Question

A baseball coach reviews the number of runs hit per game for the past several seasons. Since the team plays so many games, he selects a random sample of 10 games and records the number of runs scored in each game. The average number of runs scored is 7 with a standard deviation of 3.1 runs.

Compute the margin of error given a confidence level of 99%. (Use a table or technology. Round your answer to three decimal places.)

Answer 1

Given that,

= 7

s =3.1

n = 10

Degrees of freedom = df = n - 1 = 10- 1 = 9

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

_{t /2  df} = t_0.005, 9=3.250 ( using student t table)

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

= 3.250* ( 3.1/ 10)

E= 3.186

The 99% confidence interval estimate of the population mean is,

- E < < + E

7 - 3.186 < < 7+ 3.186

3.814< < 10.186

( 3.814,  10.186 )