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Chapter 10: Problem 45

Newly purchased automobile tires of a certain type are supposed to be filled to a pressure of 30 psi. Let \(\mu\) denote the true average pressure. Find the \(P\) -value associated with each of the following given \(z\) statistic values for testing \(H_{0}: \mu=30\) versus \(H_{a}: \mu \neq 30\) when \(\sigma\) is known: a. \(2.10\) d. \(1.44\) b. \(-1.75\) e. \(-5.00\) c. \(0.58\)

Short Answer

Expert verified
The corresponding P-values for the given Z-values are as follows: for Z=2.10 is 0.0358, for Z=1.44 is 0.1498, for Z=-1.75 is 1.9198, for Z=-5.00 is 2, and for Z=0.58 is 0.5620.

Step by step solution

01

Understanding the Terminology

The first step to solve this exercise is understanding the given values. Here, \(\mu\) is the true average pressure, \(\sigma\) is the standard deviation which is a known value. The aim is to test the null hypothesis \(H_{0}: \mu=30\) versus the alternative hypothesis \(H_{a}: \mu \neq 30\). For that, we are given z-statistic values and we need to find the corresponding P-values for each.
02

Calculating P-values using Z-table

The P-value can be calculated using a standard normal table (or Z-table), which gives the probabilities of various outcomes. For a two-tailed test, as it's in this case, the P-value is the probability that the z-score is greater than the absolute value of the given statistic, multiplied by 2. This is due to the symmetric property of the normal distribution.
03

Calculating P-value for Z=2.10

First, find the probability for Z=2.10 from the Z-table, which is 0.9821, then subtract it from 1 to get the probability which equals to 1-0.9821=0.0179. Since this is a two-tailed test, multiply this answer by 2. Therefore, P-value = 2*0.0179 = 0.0358 .
04

Calculating P-value for Z=1.44

From the Z-table, the probability for Z=1.44 is 0.9251. Subtract it from 1 to get the probability which equals to 1-0.9251=0.0749. Since it is a two-tailed test, multiply this answer by 2. Therefore, P-value = 2*0.0749 = 0.1498.
05

Calculating P-value for Z=-1.75

From the Z-table, the probability for Z=-1.75 is 0.0401. Subtract it from 1 to get the probability which equals to 1-0.0401=0.9599. Since it is a two-tailed test, multiply this answer by 2. Therefore, P-value = 2*0.9599 = 1.9198.
06

Calculating P-value for Z=-5.00

For Z=-5.00, the probability is very close to 0. However, subtract it from 1 to get the probability which equals to approximately 1. Since it is a two-tailed test, multiply this answer by 2. Therefore, P-value = 2*1 = 2.
07

Calculating P-value for Z=0.58

From the Z-table, the probability for Z=0.58 is 0.7190. Subtract it from 1 to get the probability which equals to 1-0.7190=0.2810. Since it is a two-tailed test, multiply this answer by 2. Therefore, P-value = 2*0.2810 = 0.5620.

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Most popular questions from this chapter

The authors of the article "Perceived Risks of Heart Disease and Cancer Among Cigarette Smokers" (Journal of the American Medical Association [1999]: \(1019-1021\) ) expressed the concem that a majority of smokers do not view themselves as being at increased risk of heart disease or cancer. A study of 737 current smokers selected at random from U.S. households with telephones found that of 737 smokers surveyed, 295 indicated that they believed they have a higher than average risk of cancer. Do these data suggest that \(\pi\), the true proportion of smokers who view themselves as being at increased risk of cancer is in fact less than .5, as claimed by the authors of the paper? Test the relevant hypotheses using \(\alpha=.05\).

The state of Georgia's HOPE scholarship program guarantees fully paid tuition to Georgia public universities for Georgia high school seniors who have a B average in academic requirements as long as they maintain a B average in college. Of 137 randomly selected students enrolling in the Ivan Allen College at the Georgia Institute of Technology (social science and humanities majors) in 1996 who had a B average going into college, \(53.2 \%\) had a GPA below \(3.0\) at the end of their first year ("Who Loses HOPE? Attrition from Georgia's College Scholarship Program," Southern Economic Journal [1999]: 379-390). Do these data provide convincing evidence that a majority of students at Ivan Allen College who enroll with a HOPE scholarship lose their scholarship?

The poll referenced in the previous exercise ("Military Draft Study," AP- Ipsos, June 2005) also included the following question: "If the military draft were reinstated, would you favor or oppose drafting women as well as men?" Forty-three percent of the 1000 people responding said that they would favor drafting women if the draft were reinstated. Using a \(.05\) significance level, carry out a test to determine if there is convincing evidence that fewer than half of adult Americans would favor the drafting of women.

In a study of computer use, 1000 randomly selected Canadian Internet users were asked how much time they spend using the Internet in a typical week (Ipsos Reid, August 9,2005 ). The mean of the 1000 resulting observations was \(12.7\) hours. a. The sample standard deviation was not reported, but suppose that it was 5 hours. Carry out a hypothesis test with a significance level of \(.05\) to decide if there is convincing evidence that the mean time spent using the Internet by Canadians is greater than \(12.5\) hours. b. Now suppose that the sample standard deviation was 2 hours. Carry out a hypothesis test with a significance level of . 05 to decide if there is convincing evidence that the mean time spent using the Internet by Canadians is greater than \(12.5\) hours.

Seat belts help prevent injuries in automobile accidents, but they certainly don't offer complete protection in extreme situations. A random sample of 319 front-seat occupants involved in head-on collisions in a certain region resulted in 95 people who sustained no injuries ("Influencing Factors on the Injury Severity of Restrained Front Seat Occupants in Car-to-Car Head-on Collisions," Accident Analysis and Prevention \([1995]: 143-150\) ). Does this suggest that the true (population) proportion of uninjured occupants exceeds .25? State and test the relevant hypotheses using a significance level of \(.05\).
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