Question

Systolic blood pressure is the amount of pressure that blood exerts on blood vessels while the heart is beating. The mean systolic blood pressure for people in the United States is reported to be 122 millimeters of mercury (mmHg) with a standard deviation of 15 mmHg.

The wellness department of a large corporation is investigating whether the mean systolic blood pressure of its employees is greater than the reported national mean. A random sample of 50 employees will be selected, the systolic blood pressure of each employee in the sample will be measured, and the sample mean will be calculated. Let μ represent the mean systolic blood pressure of all employees at the corporation. Consider the following hypotheses.

H₀: μ=122

H_a: μ>122

a) Assume that σ , the standard deviation of the systolic blood pressure of all employees at the corporation, is 15 mmHg and μ = 122 . Describe the shape, center, and spread of the sampling distribution of x for samples of size 50.

b) Based on the sampling distribution constructed in part a), what interval of values of x would represent sufficient evidence to reject the null hypothesis at the significance level α = .01 ?

c) It was determined that the actual mean systolic blood pressure is 125 mmHg, not the hypothesized value of 122 mmHg, and the standard deviation is 15 mmHg. Using the actual mean of 125 mmHg and the results from part b), determine the probability that the null hypothesis will be rejected?

d) What statistical term is used for the probability calculated in part c)? Explain.

e) Suppose the size of the sample of employees is greater than 50. Would the probability of rejecting the null hypothesis increase, decrease, or remain the same? Explain your reasoning.

Answer 1

(a) From Central Limit Theorem, since the sample size is greater than 30the sampling distribution of x for samples of size 50 follows Normal distribution with Mean or center = 122 and Standard error or spread = 15/√50 = 2.12

(b) For significance level of 0.01, the critical z value in this test = 2.327

Thus, the required minimum value of x to reject the null hypothesis = 122 + 2.327*2.12 = 126.94

(c) Probability that the null hypothesis will be rejected

= P {Z > (126.94 - 125)/2.12}

= P(Z > 0.915) = 0.18

(d) The statistical term used is Type I error.

Type I error is the probability of rejecting a true Null hypothesis. In this case though the null hypothesis is correct we reject it.

(e) For higher sample size, the standard error will be low.

Hence the test statistic will be comparatively higher.

Thus, the tail for rejecting the null hypothesis would be smaller and the probability of rejecting the null hypothesis decreases