Asimple random sample of 60 items resulted in a sample mean of 80. the population standard deviation is ? = 15.
a. compute the 95% confidence interval for the population mean (to 1 decimal.

Here are the solutions ...

(a)

Standard error of the mean is SE = σ/√n = 15/√60

For 95% confidence, the critical Z- score is 1.95996398454005

Width of the confidence interval is E = z * SE = 1.95996398454005 * 1.93649167310371 = 3.79545393564498

The confidence interval is [x-bar - E, x-bar + E]

= [80 - 3.79545393564498, 80 + 3.79545393564498]

= [76.20, 83.80]

(b)

Standard error of the mean is SE = σ/√n = 15/√120

For 95% confidence, the critical Z- score is 1.95996398454005

Width of the confidence interval is E = z * SE = 1.95996398454005 * 1.36930639376292 = 2.68379121557574

The confidence interval is [x-bar - E, x-bar + E]

= [80 - 2.68379121557574, 80 + 2.68379121557574]

= [77.32, 82.68]

(c)

As the sample size increases, the confidence interval narrows down.

a is 8.b is 10.c is 10.5. d is -.5

step-by-step explanation:

just minus the smallest number from the biggest.

the first option i believe

step-by-step explanation: