Related questions with answers

The heat evolved in calories per gram of a cement mixture is approximately normally distributed. The mean is thought to be 100 and the standard deviation is 2. We wish to test $H_{0} : \mu=100$ versus $H_{1} : \mu \neq 100$ with a sample of n = 9 specimens. (a) If the rejection region is defined as $\overline{x}<98.5 \text { or } \overline{x}>101.5$ find the type I error probability $\alpha.$ (b) Find $\beta$ for the case where the true mean heat evolved is 103. (c) Find $\beta$ for the case where the true mean heat evolved is 105. This value of $\beta$ is smaller than the one found in part (b). Why?

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Suppose that samples of size n = 25 are selected at random from a normal population with a mean of 100 and a standard deviation of 10. What is the probability that the sample mean falls in the interval from μ $\overline{_{x}}$ - 1.8σ $\overline{_x}$ to μ $\overline{_{x}}$ + 1.0σ $\overline{_x}$ ?

Question

The heat evolved in calories per gram of a cement mixture is approximately normally distributed. The mean is thought to be 100, and the standard deviation is 2. You wish to test H $_0$ : μ = 100 versus H $_1$ : μ ≠ 100 with a sample of n = 9 specimens.
a. If the acceptance region is defined as 98.5 ≤ $\overline{x}$ ≤ 101.5, find the type I error probability α.
b. Find β for the case in which the true mean heat evolved is 103.
c. Find β for the case where the true mean heat evolved is 105. This value of β is smaller than the one found in part (b). Why?

Solution

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The probability of making a $\textbf{type I error}$ is

\begin{align} \alpha=P(\text{type I error})=P(\text{ reject $H_0$ when $H_0$ is true}). \end{align}

Sometimes the type I error probability is called the significance level, the $\alpha$ -error, or the size of the test.

The mean is thought to be $\mu=100$ , and the standard deviation is $\sigma=2$ . You wish to test $H_0:\mu = 100$ versus $H_1:\mu\ne 100$ with a sample of $n = 9$ specimens.