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2 years ago

Which combination of factors is most likely to produce a significant value for an independent-measures t statistic?.

Mathematics

1 answer:

SashulF [63]2 years ago

8 0

Answer: A large mean difference and small sample variances 

Step-by-step explanation:

<em>Effect size (η)</em> is a statistical measure that determines the strength of the association (numerically) between two variables. For example, if we have data on the weight of male and female candidates and we realize that, on average, males are heavier than females, the difference between the weight of males and the weight of female candidates is known as the effect size.

The larger the effect size, the larger the weight difference between males and females will be.

Statistic effect size helps us in analyzing if the difference is factual or if it is affected by a change of factors.

In hypothesis testing, effect size, power, sample size, and critical significance level are related to each other.

The effect size formula for a hypothesis test of mean difference is:

<em>η = </em> $\frac{x1-x2}{\sqrt[]{8^2} }$

The denominator s² is the combined sample variance.

$8^2 =$ ⊕

The effect size is affected by two components:

Sample mean difference

Sample variance.

As the sample mean difference is directly proportional to the effect size, on increasing the sample mean difference value the effect size will also increase.

Ans the sample variance is inversely proportional to the the effect size, on decreasing the sample variance value the effect size will increase and vice-versa.

Thus, the correct option is A large mean difference and small sample variances

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3 years ago

Use any of the methods to determine whether the series converges or diverges. Give reasons for your answer.

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Answer:

It means $\sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6}$ also converges.

Step-by-step explanation:

The actual Series is::

$\sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6}$

The method we are going to use is comparison method:

According to comparison method, we have:

$\sum_{n=1}^{inf}a_n\ \ \ \ \ \ \ \ \sum_{n=1}^{inf}b_n$

If series one converges, the second converges and if second diverges series, one diverges

Now Simplify the given series:

Taking"n^2"common from numerator and "n^6"from denominator.

$=\frac{n^2[7-\frac{4}{n}+\frac{3}{n^2}]}{n^6[\frac{12}{n^6}+2]} \\\\=\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{n^4[\frac{12}{n^6}+2]}$

$\sum_{n=1}^{inf}a_n=\sum_{n=1}^{inf}\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\ \ \ \ \ \ \ \ \sum_{n=1}^{inf}b_n=\sum_{n=1}^{inf} \frac{1}{n^4}$

Now:

$\sum_{n=1}^{inf}a_n=\sum_{n=1}^{inf}\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\\ \\\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\\=\frac{7-\frac{4}{inf}+\frac{3}{inf}}{\frac{12}{inf}+2}\\\\=\frac{7}{2}$

So a_n is finite, so it converges.

Similarly b_n converges according to p-test.

P-test:

General form:

$\sum_{n=1}^{inf}\frac{1}{n^p}$

if p>1 then series converges. In oue case we have:

$\sum_{n=1}^{inf}b_n=\frac{1}{n^4}$

p=4 >1, so b_n also converges.

According to comparison test if both series converges, the final series also converges.

It means $\sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6}$ also converges.

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