Question

Suppose 4 brands of golf balls are compared by hitting a random sample of balls of each brand and recording the distance the balls travel. This is a CR design. Suppose the MSE = 24 and suppose the sample size for Brand C is 10. Further, suppose Tukey Multiple Comparison Procedure shows that Brand C's mean distance exceeds all others.

Suppose the sample mean for Brand C was 231 calculate the width of a 95% confidence interval for the true mean distance for Brand C balls. Assume that the degrees of freedom for error in the ANOVA table is 30. Give your answer to two decimal places without rounding up or down.

Answer 1

TRADITIONAL METHOD
given that,
sample mean, x =231
standard deviation, s =4.898
sample size, n =10
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 4.898/ sqrt ( 10) )
= 1.549
II.
margin of error = t α/2 * (standard error)
where,
ta/2 = t-table value
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 9 d.f is 2.262
margin of error = 2.262 * 1.549
= 3.504
III.
CI = x ± margin of error
confidence interval = [ 231 ± 3.504 ]
= [ 227.496 , 234.504 ]
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DIRECT METHOD
given that,
sample mean, x =231
standard deviation, s =4.898
sample size, n =10
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 9 d.f is 2.262
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 231 ± t a/2 ( 4.898/ Sqrt ( 10) ]
= [ 231-(2.262 * 1.549) , 231+(2.262 * 1.549) ]
= [ 227.496 , 234.504 ]
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interpretations:
1) we are 95% sure that the interval [ 227.496 , 234.504 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population mean