91影视

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 鈥渧ery familiar鈥� 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

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

From the standard normal table, the value of $z_{\frac{伪}{2}}$is given as,

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

The confidence interval is,

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

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