In an experiment involving cholesterol levels in type A and

In an experiment involving cholesterol levels in type A and type B men the following data output was reported for a Kruskal–Wallis test:

Chi Square

6.333

Df

1

Asymp. Sig

0.011853

What does this result suggest?

a.There is a significant difference between the two groups, p = .011853.

b.There is no significant difference between the groups, p > .05.

c.There is no significant difference between the groups, p = .011853.

d.None of the above.

1a.Which of the following statement about Pearson’s correlation coefficient is not true?

a.It can be used as an effect size measure.

b.It varies between –1 and +1.

c.It cannot be used with binary variables (those taking on a value of 0 or 1).

d.It can be used on ranked data.

1b.The correlation between two variables A and B is .12 with a significance of p < .01. What can we conclude?

a.That there is a substantial relationship between A and B.

b.That there is a small relationship between A and B.

c.That variable A causes variable B.

d.All of the above.

Solution

As the p-value of the test (0.011853) is less than the required significance level of 0.05, so we will reject the null hypothesis and conclude that there is significant difference between the two groups.

So, option (a) is correct.

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1a.)

Option (c) is correct.

Because we can use Pearsons Correlation coefficient for the binary variables under some conditions and its result is not easy to be interpreted.

Other options are correct because we know that -1 <= r <= 1, and it is widely used as measure of effect size. Also it can be applied to ranked data.

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1.b)

As the p-value of the correlation is < 0.01, so at significance level of alpha greater than or equal to 0.01 we would reject the null hypothesis and conclude that the correlation is significant. And as the value of correlation is very small, so we can say that there is a small correlation between the variables.

So, option (b) is correct.