Repeat Problem 4 assuming that the times spent per week playing with their children by all alcoholic and all nonalcoholic fathers both are normally distributed with unequal and unknown standard...

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Repeat Problem 4 assuming that the times spent per week playing with their children by all alcoholic and all nonalcoholic fathers both are normally distributed with unequal and unknown standard deviations.

Problem 4:

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: Repeat Problem 4 assuming that the times spent per week playing with their children by all alcoholic...

David answered on Dec 25 2021

125 Votes

transtutors.com/questions/-2086508.htm
in
A 95% CI for µ1 − µ2 without 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 not equal, and are unknown, we use 2 sample t-distribution
method:
Degrees of freedom: = df = smaller of n1 − 1 or n2 − 1 = 19
Standard Error: σx1−x2 =
√
s21
n1
+ s
2
2
n2
Critical Value = tα/2,df = t0.025,df=19 = 2.093 (from t-table, two-tails, d.f . = 19
)
Margin of Error = E = tα/2,df ×
√
s21
n1
+ s
2
2
n2
= 2.093 ×
√
0.2916
20 +
0.64
25
E = 2.093 × 0.200449 ≈ 0.420
Margin of Error,E = 0.42
Point Estimate of difference: x1 − x2 = −2.3
Limits of 95% confidence interval are given by:
Lower limit = x1 − x2 − E = −2.3 − 0.419541 = −2.719541
Upper limit = x1 − x2 + E = −2.3 + 0.419541 = −1.880459
95% confidence interval is: x1 − x2 ± E = −2.3 ± 0.419541
= (−2.719541, −1.880459)
95% CI using...

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Repeat Problem 4 assuming that the times spent per week playing with their children by all alcoholic and all nonalcoholic fathers both are normally distributed with unequal and unknown standard...

Answer To: Repeat Problem 4 assuming that the times spent per week playing with their children by all alcoholic...

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