Question

The t statistic for a test of

H0:μ=5H0:μ=5

HA:μ≠5HA:μ≠5
based on n = 6 observations has the value t = -1.93.

(a) What are the degrees of freedom for this statistic? _____

(b) Using the appropriate table in your formula packet, bound the p-value as closely as possible:
_____ < p-value < _____

Answer 1

Solution:

a)

degrees of freedom = n - 1 = 6 - 1 = 5

degrees of freedom = 5

b)

Observe that there is sign in HA.

It indicates that the test is TWO TAILED

d.f. = 5

Test statistic t = -1.93

| t | = |-1.93| = 1.93

Use t distribution table.

See the column of d.f.

Go to 5 d.f.

In the row of 5 , see where the value 1.93 fits in between.

It fits between 1.476 and 2.015

See the corresponding alpha values at TWO tailed

0.20 and 0.10

So , range of p value is 0.10 to 0.20

0.10 < p-value < 0.20