Question

select the correct answer.
the data manager for a state political party gathered data to determine how many citizens would support the partys senate candidate. he polled citizens in 30 similarly sized senate districts and found a sample mean of 70,438 citizens. statewide data shows that the population standard deviation is 645.3.
what is the approximate 90% confidence interval for this situation?

a. 70,134 and 70,742
b. 70,207 and 70,669
c. 70,403 and 70,473
d. 70,244 and 70,632

Explanation:

Step1: Identify the formula for confidence - interval

For a large - sample (n ≥ 30) confidence interval when the population standard deviation $\sigma$ is known, the formula is $\bar{x}\pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $z_{\alpha/2}$ is the z - score, $\sigma$ is the population standard deviation, and n is the sample size.

Step2: Determine the values of parameters

We are given that $\bar{x}=70438$, $\sigma = 645.3$, and $n = 30$. For a 90% confidence interval, the significance level $\alpha=1 - 0.90 = 0.10$, and $\alpha/2=0.05$. The $z$ - score $z_{\alpha/2}=z_{0.05}\approx1.645$.

Step3: Calculate the margin of error

The margin of error $E = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$. Substitute the values: $\frac{\sigma}{\sqrt{n}}=\frac{645.3}{\sqrt{30}}\approx\frac{645.3}{5.477}\approx117.82$. Then $E = 1.645\times117.82\approx193.81$.

Step4: Calculate the confidence - interval

The lower limit is $\bar{x}-E=70438 - 193.81\approx70244$. The upper limit is $\bar{x}+E=70438+193.81\approx70632$.

Answer:

D. 70,244 and 70,632