A sample of 20 alcoholic fathers showed that they spend an average of 2.3 hours per week playing with their children with a standard deviation of .54 hour. A sample of 25 nonalcoholic fathers gave a...

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Answer To: A sample of 20 alcoholic fathers showed that they spend an average of 2.3 hours per week playing...

Robert answered on Dec 25 2021
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A 95% CI for µ1 − µ2 using equal variance as-
sumption

We want to construct confidence interval of mean difference for independent
samples from data:
sample 1 sample 2
x1 = 2.3 x2 = 4.6
s1 = 0.54 s2 = 0.8
n1 = 20 n2 = 25
Significance level = α = 1 − 0.95 = 0.05
Since Variances are assumed equal, but known, we use pooled variance.
Degrees of freedom: = df = n1 + n2 − 2 = 43
The pooled estimate of Variance is:
= s2p =
(n1 − 1)s21 + (n2 − 1)s22
(n1 − 1) + (n2 − 1)
≈ 0.486056
The standard error is sx1−x2 =

s2p
n1
+
s2p
n2
≈ 0.209153
Critical Value = tα/2,df = t0.025,df=43 = 2.017 (from t-table, two-tails, d.f . = 43
)
Margin of Error = E = tα/2,df ×

s2p
n1
+
s2p
n2
= 2.017 × 0.209153
E ≈ 0.422
Margin of...
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