Question

In a sample of 57 temperature readings taken from the freezer of a restaurant, the mean is 29.6 degrees and the population standard deviation is 2.7 degrees. What would be the 80% confidence interval for the temperatures in the freezer?

Answer 1

(29.14 ; 30.06)

Explanation:

Given that'

Sample size (n) = 57

Mean (m) = 29.6

Population standard deviation (σ) = 2.7

Confidence interval = 80%

= (1 - 0.8) / 2 = 0.1

Mean ± z * σ/√n

Using the Z probability calculator : Z0. 1 = 1.28

Hence,

29.6 ± 1.28 * (2.7 / √57)

29.6 - (1.28 * 0.3576237) ; 29.6 + (1.28 * 0.3576237)

29.142241664 ; 30.057758336

(29.14 ; 30.06)