Question

The distribution of systolic blood pressure in the general population is normal with a mean of 130 mm Hg and a standard deviation of 20 mm Hg. In a special subgroup of 85 people with glaucoma, we find that the mean systolic blood pressure is 135 m Hg with a standard deviation of 20 mm Hg.

(a) Assuming that the standard deviation of the glaucoma patients in the same as that of the general population, test for an association between glaucoma and high blood pressure. Include a precise statement of your null and alternative hypothesis.

(b) If the test is to be conducted at the α = 0.05 level, what is the power of this test against the alternative that the population mean is 135 mm Hg.

(c) Suppose that you run the same test without making the assumption concerning the standard deviation. What would be the P-value associated with your test?

(d) Construct a 90% CI on the mean SBP for persons with glaucoma, assume the population s.d. is unknown.

Answer 1

(a)

Assuming that the standard deviation of the glaucoma patients in the same as that of the general population, we use one sample z test for the hypothesis test.

H0: Mean systolic blood pressure of glaucoma patients = 130 mm Hg

H1: Mean systolic blood pressure of glaucoma patients > 130 mm Hg

Standard error of mean = / = 20 / = 2.17

Test statistic, z = (Observed mean - Hypothesized mean) / standard error

= (135 - 130) / 2.17

= 2.30

For one tail test, p-value = P(z > 2.30) = 0.0107

Since p-value is less than 0.05 significance level, we reject null hypothesis H0 and conclude that there is significant evidence that Mean systolic blood pressure of glaucoma patients is greater than 130 mm Hg and thus there is an association between glaucoma and high blood pressure.

(b)

For α = 0.05 level, critical z value = 1.645

Critical value of mean  systolic blood pressure to reject H0 = 130 + 1.645 * 2.17 = 133.57

Power = P(Reject H0 | = 135)

= P(Mean > 133.57 | = 135)

= P(z > (133.57 - 135) / 2.17)

= P(z > -0.66)

= 0.7454

(c)

Assuming we do not know the population standard deviation, we will use one sample t test.

Degree of freedom = n - 1 = 85 - 1 = 84

P-value = P(t > 2.30, df = 84) = 0.012

(d)

Critical value of t at 90% CI and df = 84 is, 1.66

90% CI on the mean SBP for persons with glaucoma is,

(135 - 1.66 * 2.17, 135 + 1.66 * 2.17)

(131.4, 138.6)