91Ó°ÊÓ

An article in the Nov. 11,2005 , issue of the San Luis Obispo Tribune reported that researchers making random purchases at California Wal-Mart stores found scanners coming up with the wrong price \(8.3 \%\) of the time. Suppose this was based on 200 purchases. The National Institute for Standards and Technology says that in the long run at most two out of every 100 items should have incorrectly scanned prices. a. Develop a test procedure with a significance level of (approximately) .05, and then carry out the test to decide whether the NIST benchmark is not satisfied. b. For the test procedure you employed in (a), what is the probability of deciding that the NIST benchmark has been satisfied when in fact the mistake rate is \(5 \%\) ?

Step by step solution

01

State the Hypotheses

We need to decide upon the null and alternative hypotheses. The null hypothesis, \( H_0 \), is that the proportion of incorrect prices is \( p = 0.02 \). The alternative hypothesis, \( H_a \), is that the proportion of incorrect prices is greater than 0.02, \( p > 0.02 \).

02

Identify the Significance Level

According to the problem, we aim for a significance level of \( \alpha = 0.05 \). This indicates that we have a 5% risk of rejecting the null hypothesis if it is actually true.

03

Determine the Test Statistic

We use a one-sample Z-test for proportions to determine the test statistic. The formula for the test statistic is \( Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} \), where \( \hat{p} = \frac{x}{n} \) is the sample proportion with \( x = 8.3\% \times 200 = 16.6 \), \( n = 200 \), and \( p_0 = 0.02 \).

04

Compute the Test Statistic

First, calculate the sample proportion: \( \hat{p} = \frac{16.6}{200} = 0.083 \). Then substitute into the Z formula: \[ Z = \frac{0.083 - 0.02}{\sqrt{\frac{0.02 \cdot 0.98}{200}}} = \frac{0.063}{\sqrt{0.000098}} = \frac{0.063}{0.009899} \approx 6.364 \]

05

Make a Decision

Compare the computed Z-value with the critical value for \( \alpha = 0.05 \). From Z-tables, the critical value for a one-tailed test at \( \alpha = 0.05 \) is approximately 1.645. Since 6.364 > 1.645, we reject the null hypothesis.

06

Conclusion for Part (a)

The test statistic indicates that we reject the null hypothesis. Therefore, there is significant evidence to suggest that the NIST benchmark of 2% is not satisfied.

07

Calculate the Probability in Part (b)

For Part (b), we need the probability of a Type II error (failing to reject \( H_0 \) when the true proportion is 5%). The formula for \( \beta \) (probability of Type II error) is based on \( Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} \) with \( \hat{p} = 0.05 \). First find \( Z \) at \( \hat{p} = 0.05 \):\[ Z = \frac{0.05 - 0.02}{\sqrt{\frac{0.02 \cdot 0.98}{200}}} = \frac{0.03}{0.009899} \approx 3.030 \]Check this Z-value against the critical value 1.645: \( P(Z > 1.645 | \hat{p} = 0.05) \). Calculate \( P(Z < 3.030) - P(Z < 1.645) \). It's about 0.9987 - 0.9505 \approx 0.0482.

08

Conclusion for Part (b)

The probability of deciding that the NIST benchmark is satisfied when the true mistake rate is 5% is approximately 4.82%.

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Key Concepts

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

Null Hypothesis

The null hypothesis is the foundation of hypothesis testing. It serves as the baseline assumption that there is no effect or difference in the context we are examining.
In this problem, our null hypothesis, denoted as \( H_0 \), is that the proportion of incorrect prices at Wal-Mart is \( p = 0.02 \), aligning with the NIST scale.
This means we begin our test with the belief that Wal-Mart's incorrect pricing follows the industry benchmark of 2%.
Testing the null hypothesis involves determining whether there is enough statistical evidence to reject this initial assumption.

Alternative Hypothesis

The alternative hypothesis challenges the null hypothesis. It represents the outcome we want to test, suggesting there is a significant effect or difference.
In this case, our alternative hypothesis, denoted as \( H_a \), is that the true proportion of incorrect prices is greater than 0.02, or \( p > 0.02 \).
This hypothesis supports the notion that Wal-Mart's error rate exceeds the expected industry standard, indicating a potential problem. The goal of hypothesis testing is to determine if data provides strong enough evidence to support this alternative.

Significance Level

The significance level, represented by \( \alpha \), sets the threshold for determining when to reject the null hypothesis.
It quantifies the risk of making a Type I error - rejecting \( H_0 \) when it is actually true.
In our problem, we use a significance level of 0.05, meaning we accept a 5% risk of incorrectly dismissing the null hypothesis. This value indicates the level of evidence required to claim that the proportion of incorrect prices is significantly different from 0.02.
A lower significance level would require stronger evidence to reject \( H_0 \).
Understanding significance helps in interpreting the reliability of test results.

Z-Test for Proportions

A Z-test for proportions is a statistical method used to determine if there is a significant difference between an observed sample proportion and a hypothesized population proportion.
This test is suitable for our exercise because we are examining whether the proportion of incorrect prices at Wal-Mart exceeds the given benchmark.
To conduct this test, a test statistic \( Z \) is calculated using the formula:

• \( Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} \)
• \( \hat{p} \) is the sample proportion, \( p_0 \) is the null hypothesis proportion, and \( n \) is the total number of observations.

In the exercise, after computing, we find a Z-value that is compared to the critical value from a Z-distribution table.
If the calculated Z-value exceeds this critical threshold, it implies rejecting \( H_0 \). This method helps in determining statistical significance for proportions.

Type II Error

A Type II error occurs when we fail to reject the null hypothesis when the alternative hypothesis is true.
In other words, we wrongly conclude there's insufficient evidence to support \( H_a \), even though it is correct.
In our problem context, a Type II error implies deciding that the NIST benchmark is met when actually the error rate at Wal-Mart is higher (e.g., 5%).
We quantitatively evaluate this error using \( \beta \), the probability of making a Type II error, which is calculated for specific conditions assumed in \( H_a \).
Assessing \( \beta \) helps understand the risk of overlooking a meaningful effect, guiding towards balancing risks in hypothesis testing.