Question

A sample of the length in inches for newborns is given below. Assume that lengths are normally distributed. Find the 95% confidence interval of the mean length.

Length

21.9

20.8

18.6

15.1

16.5

19.9

15.1

21.5

16.7

22.1

< Select an answer σ² s² x̄ p̂ μ s p σ  <

Do not round in between steps. Round answers to at least 4 decimal places.

Answer 1

Sol:

sample size=n=10

sample mean=xbar

=188.2/10

=18.82

x bar=18.82

sample standard deviation=s=2.791177

alpha=0.05

alpha/2=0.05/2=0.025

t critical=2.26215716

95% confidence interval for true mean is

xbar-t *s/sqrt(n),xbar+t *s/sqrt(n)

18.82-2.26215716*2.791177/sqrt(10),18.82+2.26215716*2.791177/sqrt(10)

16.8233<μ<20.8167

ANSWER:

16.8233<μ<20.8167

method 2:

same thing can be done in R software in 2 lines of code as:

Length <- c(
21.9,
20.8,
18.6,
15.1,
16.5,
19.9,
15.1,
21.5,
16.7,
22.1
)
t.test(Length)

output:

One Sample t-test

data: Length
t = 21.322, df = 9, p-value = 5.159e-09
alternative hypothesis: true mean is not equal to 0
95 percent confidence interval:
16.82331 20.81669
sample estimates:
mean of x
18.82

From that output:

95 percent confidence interval:
16.8233 and 20.8167