Respuesta :
Answer:
(a) m = 12, n = 10, s1 = 4.0, s2 = 5.0
[tex]df=m+n-2=12+10-2=20[/tex] (equal variances)
[tex]df=\frac{[\frac{4^2}{12}+\frac{5^2}{10}]^2}{\frac{(4^2 /12)^2}{12 -1}+\frac{(5^2 /10)^2}{10 -1}}=17.16 \approx 18[/tex] (Assuming unequal variances)
(b) m = 12, n = 15, s1 = 4.0, s2 = 5.0
[tex]df=m+n-2=12+15-2=25[/tex] (equal variances)
[tex]df=\frac{[\frac{4^2}{12}+\frac{5^2}{15}]^2}{\frac{(4^2 /12)^2}{12 -1}+\frac{(5^2 /15)^2}{15 -1}}=24.998 \approx 25[/tex] (Assuming unequal variances)
(c) m = 12, n = 15, s1 = 3.0, s2 = 5.0
[tex]df=m+n-2=12+15-2=25[/tex] (equal variances)
[tex]df=\frac{[\frac{3^2}{12}+\frac{5^2}{15}]^2}{\frac{(3^2 /12)^2}{12 -1}+\frac{(5^2 /15)^2}{15 -1}} =23.4 \approx 24[/tex] (Assuming unequal variances)
(d) m = 10, n = 24, s1 = 4.0, s2 = 5.0
[tex]df=m+n-2=10+24-2=32[/tex] (equal variances)
[tex]df=\frac{[\frac{4^2}{10}+\frac{5^2}{24}]^2}{\frac{(4^2 /10)^2}{10 -1}+\frac{(5^2 /24)^2}{24 -1}}=21.04 \approx 21 [/tex] (Assuming unequal variances)
Step-by-step explanation:
1) Previous concepts
The t distribution (Student’s t-distribution) is a "probability distribution that is used to estimate population parameters when the sample size is small (n<30) or when the population variance is unknown".
The shape of the t distribution is determined by its degrees of freedom and when the degrees of freedom increase the t distirbution becomes a normal distribution approximately.
The degrees of freedom represent "the number of independent observations in a set of data. For example if we estimate a mean score from a single sample, the number of independent observations would be equal to the sample size minus one."
And if we have two sample of size m and n from two independet samples the correct formula for the degrees of freedom is:
[tex]df=m+n-2[/tex] (Assuming equal variances)
[tex]df=\frac{[\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}]^2}{\frac{(s^2_1 /n_1)^2}{n_1 -1}+\frac{(s^2_2 /n_2)^2}{n_2 -1}}[/tex] (Assuming unequal variances)
2) Solution to the problem
Using this formula we can answer all the cases:
(a) m = 12, n = 10, s1 = 4.0, s2 = 5.0
[tex]df=m+n-2=12+10-2=20[/tex] (equal variances)
[tex]df=\frac{[\frac{4^2}{12}+\frac{5^2}{10}]^2}{\frac{(4^2 /12)^2}{12 -1}+\frac{(5^2 /10)^2}{10 -1}}=17.16 \approx 18[/tex] (Assuming unequal variances)
(b) m = 12, n = 15, s1 = 4.0, s2 = 5.0
[tex]df=m+n-2=12+15-2=25[/tex] (equal variances)
[tex]df=\frac{[\frac{4^2}{12}+\frac{5^2}{15}]^2}{\frac{(4^2 /12)^2}{12 -1}+\frac{(5^2 /15)^2}{15 -1}}=24.998 \approx 25[/tex] (Assuming unequal variances)
(c) m = 12, n = 15, s1 = 3.0, s2 = 5.0
[tex]df=m+n-2=12+15-2=25[/tex] (equal variances)
[tex]df=\frac{[\frac{3^2}{12}+\frac{5^2}{15}]^2}{\frac{(3^2 /12)^2}{12 -1}+\frac{(5^2 /15)^2}{15 -1}} =23.4 \approx 24[/tex] (Assuming unequal variances)
(d) m = 10, n = 24, s1 = 4.0, s2 = 5.0
[tex]df=m+n-2=10+24-2=32[/tex] (equal variances)
[tex]df=\frac{[\frac{4^2}{10}+\frac{5^2}{24}]^2}{\frac{(4^2 /10)^2}{10 -1}+\frac{(5^2 /24)^2}{24 -1}}=21.04 \approx 21 [/tex] (Assuming unequal variances)
