The data manager for a state political party gathered data to determine how many citizens would support the party's 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?

The approximate 90% confidence interval is;

70,244 to 70,732

Step-by-step explanation:

Here, we want to calculate the approximate 90% C.I for the situation

Mathematically;

CI = mean ± (z * SD)/√n

From the question;

mean = 70,438

SD = 645.3

n = 30

we can get z from the CI table

90% CI is same as; 1.645 z-score

So , substituting these values;

CI = 70,438 ± ( 1.645 * 645.3)/√30

CI = 70,438 ± 194

So the CI = 70,438 -194 to 70,438 + 194

= 70,244 to 70,732

because they could drop even lower..

if a = 1,   and b = 2, then the answer is 47.99999999. if you round that to the nearest whole number, the answer is 48.