Question

A sample of the length in inches for newborns is given below. Assume that lengths are normally distributed. Find the 90% confidence interval of the mean length.

Length
17
15.3
22
21.3
15.9
17.8
18.3
20.1
15.3
22.1

Answer 1

To find the 90% confidence interval of the mean length, we first need to calculate the mean and standard deviation of the sample.

Mean (x̄):
x̄ = (17 + 15.3 + 22 + 21.3 + 15.9 + 17.8 + 18.3 + 20.1 + 15.3 + 22.1) / 10
= 184.1 / 10
= 18.41

Standard Deviation (s):
s = √[((17-18.41)^2 + (15.3-18.41)^2 + (22-18.41)^2 + (21.3-18.41)^2 + (15.9-18.41)^2 + (17.8-18.41)^2 + (18.3-18.41)^2 + (20.1-18.41)^2 + (15.3-18.41)^2 + (22.1-18.41)^2) / (10-1)]
= √[((0.1681 + 8.4969 + 13.6189 + 0.0849 + 5.3769 + 0.3249 + 0.0081 + 0.0576 + 8.4969 + 15.5361) / 9)]
= √[52.1739 / 9]
= √5.7971
≈ 2.408

Margin of Error (E):
E = (z * (s / √n))
= (1.645 * (2.408 / √10))
≈ 1.357

Now, we can calculate the confidence interval by adding and subtracting the margin of error from the mean.
Confidence Interval:
CI = (x̄ - E, x̄ + E)
= (18.41 - 1.357, 18.41 + 1.357)
= (17.053, 19.767)

Therefore, the 90% confidence interval of the mean length is (17.053, 19.767) inches.