What is the interpretation of the data at .01 level of significance for the null
ID: 3353776 • Letter: W
Question
What is the interpretation of the data at .01 level of significance for the null hypothesis? Assume a one-tailed negative directional test.
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.
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.
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.
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.
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 8.00Explanation / Answer
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
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