A 95% confidence interval for a population mean is given as (18.985, 21.015). This confidence interval is based on a simple random sample of 36 observations. Calculate the sample mean and standard deviation. Assume that all conditions necessary for inference are satisfied. Use the t-distribution in any calculations.

        Lower (18.985)                    Upper (21.015)
      -------------------------|-------------------------

Solution:

UpperBound of Confidence Interval = UB = 21.015

LowerBound of Confidence Interval = LB = 18.985

Sample Mean:

Sample_Mean <- (UB + LB)/2

UB <- 21.015
LB <- 18.985
Sample_Mean <- (UB + LB)/2
Sample_Mean
## [1] 20

Margin of Error:

(UB - LB)/2

Margin_of_Error <- (UB - LB)/2 
Margin_of_Error
## [1] 1.015

Degree of Freedom and T-score:

Df <- 36 - 1
T35 <- qt(.95,Df)
T35
## [1] 1.689572

Standard Deviation:

Sd <- ((Margin_of_Error/T35)*sqrt(36))
Sd
## [1] 3.604462

Validate Results:

UB_Val <- 20 + 1.689572 * 3.604462 / sqrt(36)
UB_Val
## [1] 21.015
LB_Val <- 20 - 1.689572 * 3.604462 / sqrt(36)
LB_Val
## [1] 18.985

UB = UB_Val

LB = LB_Val