Question

What is the interpretation of the data at .01 level of significance for the null hypothesis? Assume a one-tailed negative directional test.

Group 1	Group 2
Phoenix Males	Chicago Males
4.00	5.00
5.00	5.00
6.00	6.00
6.00	5.00
5.00	7.00
6.00	8.00

	a.	t(10) = 1.10, p < .10, the obtained value does not exceed the critical value, so the null hypothesis must be accepted and it must be concluded that the difference is due to chance.
	b.	t(10) = 1.10, p > .05, the obtained value does not exceed the critical value, so the null hypothesis must be accepted and it must be concluded that the difference is due to chance.
	c.	t(10) = -1.10, p > .01, the obtained value does not exceed the critical value, so the null hypothesis must be accepted and it must be concluded that the difference is due to chance.
	d.	t(10) = 1.10, p > .01, the obtained value does not exceed the critical value, so the null hypothesis must be accepted and it must be concluded that the difference is not due to chance.

Answer 1

Solution:

code in R"

phoenix_males <- c(4,5,6,6,5,6)
chicago_males <- c(5,5,6,5,7,8)
t.test(phoenix_males,chicago_males,alternative = "less")

output:

Welch Two Sample t-test

data: phoenix_males and chicago_males
t = -1.0847, df = 8.5503, p-value = 0.1539
alternative hypothesis: true difference in means is less than 0
95 percent confidence interval:
-Inf 0.4668154
sample estimates:
mean of x mean of y
5.333333 6.000000

t=-1.10

p=0.1539

p>0.01

Fail to reject Null hypothesis

Accept null hypothesis

t(10) = -1.10, p > .01, the obtained value does not exceed the critical value, so the null hypothesis must be accepted and it must be concluded that the difference is due to chance