Question

The weights of four randomly and independently selected bags of potatoes labeled 20.0 pounds were found to be 20.6, 21.4​, 20.9, and 21.2 pounds. Assume Normality. Answer parts​ (a) and​ (b) below.

a. Find a 95% confidence interval for the mean weight of all bags of potatoes

(___ , ____)

b. Can you reject the population mean of 20 pounds?

Answer 1

Part a

Confidence interval for Population mean is given as below:

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

From given data, we have

Xbar = 21.025

S = 0.35

n = 4

df = n – 1 = 3

Confidence level = 95%

Critical t value = 3.1824

(by using t-table)

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

Confidence interval = 21.025 ± 3.1824*0.35/sqrt(4)

Confidence interval = 21.025 ± 0.5569

Lower limit = 21.025 - 0.5569 = 20.4681

Upper limit = 21.025 + 0.5569 = 21.5819

Confidence interval = (20.4681, 21.5819)

Part b

Yes, we reject the population mean of 20 pounds because the value 20 is not included in the above confidence interval.