Question

simple random sample of size n= 40 is drawn from a population. The sample mean is found population mean. be x 120.6 and the sample standard deviation is found to be s 12.6. Construct a 99% confidence interval for the The lower bound is (Round to two decimal places as needed.)

Answer 1

Solutions:-

Since n>30 we use z instead of t but some professor use t whenever population standard deviation is unknown so Im solving for both cases, I will go with case 1 by the way

sample size,n = 40

sample mean,x = 120.6

standard deviation,s = 12.6

alpha,a = 1-0.99 = 0.01

Za/2 = Z0.005 = 2.575

confidence interval:

x +/- [Za/2 * (s/sqrt(n))]

120.6+/- [2.575 * (12.6/sqrt(40))]

120.6+/- 5.1300

= ( 115.47 , 125.73)

The lower bound is 115.47

The upper bound is 125.73

please like