Question

A radio talk show invites listeners to enter a dispute about a proposed pay increase for city council members. "What yearly pay do you think council members should get? Call us with your number." In all, 958 people call. The mean pay they suggest is x = $9740 per year, and the standard deviation of the responses is s = $1125. For a large sample such as this, s is very close to the unknown population. The station calculates the 95% confidence interval for the mean pay that all citizens would propose for council members to be $9669 to $9811. Is this result trustworthy? Explain your answer.

Answer 1

@jim_thompson5910

Answer 2

I found the confidence interval to be (9668.76, 9811.24)

Answer 3

one sec

Answer 4

alright

Answer 5

I'm getting

9740-1.96*1125/sqrt(958) = 9,668.75968365992


9740+1.96*1125/sqrt(958) = 9,811.24031634008

so that's the same thing you got when you round

Answer 6

Yeah, but I don't know if the result is trustworthy or not. I know their interval falls into the confidence interval but does that make it trustworthy?

Answer 7

if you round even further, you get (9669, 9811)

Answer 8

so it looks like the confidence interval we got matches up with the confidence interval they got

Answer 9

does that make it trustworthy?

Answer 10

so I'd say it's trustworthy

why? because the math confirms the answer (and it's not just one of those answers where it's a "just trust me on it" answer)

Answer 11

so it does not matter that it was not a random sample?

Answer 12

it does matter and the fact that it's random further confirms the trustworthyness

if it wasn't random, then the bias would skew the results in one way, making the trustworthyness decrease

Answer 13

oh the 958 people that called in are a random sample?

Answer 14

correct

Answer 15

okay thank you!

Answer 16

np