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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 \(5 \%\) ?

Short Answer

Expert verified
The NIST benchmark is not satisfied. Probability of Type II error is 0.9956.

Step by step solution

01

Formulate Hypotheses

We need to test if the mistake rate exceeds the NIST benchmark of 2%. Define the null hypothesis \( H_0 \) that the true mistake rate \( p = 0.02 \). The alternative hypothesis \( H_a \) is that \( p > 0.02 \), indicating a greater mistake rate.
02

Determine Test Statistic

Since the sample size \( n = 200 \) is large enough, we use the normal approximation to the binomial distribution. The test statistic is \( Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} \), where \( \hat{p} = 0.083, \ p_0 = 0.02, \) and \( n = 200 \).
03

Calculate Test Statistic

Substitute \( \hat{p} = 0.083 \), \( p_0 = 0.02 \), and \( n = 200 \) into the formula. \[ Z = \frac{0.083 - 0.02}{\sqrt{\frac{0.02 \times 0.98}{200}}} \approx 4.71 \]
04

Determine Critical Value

For a significance level of 0.05 in a right-tailed test, the critical Z value is approximately 1.645.
05

Compare Test Statistic with Critical Value

Since our calculated Z value of 4.71 is greater than the critical value of 1.645, we reject the null hypothesis \( H_0 \). The data suggests the NIST benchmark is not satisfied.
06

Calculate Probability (Type II Error)

Assuming the true mistake rate is 5% when using the test, calculate the probability of failing to reject \( H_0 \). First, update the test statistic formula with \( p = 0.05 \): \[ Z = \frac{0.05 - 0.02}{\sqrt{\frac{0.02\times0.98}{200}}} \approx 2.57 \]. Then find \( P(Z < 1.645) \) for normal distribution when the test statistic is recalculated for \( p = 0.05 \). This probability is approximately 0.9956.

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Key Concepts

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

Understanding Significance Level
In hypothesis testing, the significance level is a critical concept that determines the threshold at which you decide whether a sample result is statistically significant. It is denoted by the Greek letter \( \alpha \) and usually set at values such as 0.05, 0.01, or 0.10. The significance level indicates the probability of rejecting the null hypothesis when it is actually true.

In the context of our exercise, we used a significance level of 0.05. This means we were willing to accept a 5% chance of incorrectly rejecting the null hypothesis, which claims the error rate of scanned prices is 2%.

- A lower significance level often means you're being more conservative and less likely to make a Type I error (false positive).- Hypothesis tests can be one-tailed (checking for an effect in one direction) or two-tailed (checking both directions). Here, we conducted a right-tailed test, hypothesizing the error rate is greater than 2%, aligning with the direction of our alternative hypothesis.
Normal Approximation for Binomial Distribution
Normal approximation is a method used to simplify calculations in statistics, especially for hypothesis testing. When dealing with a large sample size, like our 200 purchase records, it is common to approximate the binomial distribution with a normal distribution.

This approximation is based on two key conditions:
  • The number of trials (\( n \)) is large enough.
  • The expected number of successes and failures (\( np \) and \( n(1-p) \)) are both greater than or equal to 5.
In our exercise, since we are dealing with a proportion (\( ychside \)), we approximated the test statistic using the normal distribution’s Z-score formula:\(Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}\).
This calculation provided a way to compare observed results to what is generally expected, determining significance.
Exploring Type II Error
A Type II error occurs in hypothesis testing when we fail to reject the null hypothesis, even though the alternative hypothesis is true. This is often denoted as \( \beta \) and demonstrates the test’s power, representing the probability of missing a real effect.

In our example, we calculated the probability of making a Type II error given that the true mistake rate is 5% and the tested rate is still assumed to be 2%. We adjusted the formula for our test statistic, recalculating with the new assumed rate:
\(Z = \frac{0.05 - 0.02}{\sqrt{\frac{0.02\times0.98}{200}}}\),and explored the area under the normal curve up to the critical value from this Z-score.

- A high Type II error probability suggests low test power, indicating the test might not be sensitive enough to detect effects if they exist in the accessed population.- Reducing Type II errors often involves increasing the sample size or selecting a significance level that effectively balances Type I and Type II risks.

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Most popular questions from this chapter

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