point estimate A sample of 81 observations is taken from a normal population with a standard deviation of 5. The sample mean is 40. Determine the 95% confidence interval for the population mean

The 95 percent Confidence Interval is for the population is (38.911 , 41.089)

Step-by-step explanation:

To solve the above question, we would be making use of the confidence interval formula:

Confidence Interval = Mean ± z score × σ/√n

In the above question,

Mean = 40

σ = Standard deviation = 5

n = number of samples = 81

Confidence Interval = 95%

The z score for a 95% confidence interval = 1.96

Therefore, the confidence interval =

= 40 ± 1.96 (5/√81)

= 40 ± 1.96(5/9)

= 40 ± 1.0888888889

Confidence Interval

a)40 + 1.0888888889

= 41.0888888889

Approximately = 41.089

b ) 40 - 1.0888888889

= 38.911111111

Approximately = 38.911

Therefore, the 95 percent Confidence Interval is for the population is (38.911 , 41.089)

step-by-step explanation:

i need to answer questions so i dont know but i think its the first one because it goes up but a consent rate

1: -2 3/4, -2 1/2, -14, 0, 3, 8 1/3

2: -4, -3, -2 1/2, -2 1/4, 3 1/4,

step-by-step explanation: