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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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 mean of 4.6 hours per week with a standard deviation of .8 hour.


a. Construct a 95% confidence interval for the difference between the mean times spent per week playing with their children by all alcoholic and all nonalcoholic fathers.


b. Test at the 1% significance level whether the mean time spent per week playing with their children by all alcoholic fathers is less than that of nonalcoholic fathers.


Assume that the times spent per week playing with their children by all alcoholic and all nonalcoholic fathers both are normally distributed with equal but unknown standard deviations.




Answered Same DayDec 25, 2021

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
129 Votes
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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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