How can they tell that this is a two-tailed test?

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How can they tell that this is a two-tailed test?

GUIDED EXERCISE 6
Testing p
A medical research team is studying the effect of a new drug on red blood cells. Let x be a random variable
representing milligrams of the drug given to a patient. Let y be a random variable representing red blood cells
per cubic milliliter of whole blood. A random sample of n =
7 volunteer patients gave the following results.
9.2
10.1
9.0
12.5
8.8
9.1
9.5
y
5.0
4.8
4.5
5.7
5.1
4.6
4.2
Use a calculator to verify that r~
0.689. Then use a 1% level of significance to test the claim that p + 0.
(a) State the null and alternate hypotheses. What is
the level of significance a?
H6: ρ-0 Η: ρ 4 0; α 0.01
rVn – 2
V1 - P
0.689 V7 – 2
V1 - 0.689?
-
1.5406
(b) Compute the t value of the sample test statistic.
2.126
f =
0.7248
(c) Use the Student's t distribution, Table 6 of
Appendix II, to estimate the P-value.
d.f. = n – 2 = 7 – 2 = 5; two-tailed test
/ two-tail area
0.100
0.050
d.f. = 5
2.015
2.571
↑
Sample t = 2.126
0.050 < P-value < 0.100
(d) Do we reject or fail to reject H,?
Since the interval containing the P-value lies to the
right of a =
P-value = 0.0866.
0.01, we do not reject H,. Technology gives
0.01
0.050
0.100
Transcribed Image Text:GUIDED EXERCISE 6 Testing p A medical research team is studying the effect of a new drug on red blood cells. Let x be a random variable representing milligrams of the drug given to a patient. Let y be a random variable representing red blood cells per cubic milliliter of whole blood. A random sample of n = 7 volunteer patients gave the following results. 9.2 10.1 9.0 12.5 8.8 9.1 9.5 y 5.0 4.8 4.5 5.7 5.1 4.6 4.2 Use a calculator to verify that r~ 0.689. Then use a 1% level of significance to test the claim that p + 0. (a) State the null and alternate hypotheses. What is the level of significance a? H6: ρ-0 Η: ρ 4 0; α 0.01 rVn – 2 V1 - P 0.689 V7 – 2 V1 - 0.689? - 1.5406 (b) Compute the t value of the sample test statistic. 2.126 f = 0.7248 (c) Use the Student's t distribution, Table 6 of Appendix II, to estimate the P-value. d.f. = n – 2 = 7 – 2 = 5; two-tailed test / two-tail area 0.100 0.050 d.f. = 5 2.015 2.571 ↑ Sample t = 2.126 0.050 < P-value < 0.100 (d) Do we reject or fail to reject H,? Since the interval containing the P-value lies to the right of a = P-value = 0.0866. 0.01, we do not reject H,. Technology gives 0.01 0.050 0.100
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