) Ten randomly selected people were asked how long thy sept at night. The mean time was 7.1 hours, and the standard deviation was 0.78 hour. Find the 95% confidence interval of the mean time. Assume the variable is normally distributed

7.1+/-0.48

= (6.62, 7.58) hours

Therefore, the 95% confidence interval (a,b)= (6.62, 7.58) hours

Step-by-step explanation:

Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.

The confidence interval of a statistical data can be written as.

x+/-zr/√n

Given that;

Mean x = 7.1 hours

Standard deviation r = 0.78 hour

Number of samples n = 10

Confidence interval = 95%

z value(at 95% confidence) = 1.96

Substituting the values we have;

7.1+/-1.96(0.78/√10)

7.1+/-1.96(0.246657657493)

7.1+/-0.483449008686

7.1+/-0.48

= (6.62, 7.58) hours

Therefore, the 95% confidence interval (a,b)= (6.62, 7.58) hours

step-by-step explanation:

step-by-step explanation:

if x = 2 you have to find x to finish the problem