a research group conducted an extensive surve...
a research group conducted an extensive survey of 2,816 wage and salaried workers on issues ranging from relationships with their bosses to household chores. the data were gathered through hour - long telephone interviews with a nationally representative sample. in response to the question, “what does success mean to you?” 1,492 responded, “personal satisfaction from doing a good job.” let (p) be the population proportion of all wage and salaried workers who would respond the same way to the stated question. find a 90% confidence interval for (p). (enter your answer in the form: lower limit, upper limit. include the word “to”. round your answers to three decimal places.)
Answer
# Explanation:
## Step1: Calculate sample proportion $\hat{p}$
The sample size $n = 2816$ and the number of successes $x=1492$. The sample - proportion $\hat{p}=\frac{x}{n}=\frac{1492}{2816}\approx0.530$.
## Step2: Determine the z - value for 90% confidence interval
For a 90% confidence interval, the significance level $\alpha = 1 - 0.90=0.10$, and $\alpha/2=0.05$. The z - value $z_{\alpha/2}=z_{0.05}\approx1.645$.
## Step3: Calculate the margin of error $E$
The formula for the margin of error for a proportion is $E = z_{\alpha/2}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$. Substitute $\hat{p}=0.530$, $n = 2816$, and $z_{\alpha/2}=1.645$ into the formula:
\[
\begin{align*}
E&=1.645\sqrt{\frac{0.530\times(1 - 0.530)}{2816}}\\
&=1.645\sqrt{\frac{0.530\times0.470}{2816}}\\
&=1.645\sqrt{\frac{0.2491}{2816}}\\
&=1.645\sqrt{0.00008846}\\
&=1.645\times0.0094\\
&=0.0154
\end{align*}
\]
## Step4: Calculate the confidence interval
The confidence interval for the population proportion $p$ is given by $\hat{p}-E<p<\hat{p} + E$.
Substitute $\hat{p}=0.530$ and $E = 0.0154$:
$0.530-0.0154 < p<0.530 + 0.0154$, which simplifies to $0.515 < p<0.545$.
# Answer:
$(0.515,0.545)$