AP Statistics 5.5 Sampling Distributions for Sample Proportions- FRQs - Exam Style Questions

(a)
A Type II error occurs when we fail to reject the null hypothesis when the alternative hypothesis is actually true. In this context, a Type II error would occur if the true mean systolic blood pressure of all employees at the corporation is greater than 122 mmHg, but the test fails to provide sufficient evidence to reject the null hypothesis that \( \mu = 122 \) mmHg.

(b)
For a one-sided test with \( \alpha = 0.05 \), the critical z-value is \( z = 1.645 \).
The test statistic is \( z = \frac{\bar{x} – 122}{1.5} \).
We reject \( H_0 \) when \( \frac{\bar{x} – 122}{1.5} > 1.645 \).
Solving for \( \bar{x} \): \( \bar{x} > 122 + 1.645 \times 1.5 = 122 + 2.4675 = 124.4675 \).
So we would reject \( H_0 \) when \( \bar{x} > 124.4675 \) mmHg.

(c)
If the true mean is 125 mmHg, then the sampling distribution of \( \bar{x} \) has mean 125 and standard deviation 1.5.
We want \( P(\bar{x} > 124.4675) \) when \( \bar{x} \sim N(125, 1.5) \).
\( z = \frac{124.4675 – 125}{1.5} = \frac{-0.5325}{1.5} \approx -0.355 \).
\( P(z > -0.355) = 1 – P(z < -0.355) = 1 – 0.3613 = 0.6387 \).
The probability is approximately \( \boxed{0.64} \).

(d)
The probability found in part (c) is called the \( \boxed{\text{power}} \) of the test.

(e)
The probability would be \( \boxed{\text{greater}} \) than the probability calculated in part (c).
With a larger sample size, the standard error \( \left( \frac{\sigma}{\sqrt{n}} \right) \) decreases. This causes two effects:
1. The critical value in terms of \( \bar{x} \) decreases (moves closer to 122), making it easier to reject \( H_0 \)
2. The sampling distribution under the alternative hypothesis (mean = 125) becomes more concentrated around 125
Both effects increase the probability that \( \bar{x} \) will exceed the critical value, thus increasing the power of the test.