/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 122 Acme Bicycle Company makes derai... [FREE SOLUTION] | 91影视

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Chapter 9: Problem 122

Acme Bicycle Company makes derailleurs for mountain bikes. Usually no more than \(4 \%\) of these parts are defective, but occasionally the machines that make them get out of adjustment and the rate of defectives exceeds \(4 \%\). To guard against this, the chief quality control inspector takes a random sample of 130 derailleurs each week and checks each one for defects. If too many of these parts are defective, the machines are shut down and adjusted. To decide how many parts must be defective to shut down the machines, the company's statistician has set up the hypothesis test $$ H_{0}: p \leq .04 \text { versus } H_{1}: p>.04 $$ where \(p\) is the proportion of defectives among all derailleurs being made currently. Rejection of \(H_{0}\) would call for shutting down the machines. For the inspector's convenience, the statistician would like the rejection region to have the form, "Reject \(H_{0}\) if the number of defective parts is \(C\) or more." Find the value of \(C\) that will make the significance level (approximately) \(.05\).

Short Answer

Expert verified
The value of C that makes the significance level approximately .05 is calculated by finding the nearest whole number greater than the observed value obtained from the z-score formula. This involves calculating the standard deviation under H0, identifying the critical z-score for a significance level of .05 in a one-tail test, and rearranging the z-score formula to find the critical value.

Step by step solution

01

Calculate the standard deviation under H0

Under the assumption of H0 being true, we calculate the standard deviation '蟽' of the proportion of defective parts using the formula \(蟽 = \sqrt{\frac{{p*(1-p)}}{n}}\), where 'p' is the proportion under H0 (0.04) and 'n' is the number of samples (130). This gives \(蟽 = \sqrt{\frac{{0.04*0.96}}{130}}\).
02

Identify the critical z-score

The next step is to find the critical z-score that corresponds to the significance level of .05 in a one-tailed test (as 'H1' is 'p > 0.04', we are considering a one-tailed test). The z-score corresponding to .05 significance level in a one-tail test is approximately 1.645.
03

Calculate the critical value

The critical value 'C' for the proportion 'p' is the value for which the z-score is 1.645. We can find 'C' by rearranging the z-score formula \(z = \frac{{\text{observed} - \text{expected}}}{\text{standard deviation}}\) as \(\text{observed} = \text{expected} + z * \text{standard deviation}\). Substituting 'expected' with the hypothesized proportion under H0 (0.04*130 = 5.2, which must be rounded down to 5), 'z' with 1.645 and '蟽' from Step 1, we should find 'C'.
04

Round to the nearest whole number

Because we can't have a fractional defective part, the value calculated in Step 3 must be round up to the nearest whole number, giving the value of 'C' as the smallest integer count of defective parts that leads to the rejection of 'H0'.

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Most popular questions from this chapter

According to the University of Wisconsin Dairy Marketing and Risk Management Program (http:// future.aae.wisc.edu/index.html), the average retail price of 1 gallon of whole milk in the United States for April 2009 was \(\$ 3.084\). A recent random sample of 80 retailers in the United States produced an average milk price of \(\$ 3.022\) per gallon, with a standard deviation of \(\$ .274 .\) Do the data provide significant evidence at the \(1 \%\) level to conclude that the current average price of 1 gallon of milk in the United States is lower than the April 2009 average of \(\$ 3.084\) ?

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.

Make the following hypothesis tests about \(p\). a. \(H_{0}: p=.45, \quad H_{1}: p \neq .45, \quad n=100, \quad \hat{p}=.49, \quad \alpha=.10\) b. \(H_{0}: p=.72, \quad H_{1}: p<.72, \quad n=700, \quad \hat{p}=.64, \quad \alpha=.05\) c. \(H_{0}: p=.30, \quad H_{1}: p>.30, \quad n=200, \quad \hat{p}=.33, \quad \alpha=.01\)

More and more people are abandoning national brand products and buying store brand products to save money. The president of a company that produces national brand coffee claims that \(40 \%\) of the people prefer to buy national brand coffee. A random sample of 700 people who buy coffee showed that 259 of them buy national brand coffee. Using \(\alpha=.01\), can you conclude that the percentage of people who buy national brand coffee is different from \(40 \%\) ? Use both approaches to make the test.

Consider \(H_{0}: \mu=45\) versus \(H_{1}: \mu<45\). a. A random sample of 25 observations produced a sample mean of \(41.8\). Using \(\alpha=.025\), would you reject the null hypothesis? The population is known to be normally distributed with \(\sigma=6\). b. Another random sample of 25 observations taken from the same population produced a sample mean of \(43.8\). Using \(\alpha=.025\), would you reject the null hypothesis? The population is known to be normally distributed with \(\sigma=6\).
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