Question

When not under undue stress, healthy 18-year-old women have systolic blood pressures that are normally distributed with a mean of 120 mm Hg and a standard deviation of 14 mm Hg. Rosaura wants to see if the stress of final exams elevates the blood pressure of freshmen women. She conducts a test of: H0: systolic blood pressures of freshmen women taking final exams are normal (i.e. they are normally distributed with µ = 120 and σ = 14). Ha: systolic blood pressures of freshmen women taking final exams are elevated (i.e. µ > 120). To conduct the test she recruits 32 women from her statistics class and measures their systolic blood pressure one hour after they complete the final exam. Working at the 5% significance level it can be shown that if the average blood pressure of women in the sample X¯ is greater than 124.1 mm Hg then the null hypothesis should be rejected. Suppose the systolic blood pressure of freshmen women taking final exams is, in fact, elevated but also has a higher variance. Indeed, suppose the systolic blood pressure has a mean of 128 mm Hg and a standard deviation of 18 mm Hg. What is the probability her test will reveal that the alternative hypothesis is true?

Answer 1

Explanation:

Hello!

The variable of interest is:

X: Systolic blood pressure of a freshmen women one hour after she completed a statistics exam.

This variable is known to have a normal distribution with mean 128 mmHg and standard deviation 18 mmHg

The researcher took a sample of n= 32 freshmen women

Her hypotheses are

H₀: The systolic blood pressures of freshmen women taking final exams are within a normal range. (μ=120)

H₁: The systolic blood pressures of freshmen women taking final exams are elevated. (μ>120)

Using an α: 0.05

If the sample average shows to be greater than 124.1 mmHg, the null hypothesis should be rejected.

Symbolically: P(X[bar]>124.1)

To calculate this probability you have to use the distribution of the population of 18-year-old women:

μ= 120mmHg and σ= 14mmHg

Z= (X[bar]-μ)/(σ/√n)

The Z value corresponding to X[bar]= 124.1 mmHg is:

Z= (124.14-120)/(14/√32)= 1.67

Then the probability of the null hypothesis being rejected is:

P(X[bar]>124.1)= P(Z>1.67)= 1 - P(Z≤1.67)= 1 - 0.95254= 0.04746

(Note, what was just calculated is the p-value of the test)

I hope this helps!