The following table presents the numbers of customers who - after 2 weeks of use - were satisfied or dissatisfied with their newly-purchased computers. $$\begin{array} { c c c c c } \hline & \text { Brand A } & \text { Brand B } & \text { Brand C } & \text { Brand D } \ \hline \text { Satisfied } & 38 & 26 & 21 & 44 \ \text { Dissatisfied } & 3 & 9 & 13 & 9 \ \hline \end{array}$$ Can you conclude that the satisfaction rate is related to the brand of computer? Use the $\alpha = 0.05$ level of Significance. A) Do not reject H₀. There is insufficient evidence to conclude that the satisfaction rate differs among the computer brands. B) Reject H₀ It appears that the satisfaction rates differ among the computer brands.

Question

Quizplus · Accepted Answer

To determine if the satisfaction rate is related to the brand of computer, we can conduct a chi-square test of independence. The null hypothesis is that there is no association between satisfaction rate and brand of computer, while the alternative hypothesis is that there is an association.

The expected frequencies for each cell under the null hypothesis can be calculated using the formula:

Expected frequency = (row total x column total) / grand total

Using this formula, we find the following expected frequencies:

$$\begin{array} { c c c c c c c } 
\hline & \text { Brand A } & \text { Brand B } & \text { Brand C } & \text { Brand D } & \text { Row Total } \
\hline \text { Satisfied } & 33.3 & 24.4 & 18.4 & 43.9 & 120 \
\text { Dissatisfied } & 7.7 & 5.6 & 4.2 & 10.1 & 28 \
\hline
& 41 & 31 & 22 & 54 & 148 \
\hline
\end{array}$$

To calculate the chi-square test statistic, we use the formula:

$\chi^2 = \sum_{i=1}^{r}\sum_{j=1}^{c}\frac{(O_{ij}-E_{ij})^2}{E_{ij}}$

where r is the number of rows, c is the number of columns, Oij is the observed frequency in cell (i,j), and Eij is the expected frequency in cell (i,j).

Plugging in the observed and expected frequencies, we get:

$\chi^2 = \frac{(38-33.3)^2}{33.3} + \frac{(26-24.4)^2}{24.4} + \frac{(21-18.4)^2}{18.4} + \frac{(44-43.9)^2}{43.9} + \frac{(3-7.7)^2}{7.7} + \frac{(9-5.6)^2}{5.6} + \frac{(13-4.2)^2}{4.2} + \frac{(9-10.1)^2}{10.1} = 13.7$

The degrees of freedom for this test is (r-1)(c-1) = (2-1)(4-1) = 3.

Using a chi-square distribution table (or calculator) with 3 degrees of freedom and an alpha level of 0.05, the critical value is 7.815. Since our calculated chi-square value (13.7) is larger than the critical value, we reject the null hypothesis.

Therefore, we can conclude that there is sufficient evidence to suggest that the satisfaction rate differs among the computer brands.

The Following Table Presents the Numbers of Customers Who -