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

step-by-step explanation:

you divide the denominator by the numerator on both sides and if the quotients are equal then the fractions are equivalent

y-y1=4(x-x1)

step-by-step explanation:

this is point-slope form since you only have 4 as the slope it goes in place of m

hope this : )