Question

A sample of size 6 is taken from a population with an unknown variance. The sample mean is 603.3 and the sample standard deviation is 189.8. Calculate a 95% confidence interval. What is the upper limit of the interval? Enter your answer to 1 decimal places.

Answer 1

Here, sample size(n) = 6

sample mean() = 603.3

sample standard deviation(s) = 189.8

Confidence level() = 0.05

The confidence interval is

(- *t/2 , + * t/2 )

Here, /2 = 0.025

t/2;n-1 = 2.571

The 95% confidence interval is

(603.3- * 2.571, 603.3+ * 2.571)

i.e, (385.070588 , 821.529412)

The upper limit of the interval is 821.5 (upto 1 decimal place)