Question

We are interested in the estimate of the average temperature, measured in degrees Celsius, in Bermuda during a day in July. In order to achieve this, we decide to build a 99% confidence interval. Provide the lower and upper bounds of this interval if the temperature in this country was noted over 22 days and a variance of 25 ° C2 and a sum of the observations of 704 were obtained. Suppose the temperature in Bermuda for a day of July follows normal law.

Answer 1

Confidence interval for Population mean is given as below:

Confidence interval = Xbar ± t*S/sqrt(n)

From given data, we have

∑X = 704

n = 22

Xbar = ∑X/n = 704/22 = 32

S^2 = 25

S = sqrt(25) = 5

df = n – 1 = 21

Confidence level = 99%

Critical t value = 2.8314

(by using t-table)

Confidence interval = Xbar ± t*S/sqrt(n)

Confidence interval = 32 ± 2.8314*5/sqrt(22)

Confidence interval = 32 ± 3.0182

Lower limit = 32 - 3.0182 = 28.98

Upper limit = 32 + 3.0182 = 35.02

Confidence interval = (28.98, 35.02)