For Problems 3–11, please provide the following information.
- (a) What is the level of significance? State the null and alternate hypotheses.
- (b) Find the value of the chi-square statistic for the sample. What are the degrees of freedom? What assumptions are you making about the original distribution?
- (c) Find or estimate the P-value of the sample test statistic.
- (d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis of independence?
- (e) Interpret your conclusion in the context of the application.
- (f) Find the requested confidence interval for the population variance or population standard deviation. Interpret the results in the context of the application.
In each of the following problems, assume a
Mountain Climbing: Accidents The following problem is based on information taken from Accidents in North American Mountaineering (jointly published by The American Alpine Club and The Alpine Club of Canada). Let x represent the number of mountain climbers killed each year. The long-term variance of x is approximately σ2 = 136.2. Suppose that for the past 8 years, the variance has been s2 = 115.1. Use a 1% level of significance to test the claim that the recent variance for number of mountain climber deaths is less than 136.2. Find a 90% confidence interval for the population variance.
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Chapter 10 Solutions
Bundle: Understandable Statistics, Loose-leaf Version, 12th + WebAssign Printed Access Card for Brase/Brase's Understandable Statistics: Concepts and Methods, 12th Edition, Single-Term
- Conduct the hypothesis test and provide the test statistic and the critical value, and state the conclusion. A person drilled a hole in a die and filled it with a lead weight, then proceeded to roll it 200 times. Here are the observed frequencies for the outcomes of 1, 2, 3, 4, 5, and 6, respectively: 28, 32, 46, 39, 29, 26. Use a 0.025 significance level to test the claim that the outcomes are not equally likely. Does it appear that the loaded die behaves differently than a fair die? Click here to view the chi-square distribution table. The test statistic is (Round to three decimal places as needed.) Chi-square distribution table Area to the Right of the Critical Value Degrees of Freedom 0.995 0.99 0.975 0.95 0.90 0.10 0.05 0.025 0.01 0.005 1 0.001 0.004 0.016 2.706 3.841 5.024 6.635 2 0.010 0.020 0.051 0.103 0.211 4.605 5.991 7.378 9.210 7.879 10.597 3 0.072 0.115 0.216 0.352 0.584 6.251 7.815 9.348 11.345 12.838 4 0.207 0.297 0.484 0.711 1.064 7.779 9.488 11.143 13.277 14.860 5…arrow_forwardThe online clothing retailer e-Parel is conducting a study to estimate the average size of the orders placed by visitors to its website. The project manager desires a $60 bound on the error of estimation at 90% confidence. The population standard deviation is unknown, and a “best guess” of $175 is used as the planning value for σ. Use the Distributions tool to help you answer the questions that follow. 0123 Select a Distribution The z-value for a 90% confidence interval of the population mean is . In order to satisfy the requirement of a $60 bound on the error of estimation, a sample size no smaller than is needed.arrow_forwardA local electronics store just received a shipment of 620 HDMI cables. The manager wants to estimate the number of defective HDMI cables in the shipment. Rather than checking every HDMI cable, the manager plans to take a simple random sample of size 62 in order to estimate the proportion of defective HDMI cables in the shipment. If the sample proportion of defective HDMI cables, p̂p̂, is greater than 0.0323 (there are more than two defective HDMI cables in the sample), the manager will file a complaint and request a new shipment. Suppose that the true proportion of defective HDMI cables in the shipment is approximately p = 0.02. What is the expected value of the sample proportion? E(Pˆ)E(P^)= Since the sample is to be drawn from a finite population, and since the sample is 5% of the population size, the finite population correction factor needed when you calculate the standard deviation of the sampling distribution. What is the standard deviation of the…arrow_forward
