Question

Let \(\mu\) denote the true average lifetime (in hours) for a certain type of battery under controlled laboratory conditions. A test of \(H_{0}: \mu=10\) versus \(H_{a}\) : \(\mu<10\) will be based on a sample of size \(36 .\) Suppose that \(\sigma\) is known to be 0.6 , from which \(\sigma_{\bar{x}}=.1\). The appropriate test statistic is then $$ z=\frac{\bar{x}-10}{0.1} $$ a. What is \(\alpha\) for the test procedure that rejects \(H_{0}\) if \(z \leq-1.28 ?\) b. If the test procedure of Part (a) is used, calculate \(\beta\) when \(\mu=9.8\), and interpret this error probability. c. Without doing any calculation, explain how \(\beta\) when \(\mu=9.5\) compares to \(\beta\) when \(\mu=9.8\). Then check your assertion by computing \(\beta\) when \(\mu=9.5\). d. What is the power of the test when \(\mu=9.8\) ? when \(\mu=9.5 ?\)

Short answer

\(\alpha\) is the probability of the observing Z-value less than or equal to -1.28. To calculate \(\beta\) for \(\mu=9.8\) and \(\mu=9.5\), subtract the calculated z-scores from -1.28 and find the corresponding probability. As expected, \(\beta\) decreases with a decrease in \(\mu\). The power of the test is \(1 - \beta\), for \(\mu = 9.8\) and \(\mu = 9.5\).

Step-by-step solution

Calculating \(\alpha\)

To calculate \(\alpha\), determine the area to the left of the test statistic value under the standard normal curve. The value where the null hypothesis is rejected is \(z \leq -1.28\). As \(\alpha\) represents the probability of rejecting the null hypothesis when it is true, use the standard normal distribution to find the cumulative probability at \(z = -1.28\). This is given by \(\alpha = P(Z \leq -1.28)\).

Calculating \(\beta\) when \(\mu = 9.8\)

To calculate \(\beta\) at \(\mu = 9.8\), first calculate the z-score at \(\mu = 9.8\), then use the standard normal distribution to find the probability of not rejecting the null hypothesis when it is false. The z-score is given by \(z = \frac{(9.8 - 10)}{0.1} = -2\), and so \(\beta = P(Z > -1.28 | \mu = 9.8) = P(Z > -2 + 1.28) = P(Z > -0.72)\).

Comparing \(\beta\)s

The value of \(\beta\) decreases with a decrease in the value of \(\mu\) because for smaller values of \(\mu\), it's less likely to not reject \(H_0\) when it is false.

Computing \(\beta\) when \(\mu=9.5\)

Calculate the z-score at \(\mu = 9.5\), then use the standard normal distribution to find the probability of not rejecting the null hypothesis when it is false. The z-score is given by \(z = \frac{(9.5 - 10)}{0.1} = -5\), and so \(\beta = P(Z > -1.28 | \mu = 9.5) = P(Z > -5 + 1.28) = P(Z > -3.72)\). As expected, \(\beta\) when \(\mu = 9.5\) is smaller than \(\beta\) when \(\mu = 9.8\).

Calculating Power of Test

The power of the test is \(1 - \beta\). So, for \(\mu = 9.8\), the power of the test is \(1 - \beta_{\mu=9.8}\) and for \(\mu = 9.5\), it is \(1 - \beta_{\mu=9.5}\).

Key concepts

Null Hypothesis

Understanding the concept of the null hypothesis is crucial in hypothesis testing. It is a statement that there is no effect or no difference, and it often represents a position of skepticism or the status quo. For example, in the given exercise, the null hypothesis (\( H_0 \)) is that the true average lifetime (\( \mu \) for a certain type of battery is 10 hours (\( \mu = 10 \) hours).

When conducting a hypothesis test, you aim to determine if there is enough evidence to reject the null hypothesis. The test is set up to control the rate at which we are willing to risk a type I error, which is rejecting the null hypothesis when it is true. This is signified by \( \alpha \) in the exercise, and its value is determined based on the chosen significance level for the test.

Test Statistic

The test statistic is a standardized value that is calculated from sample data during a hypothesis test. It is used to decide whether to reject the null hypothesis. The formula for the test statistic takes into account the sample mean, the hypothesized population mean, and the standard error of the mean.

In our given problem, the test statistic is a z-score calculated by \( z = \frac{\bar{x} - 10}{0.1} \) which measures the number of standard errors the sample mean (\( \bar{x} \) is from the hypothesized population mean of 10 hours. The smaller the test statistic, the less likely the sample mean is drawn from a population where the true mean is 10 hours, providing evidence against the null hypothesis.

Type II Error

A Type II error, represented by \( \beta \), occurs when a hypothesis test fails to reject a false null hypothesis. In other words, it's the error of not detecting an effect that is actually there. The probability of making a Type II error depends on the true population mean. The further the true mean is from the hypothesized mean under the null hypothesis, the lower the probability of making a Type II error.

In the given exercise, we're calculating \( \beta \) when \( \mu = 9.8 \) hours, which indicates the probability of erroneously not rejecting the null hypothesis when the true average lifetime of the battery is actually 9.8 hours. Students are often confused about calculating \( \beta \), so emphasizing that it is the probability of the test statistic being less extreme than the critical value, given that the alternative hypothesis is actually true, can help clarify.

Power of a Test

The power of a test, which is \( 1 - \beta \) , measures the test's ability to correctly reject a false null hypothesis. It is directly related to Type II error (\( \beta \) in that increasing power decreases the probability of a Type II error. It is affected by several factors including the significance level (\( \alpha \), sample size, effect size, and population variance.

In the context of the exercise, students are asked to calculate the power of the test for two different true means of the battery life (\( \mu \) . As one moves further away from the null hypothesis value, the power of the test increases because the likelihood of correctly rejecting the false null hypothesis rises. Hence, when the true mean battery life (\( \mu \) decreases from 9.8 to 9.5 hours, the power of the test increases, indicating a higher chance of detecting the actual difference.