Question

A) Sample of size 15 is taken from among a population of egrets and used to calculate a 99% confidence interval for the true average wingspan in the population. Suppose the sample standard deviation is calculated to be s =6.2. What is the margin of error for the confidence interval? Round your answer to 3 decimal places.

B) Suppose you are asked to construct a 99% confidence interval for an unknown population mean μ. The population standard deviation is unknown and the sample used to construct the interval has sample size n =16. What critical value t ∗should be used in the formula?

Answer 1

)solution

Given that,

s =6.2

n = 15

Degrees of freedom = df = n - 1 = 15- 1 = 14

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

t /2  df = t0.005,14 =2.977 ( using student t table)

Margin of error = E = t/2,df * (s /n)

= 2.977* ( 6.2/ 15) = 4.766

b.

df=n-1=16-1=15

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

t /2  df = t0.005,15 =2.947 ( using student t table)