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 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 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,

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

Thus, the estimate of the standard error of$\hat{p}$ 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 $z_{\frac{\alpha }{2}}$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).