Question: Furniture brand familiarity. A brand name that consumers recognize is a highly valued commodity in any industry. To assess brand familiarity in the furniture industry, NPD (a market research firm) surveyed 1,333 women who head U.S. households that have incomes of $25,000 or more. The sample was drawn from a database of 25,000 households that match the criteria listed above. Of the 10 furniture brands evaluated, La-Z-Boy was the most recognized brand; 70.8% of the respondents indicated they were “very familiar” with La-Z-Boy.

a. Describe the population being investigated by NPD.

b. In constructing a confidence interval to estimate the proportion of households that are very familiar with the La-Z-Boy brand, is it necessary to use the finite population correction factor? Explain.

c. What estimate of the standard error of$\hat{p}$ should be used in constructing the confidence interval of part b?

d. Construct a 90% confidence interval for the true proportion and interpret it in the context of the problem.

To assess brand familiarity in the furniture industry, surveyed 1,333 women with incomes of $25,000 or more.

The sample was drawn from a database of 25,000 households. It is also given that 70.8% of the respondents indicated they were “very familiar” with La-Z-Boy.

In the study, the population of interest is all women who head U.S. households that have incomes of $25,000 or more in the database.

Yes, it is necessary to use the finite population correction factor.

If $\frac{n}{N}>0.05$, then use the finite population correction factor.

The value of$\frac{n}{N}$ is

$$\begin{gathered} \frac{n}{N}=\frac{1333}{25000} \\ =0.053 \end{gathered}$$

Here, the value of $\frac{n}{N}$is greater than 0.05. Hence, the finite population correction factor is necessary to use in the study

The standard error is calculated using the following formula,

$$\begin{gathered} \hat{\sigma }=\sqrt{\frac{0.708\left(1-0.708\right)}{1333}\left(\frac{25000-1333}{25000}\right)} \\ =\sqrt{\frac{0.2067}{1333}\left(\frac{23667}{25000}\right)} \\ =0.012 \end{gathered}$$

Thus, the estimate of the standard error to vector p is 0.012.

For the 90% confidence level, the level of significance is 0.90.

For,

$$\begin{gathered} \left(1-\alpha \right)=0.90 \\ \alpha =0.1 \\ \frac{\alpha }{2}=0.05 \end{gathered}$$

From the standard normal table, the value of is given as

$$\begin{gathered} z_{\frac{\alpha }{2}}=z_{0.05} \\ =1.645 \end{gathered}$$

The confidence interval is,

$$\begin{gathered} \hat{p}\pm z_{\frac{\alpha }{2}}{\sigma }_{\hat{p}}=0.708\pm 1.645\left(0.012\right) \\ =0.708\pm 0.0197 \\ =\left(0.6883,0.7277\right) \end{gathered}$$

Thus, the 90% confidence interval for the proportion of households that are very familiar with the La-Z-brand is (0.6883, 0.7277).