Question

W1, W2, W3…W9 represent the holding times of 9 water samples, which follow a normal distribution with mean = 8.33 and standard deviation = 4.472. M represents the sample mean value of holding times, which also has a normal distribution. Assuming $Z$ has a standard normal distribution (mean = 0 and standard deviation = 1), select the correct statement which describes the expression for calculating the value of Type I error where: \[ H_0: M>6 \text{(null hypothesis)}, H_a: M \leq 6 \text{(alternate hypothesis)} \]

• A. $P\{Z<-1.565\}$
• B. $P\{Z<1.565\}$
• C. $P\{Z>-1.565\}$
• D. $P\{Z>1.565\}$

Hint:

Always compute Type I error as the probability of falling into the rejection region when the null hypothesis is true. Here it corresponds to the lower tail of the Z-distribution.

Answer 1

The Correct Option is A

Step 1: Standard error of the mean (SEM).
For sample size $n = 9$, \[ \sigma_M = \frac{\sigma}{\sqrt{n}} = \frac{4.472}{\sqrt{9}} = \frac{4.472}{3} \approx 1.491 \]
Step 2: Test statistic (Z).
\[ Z = \frac{M - \mu}{\sigma_M} \] Here $\mu = 8.33$, threshold = 6. \[ Z = \frac{6 - 8.33}{1.491} = \frac{-2.33}{1.491} \approx -1.565 \]
Step 3: Interpret Type I error.
Type I error = rejecting $H_0$ when $H_0$ is true. Since $H_0: M>6$, Type I error occurs when $M \leq 6$. This corresponds to \[ P(Z<-1.565) \] Final Answer: \[ \boxed{P(Z<-1.565)} \]