You have an SRS of 23 observations from a large population. The distribution of sample values is roughly symmetric with no outliers. What critical value would you use to obtain a 95% confidence interval for the mean of the population?

Therefore the critical value= 2.073

Step-by-step explanation:

The number of observation = 23.

For the mean of the population the confidence interval  = 95%.

Here mean and stander deviation of the distribution is not given. So we use t distribution.

T distribution is called as student's t distribution.

Sample number =n =23.

Confidence level = c= 95% =0.95

The degree of freedom is sample size decreased by 1

df=n-1 = 23-1 =22.

The critical value can be found in the row df = 22 and column with 0.95 of the T distribution table .

= 2.073

Therefore the critical value= 2.073

area = length x width

length = 2 x width

32 = 2 x width x width

32 = 2 x width^2

36 = width ^2  

width = 6

length = 12

area is 32

perimeter = 2 x 12 + 2 x 6 = 24 + 12 = 36

step-by-step explanation: