Question

Suppose that a customer service center claims that the telephone calls it receives last 114 seconds on average, with a standard deviation of 9 seconds. If you took a sample of 324 telephone calls, which of the following mean times would be within the 95% confidence interval?

A. 114.5 seconds
B. 115.5 seconds
C. 117.5 seconds
D. 116.5 seconds

So I set up a 95% confidence interval and got 114-.98= 113.02 and .98+114= 114.98 but I don't really know where to go from there, can someone help me please?

Answer 1

I used the formula (114- 1.96*9/sqrt324), (114+1.96*9/sqrt324)

Answer 2

\[Z = (x' - \mu)/ (\sigma / \sqrt{n})\]

Answer 3

Z bar value for confidence level is of 95% is 1.96...( Google it :p)

Answer 4

i did use 1.96 though

Answer 5

and,
x' (suppose to x bar) = population mean = 114
u (MU) = sample mean = let it be x
Sigma = standard deviation = 9
n = sample size = 324

1.96 = (114-u) / (9/sq rt 324)
1.96 = (114 - u) / (9/18)

1.96 * 2 = 114 - u

u = 110.08.... which is not an option !! :D

Answer 6

So definitely the approach is wrong.. or something is missing.. lets ask experts to help !
Guys if you can please help... @ganeshie8 @mathslover

Answer 7

\[\large \left( 114- 1.96\times \frac{9}{\sqrt{324}},~~~ 114+1.96\times \frac{9}{\sqrt{324}}\right)\]

I think we need to tick the option that falls with in this interval