Suppose heights of dogs, in inches, in a city are normally distributed and have a known population standard deviation of 7 inches and an unknown population mean. A random sample of 15 dogs is taken and gives a sample mean of 34 inches. Find the confidence interval for the population mean with a 99% confidence level.

Step-by-step explanation:

We would use the t- distribution.

From the information given,

Mean, μ = 34 inches

Standard deviation, σ = 7 inches

number of sample, n = 15

Degree of freedom, (df) = 15 - 1 = 14

Alpha level,α = (1 - confidence level)/2

α = (1 - 0.99)/2 = 0.005

We will look at the t distribution table for values corresponding to (df) = 14 and α = 0.005

The corresponding z score is 2.977

We will apply the formula

Confidence interval

= mean ± z ×standard deviation/√n

It becomes

34 ± 2.977 × 7/√15

= 34 ± 2.977 × 1.81

= 34 ± 5.39

The lower end of the confidence interval is 34 - 5.39 =28.61

The upper end of the confidence interval is 34 + 5.39 =39.39

the one on the bottom left

step-by-step explanation:

all of the other ones have a specific pattern