Concept explainers
Dosage Calculations. In Exercises 21–26, use the data in the accompanying table, which lists the numbers of correct and wrong dosage amounts calculated by physicians. In a research experiment, one group of physicians was given bottles of epinephrine labeled with a concentration of “1 milligram in 1 milliliter solution,” and another group of physicians was given bottles labeled with a ratio of “1 milliliter of a 1:1000 rotation.” The two labels describe the exact same amount, and the physicians were instructed to administer 0.12 milligrams of epinephrine. The results were reported in The New York Times.
Correct Dosage Calculation | Wrong Dosage Calculation | |
Concentration Label (“1 milligram in 1 milliliter solution”) |
11 | 3 |
Ratio Label (“1 milliliter of a 1:1000 solution”) |
2 | 12 |
23 Correct or Concentration If one of the physicians is randomly selected, find the
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Elementary Statistics
- A case-control (or retrospective) study was conducted to investigate a relationship between the colors of helmets worn by motorcycle drivers and whether they are injured or killed in a crash. Results are given in the accompanying table. Using a 0.01 significance level, test the claim that injuries are independent of helmet color. Color of Helmet Black White Yellow Red Blue Controls (not injured) 499 373 32 159 79 Cases (injured 221 108 8 66 38 or killed) Click here to view the chi-square distribution table. 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 C. Ho: Injuries and neimet color are dependent H₁: Injuries and helmet color are independent D. Ho: Whether a crash occurs and helmet color are dependent 1 0.001 0.004 0.016 2.706 3.841 5.024 6.635 7.879 2 0.010 0.020 0.051 0.103 0.211 4.605 5.991 7.378 9.210 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…arrow_forwardConduct 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_forward
- A 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_forwardAn 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_forward
- Can 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_forwardThe 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_forward
- Consider 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_forwardplease solve this problem step by step and make it quick pleasearrow_forwardWHAT IS THE CORRECT ANSWER AND WHY?arrow_forward
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