Week 13 Worksheet 23 MATH 10B
Thu 4/18/19
1 Hypothesis testing
1. Let Xbe a measurement of a patient’s blood pressure, which follows the normal distribution
with σ= 5 mmHg center around the true blood pressure. Let the null hypothesis be that
the patient’s blood pressure is µ= 120, and the alternative hypothesis H1:µ > 120 with
significance level α= 0.05? Supposed that the patient’s blood pressure is measure to be 135
mmHg.
(a) Is it a one-sided test?
(b) Calculate the p-value.
(c) Calculate critical value and find the rejection region.
(d) Draw a conclusion.
Solution:
i. It is a one sided test.
ii. Assume the null is true, z=X−µ
σ=135−120
5= 3. Looking up the table, Three
SD corresponds to an area of 0.4987. Since it is a one-sided test, the p-value is
0.5−0.4987 ≈0.0013.
iii. Let kbe the critical value. Then, assuming the null is true, P(X > k) = 0.05 =⇒
P(X−120
5>k−120
5) = P(Z > k−120
5) = 0.05. Thus, k−120
5>1.65 =⇒k > 128.25,
and the rejection region is [128.25,∞).
iv. Since the p-value is 0.0013 <0.05 or X= 135 >128.25, we can reject the null and
conclude that µ > 120.
2. Let Xbe a measurement of a patient’s blood pressure, which follows the normal distribution
with σ= 5 mmHg center around the true blood pressure. Let the null hypothesis be that
the patient’s blood pressure is µ= 120, and the alternative hypothesis H1:µ6= 120 with
significance level α= 0.05? Supposed that the patient’s blood pressure is measure to be 130
mmHg.
(a) Is it a one-sided test?
(b) Calculate the p-value.
(c) Draw a conclusion.
Solution:
i. It is a two sided test.
ii. Assume the null is true, z=X−µ
σ=130−120
5= 2. Looking up the table, Three
SD corresponds to an area of 0.4772. Since it is a one-sided test, the p-value is
2(0.5−0.4772) ≈0.0456.
iii. Since the p-value is 0.0456 <0.05, we can reject the null and conclude that µ6= 120.
3. You have a coin that you suspect is biased toward coming up heads. You let pbe the
probability that the coin will give heads on any one flip. Can you reject the null hypothesis
H0:p=1
2in favor of the alternative hypothesis H1:p > 1
2with significance level α= 0.05?
