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Chapter 9: Problem 43

Records in a three-county area show that in the last few years, Girl Scouts sell an average of \(47.93\) boxes of cookies per year, with a population standard deviation of \(8.45\) boxes per year. Fifty randomly selected Girl Scouts from the region sold an average of \(46.54\) boxes this year. Scout leaders are concerned that the demand for Girl Scout cookies may have decreased. a. Test at the \(10 \%\) significance level whether the average number of boxes of cookies sold by all Girl Scouts in the three-county area is lower than the historical average. b. What will your decision be in part a if the probability of a Type I error is zero? Explain.

Short Answer

Expert verified
The solution requires you to first state the null and alternative hypotheses, compute the test statistic (z-value) using the given data, find the corresponding P-value, and then make a decision based on the P-value compared to the significance level. For part b, absolutes are rarely possible in statistics and there is always a risk of Type I error except in cases of complete certainty.

Step by step solution

01

Stating Hypotheses

First, clearly state the null hypothesis \((H_0)\) and the alternative hypothesis \((H_a)\). Here, we are testing if the average number of boxes sold is lower than the historical average. Therefore, \(H_0: \mu = 47.93\) and \(H_a: \mu < 47.93\). The null hypothesis assumes there is no change while the alternative hypothesis suggests a decrease.
02

Computing Test Statistic

Compute the test statistic using the formula for a single sample z statistic: \(z = \frac{(\bar{x} - \mu_0)}{\frac{s}{\sqrt{n}}}\). Here, \(\bar{x} = 46.54\), \(\mu_0 = 47.93\), \(s = 8.45\), and \(n = 50\). Plug in these values and calculate the z-value.
03

Finding the P-value

Once the z-statistic is computed, the P-value can be found from the standard normal (z) distribution table or using software tools.
04

Making Decision

If the computed P-value is less than the given significance level (\(0.10\)), reject the null hypothesis favoring the alternative hypothesis. If it's greater, fail to reject the null hypothesis.
05

Answering part b

If the probability of a Type I error is zero, it means there is no chance of incorrectly rejecting the null hypothesis when it is true. This would only happen if there was absolute certainty of a decrease (i.e., the P-value was exactly zero), which is practically impossible from a statistical perspective.

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Key Concepts

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

z-test
The z-test is a statistical procedure used to determine whether there is a significant difference between the mean of a sample and the mean of a population. It is particularly applicable when the population variance is known and the sample size is large (typically n > 30).
When conducting a z-test, you follow these steps:
  • State the null hypothesis (\( H_0 \)) and the alternative hypothesis (\( H_a \)).
  • Calculate the test statistic using the formula:\[ z = \frac{(\bar{x} - \mu_0)}{\frac{s}{\sqrt{n}}} \] where\( \bar{x} \) is the sample mean,\( \mu_0 \) is the population mean,\( s \) is the population standard deviation, and\( n \) is the sample size.
  • Determine the P-value or compare the z-statistic to critical values from the z-distribution.
p-value
The P-value is a measure of the strength of the evidence against the null hypothesis provided by the sample data. It is obtained after calculating the test statistic, such as the z-score in a z-test.
A low P-value indicates stronger evidence against the null hypothesis, suggesting that the observed data is unlikely under the assumption that the null hypothesis is true.
  • If the P-value is less than or equal to the pre-determined significance level (\( \alpha \)), you reject\( H_0 \).
  • If the P-value is greater, you fail to reject\( H_0 \).
The P-value helps quantify the probability of observing data as extreme as, or more extreme than, the observed data, assuming that the null hypothesis is true.
type I error
A Type I error occurs when a true null hypothesis is incorrectly rejected. In simpler terms, it's a false positive finding in statistical hypothesis testing - concluding there is an effect when none exists.
Type I error, denoted as \( \alpha \), is directly related to the significance level of a test:
  • Choosing a lower \( \alpha \) (like 0.01 instead of 0.05) reduces the chance of making this error but increases the risk of a Type II error (failing to reject a false null hypothesis).
  • The probability of a Type I error is equivalent to the significance level of the test.
The consequences of a Type I error depend on the context of the test and its important to balance minimizing this error while conducting hypothesis testing.
significance level
The significance level, denoted by \( \alpha \), is the probability of rejecting the null hypothesis when it is actually true. It sets the threshold for determining statistical significance in hypothesis testing.
Commonly used significance levels are 0.05, 0.01, and 0.10. If the P-value obtained from testing is less than \( \alpha \), the result is considered statistically significant.
  • Choosing \( \alpha = 0.10 \) means you are allowing a 10% risk of committing a Type I error.
  • The choice of significance level depends on the context and effect of potential errors.
A lower significance level means you require stronger evidence to reject the null hypothesis, making it more difficult to claim a significant effect when doing hypothesis testing.

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Most popular questions from this chapter

A May 8,2008 , report on National Public Radio (www.npr.org) noted that the average age of firsttime mothers in the United States is slightly higher than 25 years. Suppose that a recently taken random sample of 57 first-time mothers from Missouri produced an average age of \(23.90\) years and that the population standard deviation is known to be \(4.80\) years. a. Find the \(p\) -value for the test of hypothesis with the alternative hypothesis that the current mean age of all first-time mothers in Missouri is less than 25 years. Will you reject the null hypothesis at \(\alpha=.025\) ? b. Test the hypothesis of part a using the critical-value approach and \(\alpha=.025\).

A random sample of 14 observations taken from a population that is normally distributed produced a sample mean of \(212.37\) and a standard deviation of \(16.35 .\) Find the critical and observed values of \(t\) and the ranges for the \(p\) -value for each of the following tests of hypotheses, using \(\alpha=.10\). a. \(H_{0}: \mu=205\) versus \(H_{1}: \mu \neq 205\) b. \(H_{0}: \mu=205\) versus \(H_{1}: \mu>205\)

Make the following hypothesis tests about \(p\). a. \(H_{0}: p=.57, \quad H_{1}: p \neq .57, \quad n=800, \quad \hat{p}=.50, \quad \alpha=.05\) b. \(H_{0}: p=.26, \quad H_{1}: p<.26, \quad n=400, \quad \hat{p}=.23, \quad \alpha=.01\) c. \(H_{0}: p=.84, \quad H_{1}: p>.84, \quad n=250, \quad \hat{p}=.85, \quad \alpha=.025\)

As noted in U.S. Senate Resolution \(28,9.3 \%\) of Americans speak their native language and another language fluently (Source: www.actfl.org/i4a/pages/index.cfm?pageid=3782). Suppose that in a recent sample of 880 Americans, 69 speak their native language and another language fluently. Is there significant evidence at the \(10 \%\) significance level that the percentage of all Americans who speak their native language and another language fluently is different from \(9.3 \%\) ? Use both the \(p\) -value and the critical-value approaches.

Consider the following null and alternative hypotheses: $$ H_{0}: p=.82 \text { versus } H_{1}: p \neq .82 $$ A random sample of 600 observations taken from this population produced a sample proportion of \(.86 .\) a. If this test is made at the \(2 \%\) significance level, would you reject the null hypothesis? Use the critical-value approach. b. What is the probability of making a Type I error in part a? c. Calculate the \(p\) -value for the test. Based on this \(p\) -value, would you reject the null hypothesis if \(\alpha=.025 ?\) What if \(\alpha=.005 ?\)
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