Question

Question 4 1 pts A recent study of 25 students showed that they spent an average of $18.56 on gasoline with a standard deviation of $2.80. Find a 99% confidence interval for the population mean using interval notation.

18.56 ± 0.958

(16.994, 20.126)

18.56 ± 1.156

(17.602, 19.518)

18.56 ± 1.024

Answer 1

Solution:

We have to find 99% confidence interval for the population mean.

We are given

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

We are given

Xbar = 18.56,

S = 2.80,

n = 25,

df = n – 1 = 25 – 1 = 24

c = 99% = 0.99

t = 2.7969 (by using t-table or excel)

Confidence interval = 18.56 -/+ 2.7969*2.80/sqrt(25)

Confidence interval = 18.56 -/+ 2.7969*2.80/5

Confidence interval = 18.56 -/+ 2.7969* 0.56

Confidence interval = 18.56 -/+ 1.5663

Confidence interval = 18.56 -/+ 1.566

Lower limit = 18.56 - 1.566 = 16.994

Upper limit = 18.56 + 1.566 = 20.126

Confidence interval is given as below:

(16.994, 20.126)