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A university library ordinarily has a complete shelf inventory done once every year. Because of new shelving rules instituted the previous year, the head librarian believes it may be possible to save money by postponing the inventory. The librarian decides to select at random 1000 books from the library's collection and have them searched in a preliminary manner. If evidence indicates strongly that the true proportion of misshelved or unlocatable books is less than \(.02\), then the inventory will be postponed. a. Among the 1000 books searched, 15 were misshelved or unlocatable. Test the relevant hypotheses and advise the librarian what to do (use \(\alpha=.05\) ). b. If the true proportion of misshelved and lost books is actually .01, what is the probability that the inventory will be (unnecessarily) taken? c. If the true proportion is \(.05\), what is the probability that the inventory will be postponed?

Short Answer

Expert verified
a: Do not postpone the inventory. b: 0.943. c: Essentially 0.

Step by step solution

01

Define Hypotheses

We need to set up the null and alternative hypotheses. The null hypothesis, denoted as \( H_0 \), is that the true proportion of misshelved books, \( p \), is greater than or equal to 0.02. The alternative hypothesis, \( H_a \), is that \( p < 0.02 \). These are represented as:\[ H_0: p \geq 0.02 \]\[ H_a: p < 0.02 \]
02

Calculate Sample Proportion

The sample size \( n \) is 1000, and the number of misshelved books is 15. The sample proportion \( \hat{p} \) is calculated by dividing the number of misshelved books by the total sample size: \[ \hat{p} = \frac{15}{1000} = 0.015 \]
03

Use a Z-Test for Proportions

We will use a Z-test for the proportion to determine if \( \hat{p} \) significantly differs from the hypothesized proportion. The test statistic \( Z \) is calculated as follows:\[ Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0 (1-p_0)}{n}}} \]where \( p_0 = 0.02 \). Substituting the values, we have:\[ Z = \frac{0.015 - 0.02}{\sqrt{\frac{0.02 \times (1-0.02)}{1000}}} \]
04

Calculate Z-Value

Calculate the Z-value from the formula in Step 3:\[ Z = \frac{-0.005}{\sqrt{\frac{0.02 \times 0.98}{1000}}} = \frac{-0.005}{0.004427} = -1.1294 \]
05

Determine Critical Value and Conclusion

Using the significance level \( \alpha = 0.05 \), the critical Z-value for a one-tailed test is approximately -1.645. Since the computed \( Z \) value of -1.1294 is greater than -1.645, we fail to reject the null hypothesis. Therefore, there is not sufficient evidence to suggest that the true proportion is less than 0.02, so the inventory should not be postponed.
06

Calculate Type II Error Probability (Part b)

If the true proportion is 0.01 and we need the probability that the inventory will unnecessarily be taken (Type II error), we recalculate the \( Z \) with \( p = 0.01 \) and the same decision limit (critical Z-value): \[ Z = \frac{0.015 - 0.01}{0.003146} = 1.5873 \]Now find the probability, which is 1 minus the cumulative distribution function of Z, i.e., the probability that the Z calculated is more extreme than this, approximately 0.943 (for Z = 1.5873). So, the probability that the inventory will be unnecessarily taken is about 0.943.
07

Calculate Probability of Postponing (Part c)

If the true proportion is 0.05, calculate the probability that the Z is less than -1.645 (our test in Step 5): \[ Z = \frac{0.015 - 0.05}{0.006892} = -5.0837 \]This corresponds to a very low probability, essentially zero, indicating that if the true proportion is really 0.05, the probability of incorrectly postponing the inventory is practically 0.

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Key Concepts

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

Z-test for Proportions
The Z-test for proportions is a statistical method used to determine whether there is a significant difference between an observed sample proportion and a hypothesized population proportion. This test is particularly useful when dealing with categorical data, such as the proportion of items in specific categories. In this context, we are interested in testing whether the proportion of misshelved books is below a certain threshold.

The calculation involves determining a Z-score with the formula:\[Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} \]where \( \hat{p} \) is the observed sample proportion, \( p_0 \) is the hypothesized population proportion, and \( n \) is the sample size. This formula helps us measure how far the sample proportion is from the hypothesized one, adjusted for the number of observations.

If the calculated Z-score falls within the critical region, defined by the significance level, we reject the null hypothesis in favor of the alternative. Essentially, this method gives a statistical basis for deciding on actions based on sample data.
Type I and Type II Errors
In hypothesis testing, Type I and Type II errors are potential mistakes that can occur. Understanding these errors is crucial when making decisions based on statistical tests.

  • **Type I Error:** Occurs if we reject the null hypothesis when it is actually true. In this exercise, it would mean deciding not to conduct the inventory when the proportion of misshelved books is actually above the acceptable threshold. The probability of this error is denoted by \( \alpha \), known as the significance level, which is set at 0.05.
  • **Type II Error:** Happens if we fail to reject the null hypothesis when the alternative hypothesis is true. This would be the case if we went ahead with the inventory when the true proportion was below 0.02. The chance of this error, denoted by \( \beta \), is related to the power of the test.

Effective hypothesis testing requires balancing these risks, often by selecting an appropriate \( \alpha \) level that minimizes the chance of a Type I error while being mindful of the implications of a Type II error.
Critical Values
Critical values are thresholds in hypothesis testing that determine the boundary of rejecting the null hypothesis. They mark the point where the test statistic's value leads us to accept or reject our hypothesis.

For a Z-test, critical values are derived from the standard normal distribution. They correspond to the significance level \( \alpha \), which is the probability of committing a Type I error. In this case, for a one-tailed test with \( \alpha = 0.05 \), the critical value is approximately -1.645.

This means that if the calculated Z-statistic is smaller than -1.645, it falls into the critical region, leading us to reject the null hypothesis. However, if it is greater, as in this exercise (with a Z of -1.1294), we do not have enough evidence to reject it, thus maintaining the original hypothesis.
Null and Alternative Hypotheses
Setting up null and alternative hypotheses is the first step in hypothesis testing. These hypotheses are foundational elements that help clarify the research question and the statistical test's direction.

  • **Null Hypothesis (\(H_0\)):** Represents the status quo or a statement of no effect. In this scenario, \( H_0 \) asserts that the proportion of misshelved books is at least 0.02, suggesting no need to postpone the inventory.
  • **Alternative Hypothesis (\(H_a\)):** This is what we aim to support through our test; it indicates a new effect or difference. Here, \( H_a \) posits that the proportion of misshelved books is less than 0.02, supporting the potential postponement of inventory.

The choice between these hypotheses directs the statistical analysis. It sets the stage for evaluating evidence from the data, testing if the observed sample supports a shift from the null towards the alternative.

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Most popular questions from this chapter

The recommended daily dietary allowance for zinc among males older than age 50 years is \(15 \mathrm{mg} /\) day. The article "Nutrient Intakes and Dietary Patterns of Older Americans: A National Study" (J. Gerontology, 1992: M145-150) reports the following summary data on intake for a sample of males age \(65-74\) years: \(n=115, \bar{x}=11.3\), and \(s=6.43\). Does this data indicate that average daily zinc intake in the population of all males age 65-74 falls below the recommended allowance?

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