Mathematics, 21.12.2019 05:31, nerdykitty2927

Determine the number of degrees of freedom for the two-sample t test or ci in each of the following situations. (round your answers down to the nearest whole number.)

(a) m = 12, n = 10, s1 = 4.0, s2 = 5.0
(b) m = 12, n = 15, s1 = 4.0, s2 = 5.0
(c) m = 12, n = 15, s1 = 3.0, s2 = 5.0
(d) m = 10, n = 24, s1 = 4.0, s2 = 5.0

Answers: 3

Show answers

Answers

answered: boonkgang6821

(a) m = 12, n = 10, s1 = 4.0, s2 = 5.0

df=m+n-2=12+10-2=20 (equal variances)

$df=\frac{[\frac{4^2}{12}+\frac{5^2}{10}]^2}{\frac{(4^2 /12)^2}{12 -1}+\frac{(5^2 /10)^2}{10 -1}}=17.16 \approx 18$ (Assuming unequal variances)

(b) m = 12, n = 15, s1 = 4.0, s2 = 5.0

df=m+n-2=12+15-2=25 (equal variances)

$df=\frac{[\frac{4^2}{12}+\frac{5^2}{15}]^2}{\frac{(4^2 /12)^2}{12 -1}+\frac{(5^2 /15)^2}{15 -1}}=24.998 \approx 25$ (Assuming unequal variances)

df=m+n-2=12+15-2=25 (equal variances)

$df=\frac{[\frac{3^2}{12}+\frac{5^2}{15}]^2}{\frac{(3^2 /12)^2}{12 -1}+\frac{(5^2 /15)^2}{15 -1}} =23.4 \approx 24$ (Assuming unequal variances)

(d) m = 10, n = 24, s1 = 4.0, s2 = 5.0

df=m+n-2=10+24-2=32 (equal variances)

$df=\frac{[\frac{4^2}{10}+\frac{5^2}{24}]^2}{\frac{(4^2 /10)^2}{10 -1}+\frac{(5^2 /24)^2}{24 -1}}=21.04 \approx 21$ (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:

df=m+n-2 (Assuming equal variances)

$df=\frac{[\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}]^2}{\frac{(s^2_1 /n_1)^2}{n_1 -1}+\frac{(s^2_2 /n_2)^2}{n_2 -1}}$ (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

df=m+n-2=12+10-2=20 (equal variances)

$df=\frac{[\frac{4^2}{12}+\frac{5^2}{10}]^2}{\frac{(4^2 /12)^2}{12 -1}+\frac{(5^2 /10)^2}{10 -1}}=17.16 \approx 18$ (Assuming unequal variances)

(b) m = 12, n = 15, s1 = 4.0, s2 = 5.0

df=m+n-2=12+15-2=25 (equal variances)

$df=\frac{[\frac{4^2}{12}+\frac{5^2}{15}]^2}{\frac{(4^2 /12)^2}{12 -1}+\frac{(5^2 /15)^2}{15 -1}}=24.998 \approx 25$ (Assuming unequal variances)

df=m+n-2=12+15-2=25 (equal variances)

$df=\frac{[\frac{3^2}{12}+\frac{5^2}{15}]^2}{\frac{(3^2 /12)^2}{12 -1}+\frac{(5^2 /15)^2}{15 -1}} =23.4 \approx 24$ (Assuming unequal variances)

(d) m = 10, n = 24, s1 = 4.0, s2 = 5.0

df=m+n-2=10+24-2=32 (equal variances)

$df=\frac{[\frac{4^2}{10}+\frac{5^2}{24}]^2}{\frac{(4^2 /10)^2}{10 -1}+\frac{(5^2 /24)^2}{24 -1}}=21.04 \approx 21$ (Assuming unequal variances)

answered: Guest

the answer is; biodiversity resulting from few ancestors

adaptive radiation occurs when a population is subjected to a large natural environment with spatial ecological variations. the parts of the population in the different ecologies begin to adapt to their local environments. due to non-random mating across the large populations, this causes the different populations from the large population to speciate within their local environments. they, therefore, diverge to different species sharing a common ancestor.

answered: Guest

sorey ik so sorey i need and need to answer a few