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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

In a 2009 nonscientific poll on the Web site of the Daily Gazette of Schenectady, New York, readers were asked the following question: "Are you less inclined to buy a General Motors or Chrysler vehicle now that they have filed for bankruptcy?" Of the respondents, \(56.1 \%\) answered "Yes" (http://www. dailygazette.com/polls/2009/jun/Bankruptcy/). In a recent survey of 1200 adult Americans who were asked the same question, 615 answered "Yes." Can you reject the null hypothesis at the \(1 \%\) significance level in favor of the alternative that the percentage of all adult Americans who are less inclined to buy a General Motors or Chrysler vehicle since the companies filed for bankruptcy is different from \(56.1 \%\) ? Use both the \(p\) -value and the critical-value approaches.

The manager of a restaurant in a large city claims that waiters working in all restaurants in his city earn an average of \(\$ 150\) or more in tips per week. A random sample of 25 waiters selected from restaurants of this city yielded a mean of \(\$ 139\) in tips per week with a standard deviation of \(\$ 28\). Assume that the weekly tips for all waiters in this city have a normal distribution. a. Using the \(1 \%\) significance level, can you conclude that the manager's claim is true? Use both approaches. b. What is the Type I error in this exercise? Explain. What is the probability of making such an error?

The past records of a supermarket show that its customers spend an average of \(\$ 95\) per visit at this store. Recently the management of the store initiated a promotional campaign according to which each customer receives points based on the total money spent at the store, and these points can be used to buy products at the store. The management expects that as a result of this campaign, the customers should be encouraged to spend more money at the store. To check whether this is true, the manager of the store took a sample of 14 customers who visited the store. The following data give the money (in dollars) spent by these customers at this supermarket during their visits. \(\begin{array}{rrrrrrr}109.15 & 136.01 & 107.02 & 116.15 & 101.53 & 109.29 & 110.79 \\ 94.83 & 100.91 & 97.94 & 104.30 & 83.54 & 67.59 & 120.44\end{array}\) Assume that the money spent by all customers at this supermarket has a normal distribution. Using the \(5 \%\) significance level, can you conclude that the mean amount of money spent by all customers at this supermarket after the campaign was started is more than \(\$ 95 ?\) (Hint: First calculate the sample mean and the sample standard deviation for these data using the formulas learned in Sections \(3.1 .1\) and \(3.2 .2\) of Chapter \(3 .\) Then make the test of hypothesis about \(\mu .\) )

A company claims that the mean net weight of the contents of its All Taste cereal boxes is at least 18 ounces. Suppose you want to test whether or not the claim of the company is true. Explain briefly how you would conduct this test using a large sample. Assume that \(\sigma=.25\) ounce.

According to the American Diabetes Association (www.diabetes.org), \(23.1 \%\) of Americans aged 60 years or older had diabetes in 2007. A recent random sample of 200 Americans aged 60 years or older showed that 52 of them have diabetes. Using a \(5 \%\) significance level, perform a test of hypothesis to determine if the current percentage of Americans aged 60 years or older who have diabetes is higher than that in 2007 . Use both the \(p\) -value and the critical-value approaches.
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