Question

A physician wants to test if temperature has an effect on heart rate. In order to do this, she compares the heart rates in beats per minute of several random volunteers after a period of time in a room with a temperature of 50∘F and after a period of time in a room with a temperature of 75∘F. Suppose that data were collected for a random sample of 11volunteers, where each difference is calculated by subtracting the heart rate in beats per minute in the 50∘F room from the heart rate in beats per minute in the 75∘F room. Assume that the populations are normally distributed. The physician uses the alternative hypothesis Ha:μd≠0. Suppose the test statistic t is computed as t≈5.627, which has 10 degrees of freedom. What range contains the p-value?

Probability	0.10	0.05	0.025	0.01	0.005
Degrees of Freedom
5	1.476	2.015	2.571	3.365	4.032
6	1.440	1.943	2.447	3.143	3.707
7	1.415	1.895	2.365	2.998	3.499
8	1.397	1.860	2.306	2.896	3.355
9	1.383	1.833	2.262	2.821	3.250
10	1.372	1.812	2.228	2.764	3.169
11	1.363	1.796	2.201	2.718	3.106
12	1.356	1.782	2.179	2.681	3.055
13	1.350	1.771	2.160	2.650	3.012
14	1.345	1.761	2.145	2.624	2.977
15	1.341	1.753	2.131	2.602	2.947

Select the correct answer below:

A. p-value>0.10

B. 0.05< p-value <0.10

C. 0.01< p-value <0.05

D. p-value <0.01

Answer 1

In the question we have given

T computed = 5.627

df = degree of freedom = 10

We have two tailed test because H_a Alternative hypothesis has not equal to sign

We can use excel to find the actual p value

Select empty box

=TDIST( t computed , df , tail )

=TDIST(5.627 , 10 , 2 )

Press enter

Do not forget to type the = sign in excel

P value = 0.000219321

Correct option is

P value < 0.001

Look the above t table the above table only provide the one tailed p value only

We look for the row of df = 10

Look as the value increases from   the row of 10   p value decreases for two tailed

probability	0.10	0.05	0.025	0.01	0.005
2 tailed probability i.e p value	=2*0.10=0.20	=2*0.05 = 0.10	=2*0.025=0.05	= 2*0.01 = 0.02	= 2*0.005 = 0.01
Df
10	1.372	1.812	2.228	2.764	3.169

Look the row of df 10 and p value carefully

as the value from the row of df 10 increases then p value also decreases

P value for 3.169 = 0.01

Given t= 5.627 is more than 3.169

P value < 0.01

I hope this will help you :)