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Chapter 9: Problem 24

For each of the following significance levels, what is the probability of making a Type I error? a. \(\alpha=.025\) b. \(\alpha=.05\) c. \(\alpha=.01\)

Short Answer

Expert verified
The probability of making a Type I error for a) \(\alpha = .025\) is .025, b) \(\alpha = .05\) is .05, and c) \(\alpha = .01\) is .01.

Step by step solution

01

Understand the Significance Level

Note that \(\alpha\) is the symbol for the significance level, which is the probability of making a Type I error. A Type I error occurs when a true null hypothesis is rejected. So the task is essentially giving the probability of making a Type I error directly as the value of alpha in each part.
02

Identify the Probability for \(\alpha = .025\)

For \(\alpha = .025\), the probability of making a Type I error is .025.
03

Identify the Probability for \(\alpha = .05\)

For \(\alpha = .05\), the probability of making a Type I error is .05.
04

Identify the Probability for \(\alpha = .01\)

For \(\alpha = .01\), the probability of making a Type I error is .01.

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Most popular questions from this chapter

Briefly explain the conditions that must hold true to use the \(t\) distribution to make a test of hypothesis about the population mean.

A 2008 AARP survey reported that \(85 \%\) of U.S. workers aged 50 years and older with at least one 4-year college degree had taken employer-based training within the previous 2 years, compared to only \(50 \%\) of workers aged 50 years and older with a high school degree or less. In a current survey of \(640 \mathrm{U.S}\). workers aged 50 years and older with a high school degree or less, 341 had taken employer-based training within the previous 2 years. a. Using the critical-value approach and \(\alpha=.05\), test whether the current percentage of all U.S. workers aged 50 years and older with a high school degree or less who have taken employerbased training within the previous 2 years is different from \(50 \%\). b. How do you explain the Type I error in part a? What is the probability of making this error in part a? c. Calculate the \(p\) -value for the test of part a. What is your conclusion if \(\alpha=.05 ?\)

A food company is planning to market a new type of frozen yogurt. However, before marketing this yogurt, the company wants to find what percentage of the people like it. The company's management has decided that it will market this yogurt only if at least \(35 \%\) of the people like it. The company's research department selected a random sample of 400 persons and asked them to taste this yogurt. Of these 400 persons, 112 said they liked it. a. Testing at the \(2.5 \%\) significance level, can you conclude that the company should market this yogurt? b. What will your decision be in part a if the probability of making a Type I error is zero? Explain. c. Make the test of part a using the \(p\) -value approach and \(\alpha=.025\).

The mean balance of all checking accounts at a bank on December 31,2009, was \(\$ 850 .\) A random sample of 55 checking accounts taken recently from this bank gave a mean balance of \(\$ 780\) with a standard deviation of \(\$ 230 .\) Using the \(1 \%\) significance level, can you conclude that the mean balance of such accounts has decreased during this period? Explain your conclusion in words. What if \(\alpha=.025\) ?

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 \(\mu .\) )
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