- An automobile battery manufacturer offers a 39/50 warranty on its batteries. The first number in the warranty code is the free-replacement period; the second number is the prorated-credit period. Under this warranty, if a battery fails within 39 months of purchase, the manufacturer replaces the battery at no charge to the consumer. If the battery fails after 39 months but within 50 months, the manufacturer provides a prorated credit toward the purchase of a new battery. The manufacturer assumes that X, the lifetime of its auto batteries, is normally distributed with a mean of 44 months and a standard deviation of 3.6 months. Use the following Distributions tool to help you answer the questions that follow. (Hint: When you adjust the parameters of a distribution, you must reposition the vertical line (or lines) for the correct areas to be displayed.) 0123 Select a Distribution If the manufacturer’s assumptions are correct, it would need to replace of its…arrow_forwardIn regards to conducting a linear contrast after a one-way ANOVA, can you explain how seemingly arbitrary weights that "emphasize or de-emphasize" certain variables in a linear combination and sum to zero are able to provide information about how certain groups differ from each other? For example, if we havethree groups A, B, and C, and we want tocompare the mean of group A with theaverage of groups B and C, the weights inthis case are 1 for group A, and -0.5 for groupsB and C, which sum to zero. But how do these numbers model the relationship of comparing one group to the average of the other two? Does it have to do with how the math is carried out, such as how the test statistic is created?arrow_forwardCan you simply and intuitively explain the purpose of a contrast to the treatment sum of squares? For example, do orthogonal contrasts partition the treatment sum of squares into additive components that represent the variation due to each contrast? If so, what would be the purpose of this?arrow_forward
- The height of the graph of the probability density function f(x) varies with X as follows (round to four decimal places): X 16 Height of the Graph of the Probability Density Function You are flying out of Terminal 3 at JFK on a Wednesday afternoon between 3:00 and 4:00 PM. You get stuck in a traffic jam on the way to the airport, and if it takes you longer than 12 minutes to clear security, you'll miss your flight. The probability that you'll miss your flight is You have arrived at the airport and have been waiting 10 minutes at the security checkpoint. Recall that if you spend more than 12 minutes clearing security, you will miss your flight. Now what is the probability that you'll miss your flight? ○ 0.5 O 0.25 ○ 0.8333 ○ 0.6667arrow_forwardonsider a random variable x that follows a uniform distribution, with a = 2 and b = 9. What is the probability that x is less than 6? P(x < 6) = 0.2857 P(x < 6) = 0.5714 P(x < 6) = 0.17142 P(x < 6) = 0.4286 What is the probability that x is between 4 and 6? P(4 ≤ x ≤ 6) = 0.2857 P(4 ≤ x ≤ 6) = 0.157135 P(4 ≤ x ≤ 6) = 0.0928525 P(4 ≤ x ≤ 6) = 0.11428arrow_forwardConsider a random variable x that follows a uniform distribution, with a = 8 and b = 14. What is the probability that x is less than 13? P(x < 13) = 0.1667 P(x < 13) = 0.41665 P(x < 13) = 0.24999 P(x < 13) = 0.8333 What is the probability that x is between 11 and 12? P(11 ≤ x ≤ 12) = 0.0541775 P(11 ≤ x ≤ 12) = 0.1667 P(11 ≤ x ≤ 12) = 0.06668 P(11 ≤ x ≤ 12) = 0.091685arrow_forward
- please solve this problem step by step and make it quick pleasearrow_forwardWHAT IS THE CORRECT ANSWER AND WHY?arrow_forwardA common way for two people to settle a frivolous dispute is to play a game of rock-paper-scissors. In this game, each person simultaneously displays a hand signal to indicate a rock, a piece of paper, or a pair of scissors. Rock beats scissors, scissors beats paper, and paper beats rock. If both players select the same hand signal, the game results in a tie. Two roommates, roommate A and roommate B, are expecting company and are arguing over who should have to wash the dishes before the company arrives. Roommate A suggests a game of rock-paper-scissors to settle the dispute. Consider the game of rock-paper-scissors to be an experiment. In the long run, roommate A chooses rock 21% of the time, and roommate B chooses rock 61% of the time; roommate A selects paper 39% of the time, and roommate B selects paper 21% of the time; roommate A chooses scissors 40% of the time, and roommate B chooses scissors 18% of the time. (These choices are made randomly and independently of each…arrow_forward
- Glencoe Algebra 1, Student Edition, 9780079039897...AlgebraISBN:9780079039897Author:CarterPublisher:McGraw HillCollege Algebra (MindTap Course List)AlgebraISBN:9781305652231Author:R. David Gustafson, Jeff HughesPublisher:Cengage Learning