Question

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.

Short answer

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

b.The finite population correction factor is necessary to use in the study.

c.The estimate of the standard error of is 0.012.

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

Step-by-step solution

Given information

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.

a) Describing the population

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.

(b) Calculating finite population correction factor

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

(c) Calculating an estimate of the standard error of    

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.

(d) Calculating a confidence interval

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).