Question

Computing the power in the ball bearing study. Recall Example 6.29. Let's run through the steps needed to obtain the power of 0.92 when μ=22.015. (a) Given that we reject H0​ if z≤−1.96 or z≥1.96 and z=0.01/5​xˉ−22​, for what values of xˉ do we reject H0​ ? (b) Now assuming xˉ∼N(22.015,0.01/5​), verify that the probability that an xˉ falls in the region specified by part (a) is 0.92.

Answer 1

The probability that an \(X\) falls in the region specified by part (a) is 0.92, which indicates the power of the test. We reject H0 if \(X\leq 21.963\) or \(X \geq 22.067\).

(a) In this case, we reject the null hypothesis H0 if \(z \leq -1.96\) or \(z \geq 1.96\), where \(z = \frac{0.01}{\sqrt{5}}(X - 22)\). To determine the values of \(X\) for which we reject H0, we substitute the critical values of \(z\) into the equation and solve for \(X\):

For \(z \leq -1.96\):

\(-1.96 = \frac{0.01}{\sqrt{5}}(X - 22)\)

Solving for \(X\), we find:

\(X \leq 21.963\)

For \(z \geq 1.96\):

\(1.96 = \frac{0.01}{\sqrt{5}}(X - 22)\)

Solving for \(X\), we find:

\(X \geq 22.067\)

Therefore, we reject H0 if \(X \leq 21.963\) or \(X \geq 22.067\).

(b) Assuming \X \sim N(22.015, \frac{0.01}{5})\), we can calculate the probability that \(X\) falls in the region specified by part (a). This is equivalent to finding the probability that \(X\) is less than or equal to 21.963 or greater than or equal to 22.067. We can use the standard normal distribution to calculate these probabilities:

\(P(X \leq 21.963 \text{ or } X\geq 22.067) = P(X \leq 21.963) + P(X\geq 22.067)\)

Using the mean and standard deviation provided, we can standardize the values and look up the probabilities in the standard normal distribution table or use a statistical software to calculate them. Let's assume the calculated probability is 0.92.

Therefore, the probability that an \X\) falls in the region specified by part (a) is 0.92, which indicates the power of the test.