Question

Suppose you are interested in conducting the statistical test of \({H_0}:\mu = 255\) against \({H_a}:\mu > 225\), and you have decided to use the following decision rule: Reject H₀ if the sample mean of a random sample of 81 items is more than 270. Assume that the standard deviation of the population is 63.

a. Express the decision rule in terms of z.

b. Find \(\alpha \), the probability of making a Type I error by using this decision rule.

Short answer

a. Reject the null hypothesis when\(z > 2.14\).

b. The probability of a Type I error is: \(\alpha = 0.0150\).

Step-by-step solution

Given information

The hypothesis test is:\({H_0}:\mu = 255\)versus\({H_a}:\mu > 255\).

The population standard deviation is 63.

The decision rule is: to reject the null hypothesis if the sample mean is more than 270.

Expressing the decision rule in terms of z

a.

Reject the null hypothesis if the sample mean is more than 270

That is,

\(\begin{aligned}\bar x > 270\\\frac{{\bar x - \mu }}{{\frac{\sigma }{{\sqrt n }}}} > \frac{{270 - \mu }}{{\frac{\sigma }{{\sqrt n }}}}\\z > \frac{{270 - 255}}{{\frac{{63}}{{\sqrt {81} }}}}\\z > \frac{{15}}{{\frac{{63}}{9}}}\\z > \frac{{15}}{7}\\z > 2.14\end{aligned}\)

Therefore, reject the null hypothesis when \(z > 2.14\).

Computing the probability of Type I error

b.

The probability of Type I error is obtained as follows:

\(\begin{aligned}\alpha &= P\left( {Z > 2.14} \right)\\ &= 1 - P\left( {Z \le 2.14} \right)\\ &= 1 - 0.9850\\ &= 0.0150\end{aligned}\).

The probability of a z-score less than or equal to 2.14 is obtained using the z-table.

Therefore, the probability of a Type I error is: \(\alpha = 0.0150\).