Question

(1 point) A random sample of n measurements was selected from a population with unknown mean u and standard deviation o. Calc


just got b wrong

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Answer #1

The formula for estimation is:

μ = M ± t(sM)

where:

M = sample mean
t = t statistic determined by confidence level
sM = standard error = √(s2/n)

a)

sample mean (x-) : 102.2

sample size (n) : 100

standard deviation(s) : 2.28

Calculation

x- = 102.2
t = 1.66
sM = √(2.282/100) = 0.23

μ = x-± t*(sM)
μ = 102.2 ± 1.66*0.23
μ = 102.2 ± 0.379

90% CI [101.821, 102.579].

b)

sample mean (x-) : 84.8

sample size (n) : 90

standard deviation(s) : 2.19

Calculation

x- = 84.8
t = 1.66
sM = √(2.192/90) = 0.23

μ = x- ± t*(sM)
μ = 84.8 ± 1.66*0.23
μ = 84.8 ± 0.384

= [84.416, 85.184].

90% CI [84.416, 85.184].

c)

sample mean (x-) : 55.8

sample size (n) : 80

standard deviation(s) : 2.48

Calculation

x- = 55.8
t = 1.66
sM = √(2.482/80) = 0.28

μ = x- ± t*(sM)
μ = 55.8 ± 1.66*0.28

=  [55.339, 56.261]

90% CI [55.339, 56.261].

d)

sample mean (x-) : 76.3

sample size (n) : 90

standard deviation(s) : 2.68

Calculation

x- = 76.3
t = 1.66
sM = √(2.682/90) = 0.28

μ = x- ± t * (sM)
μ = 76.3 ± 1.66*0.28
μ = 76.3 ± 0.47
μ = 55.8 ± 0.46

=  [75.83, 76.77].

90% CI [75.83, 76.77].

Thank You...|||

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