Question

The following measurements (in picocuries per liter) were recorded by a set of ozone detectors installed in a laboratory facility:

267.7,294.8,296.5,246.2,232.2,269.3

Using these measurements, construct a 80%

confidence interval for the mean level of ozone present in the facility. Assume the population is approximately normal.

Step 1 of 4 :

Calculate the sample mean for the given sample data. Round your answer to two decimal places.

Answer 1

A) Given data

We need to determine the 80% confidence interval

Sample size (n)=6

267.7,294.8,296.5,246.2,232.2,269.3

Sample mean(X')=(267.7+294.8+296.5+246.2+232.2+269.3)/6

X'=1606.7/6=267.7833

Variance=Σ(X-X')^2/(n-1)

=[(267.7-267.7833)^2 +(294.8-267.7833)^2 +(296.5-267.7833)^2 +(246.2-267.7833)^2 + (232.2-267.7833)^2+(269.3-267.7833)^2]/(6-1)

=(3288.8683)/5 =657.77

Standard deviation (s)=√(variance)

=√(657.77) =25.647

Formula used for 80% confidence interval is

CI =X' ± t×s/(√n)

Now t value corresponding to significance level=0.2 and DOf=6-1=5 is

t value=1.4758

CI =267.7833 ± (1.4758×25.647)/√6

=267.7833 ± 15.4521

Lower limit=267.7833-15.4521=252.3312

Upper limit=267.7834+15.4521=283.2355