to market, firms carry out extensive studies to learn how consumers react to the product and how best to advertise its advantages. Here are data from a study of a new laundry detergent. The participants are a random sample of people who don’t currently use the established brand that the new product will compete with. Give subjects free samples of both detergents. After they have tried both for a while, ask which they prefer. The answers may depend on other facts about how people do laundry.

(a) How are laundry practices (water hardness and wash temperature) related to the choice of detergent? Make an appropriate graph to display this relationship. Describe what you see. 

(b) Determine whether or not the sample provides convincing evidence that laundry practices and product preference are independent in the population of interest 

Step-by-Step Solution

Answer

a). The people in each category are preferring the new product and it is the highest in the "Hard-Warm" group and the lowest in the "Soft-Hot: group. 

b). There is insufficient evidence to conclude that there is an association between two variables. 

Part (a) Step 1: Given Information

Part (a) Step 2: Explanation

The figure could be constructed:

Interpretation:

From the above figure, it is clear that most of the people in each category are preferring the new product and it is the highest in the "Hard-Warm" group and the lowest in the "Soft-Hot: group.

Part (b) Step 3: Given Information

Part (b) Step 4: Explanation

The appropriate null and alternative hypotheses for the provided test could be written as:

$H_0$ : Two variables are independent

$H_a$ : Two variables are dependent

The expected count is computed as:

Part (b) Step 5: Explanation

The test statistic is:

${\chi }^2=\sum \frac{(O-E{)}^2}{E}$

$=\frac{(53-49.8079{)}^2}{49.8079}+\frac{(27-24.0452{)}^2}{24.0452}+\dots \dots +\frac{(42-41.0847{)}^2}{41.0847}$

$=2.058$

The degree of freedom is calculated as:

Degree of freedom

$=(Number of rows-1)(Number of column-1)$

$=(4-1)(2-1)$

$=3$

The p-value using chi-square table at $3$ degree of freedom is $0.560$.

The p -value is above significance level. The null hypothesis does not get rejected. Thus, there is insufficient evidence to conclude that there is an association between two variables.