91Ó°ÊÓ

Let \(\mu\) denote the true average lifetime for a certain type of pen 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 \(\sigma_{x}=0.1\). 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 \(\mu=9.5 ?\)

Step by step solution

01

Calculation of \(\alpha\)

This step is important for getting to know if the hypothesis gets rejected. \(\alpha\) is the probability of rejecting the null hypothesis when it is true. Using the standard z-table, find the probability corresponding to the z-score -1.28. The probability, which represents \(\alpha\), is 0.1003.

02

Calculation of \(\beta\) for \(\mu=9.8\)

In this step, we calculate \(\beta\) when \(\mu=9.8\), which is the probability of accepting the null hypothesis when it should be rejected. First, calculate the z-score as \(\frac{9.8-10}{0.1} = -2\). Now, by observing the z-table, the probability corresponding to z=-2 is 0.0228. Thus, \(\beta\) when \(\mu=9.8\) is 0.0228.

03

Comparison of \(\beta\) for \(\mu=9.5\) and \(\mu=9.8\)

By comparing the two values obtained from step 2, we would expect \(\beta\) for \(\mu=9.5\) to be smaller than \(\beta\) for \(\mu=9.8\) because a smaller value of \(\mu\) leads to a higher Z-score, which in turn gives a smaller \(\beta\). It is lower because a smaller \(\mu\) increases the likelihood that the alternative hypothesis \(H_a: \mu<10\) is true. This can be validated by computing \(\beta\) when \(\mu=9.5\). The z-score for \(\mu=9.5\) is \(\frac{9.5-10}{0.1} = -5\). By looking at the z-table, the probability corresponding to z=-5 is near to zero. Thus, \(\beta\) when \(\mu=9.5\) is approximately 0.

04

Computing the Power of the Test

The power of the test is calculated as 1-\(\beta\). Therefore, the power of the test when \(\mu=9.8\) is 1-0.0228 = 0.9772 and when \(\mu=9.5\) is 1 - 0 = 1.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Type I Error (Alpha)

In hypothesis testing, a Type I error, commonly represented by \( \alpha \), occurs when we reject the null hypothesis \( H_0 \) when it is actually true. This error essentially measures the probability of making an incorrect decision about rejecting \( H_0 \). Imagine you're trying to decide if a pen's average lifetime is less than 10 years, based on a sample.

• If you decide to reject the null hypothesis (that the average is indeed 10) in favor of the alternative (average is less than 10), but the true average is actually 10, a Type I error occurs.
• In this exercise, the test statistic has a z-score cut-off of -1.28 for rejection of \( H_0 \).
• The probability of observing a z-score of -1.28 or less corresponds to \( \alpha = 0.1003 \).

This means there is a 10.03% chance of incorrectly rejecting the null hypothesis when it is true.

Type II Error (Beta)

A Type II error, denoted by \( \beta \), occurs when we fail to reject the null hypothesis \( H_0 \) when the alternative hypothesis \( H_a \) is true. In simpler terms, we miss detecting an effect that actually exists.
For this exercise:

• To calculate \( \beta \) when the true mean \( \mu \) is 9.8, first find the z-score given the difference between \( \mu \) and hypothesized mean: \( \frac{9.8 - 10}{0.1} = -2 \).
• From the z-table, the corresponding probability (\( \beta \)) is 0.0228.
• This indicates a 2.28% chance of incorrectly accepting the null hypothesis \( \mu=10 \) when \( \mu \) is actually 9.8.

With a smaller \( \mu \) such as 9.5, \( \beta \) further lessens, suggesting higher detection accuracy for the alternative hypothesis.

Power of a Test

The power of a test is the ability to correctly reject the null hypothesis when the alternative hypothesis is true. It is represented by \( 1 - \beta \). The higher the power, the more reliable the test is.
For the given scenarios in the problem:

• The power of the test when \( \mu = 9.8 \) is \( 1 - 0.0228 = 0.9772 \), meaning a 97.72% probability exists to correctly reject \( H_0 \) when \( H_a \) is valid.
• When \( \mu = 9.5 \), \( \beta \) is approximately 0—resulting in a power near 1 (or 100%).

This demonstrates that as the true mean decreases, signals supporting \( H_a \) grow stronger, enhancing the test's ability to identify a true effect.

Z-Score Calculation

Z-score calculation in hypothesis testing helps determine how far a sample statistic is from the hypothesized parameter, in terms of standard errors. It allows you to assess whether a sample result falls within a certain probability under the null hypothesis.
In this problem:

• The z-score formula used is \( z = \frac{\bar{x} - 10}{0.1} \). This compares the sample mean \( \bar{x} \) with the hypothesized population mean \( 10 \), scaled by the standard error \( 0.1 \).
• When the calculated z-score like \( -2 \) (for \( \mu = 9.8 \)) or \( -5 \) (for \( \mu = 9.5 \)) is derived, you can match these against z-table values to obtain probabilities.
• This supports the evaluation of hypothesis tests and potential error rates (\( \alpha \) and \( \beta \)).

Using z-scores ensures that statistical conclusions drawn have a known probability backing.