The maximum allowable mean level of a toxin in smelters, herbicide production facilities, and other places where the toxin is used is 0.005 milligrams per cubic meter of air. Suppose smelters at two plants are being investigated to determine whether they are meeting standards. Two analyses of the air are made at each plant, and the results are shown below. A claim is made that the standard is being violated at Plant 2 but not at Plant 1. Do you agree? Assume α=0.05. Plant 1 Plant 2 Observation Toxin Level Observation Toxin Level 1 0.01 1 0.08 2 0.006 2 0.06 Find the test statistic for Plant 1. t=enter your response here (Round to two decimal places as needed.)

The maximum allowable mean level of a toxin in smelters, herbicide production facilities, and other places where the toxin is used is 0.005 milligrams per cubic meter of air. Suppose smelters at two plants are being investigated to determine whether they are meeting standards. Two analyses of the air are made at each plant, and the results are shown below. A claim is made that the standard is being violated at Plant 2 but not at Plant 1. Do you agree? Assume α=0.05. Plant 1 Plant 2 Observation Toxin Level Observation Toxin Level 1 0.01 1 0.08 2 0.006 2 0.06 Find the test statistic for Plant 1. t=enter your response here (Round to two decimal places as needed.)

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018

18th Edition

ISBN:9780079039897

Author:Carter

Publisher:Carter

Chapter4: Equations Of Linear Functions

Section: Chapter Questions

Problem 8SGR

See similar textbooks

Plant 1	Plant 2
Observation	Toxin Level	Observation	Toxin Level
1	0.01	1	0.08
2	0.006	2	0.06

Similar questions

For each of the questions 1-4, answer the following (a-h are steps in a Hypothesis Test, see page 376 of your ebook). For steps f and g do both the p-value method and the critical value method: a) Claim in English: b) Claim in Symbolic Form: c) Null and Alternate Hypothesis: d) Significance Level: e) Test Statistic: f) Find Values: g) Make a Decision: h) Restate the Decision in Nontechnical Terms: i) If this is a statistically significant result, is it also a practically significant result?
The "spring-like effect" in a golf club could be determined by measuring the coefficient of restitution (the ratio of the outbound velocity to the inbound velocity of a golf ball fired at the clubhead). Twelve randomly selected drivers produced by two clubmakers are tested and the coefficient of restitution measured. The data follow: Club 1: 0.8406, 0.8104, 0.8234, 0.8198, 0.8235, 0.8562, 0.8123, 0.7976, 0.8184, 0.8265, 0.7773, 0.7871 Club 2: 0.8305, 0.7905, 0.8352, 0.8380, 0.8145, 0.8465, 0.8244, 0.8014, 0.8309, 0.8405, 0.8256, 0.8476 Test the hypothesis that both brands of ball have equal mean overall distance. Use α = 0.05 and assume equal variances. Question: Reject H0 if t0 < ___ or if t0 > ___.
The National Institute of Standards and Technology (NIST) offers a wide variety of “standard reference materials” (SRMs) with accurately specified analyte concentrations. SRM 1951c is intended in part for “evaluating the accuracy of clinical procedures for the determination of total cholesterol . . . in human serum.” The certified total cholesterol is 6.244 ± 0.072 mmol/L. For the sake of this question, let’s treat 6.244 mmol/L as its “true” value (μ). We will use this standard to test a method with an intrinsic variability (for an experienced analyst) of σ = 0.131 mmol/L. 1. For what value of z does exactly 95% of the area under a Gaussian curve lie between ± z? Use this multiplier to find the range of measured cholesterol concentrations that would include 95% of the trials from the “parent population” (μ and σ) given above. 2. Calculate z for a trial in which the measured total cholesterol concentration is 6.302 mmol/L. 3. What is the probability that a trial will give a cholesterol…
The National Institute of Standards and Technology (NIST) offers a wide variety of “standard reference materials” (SRMs) with accurately specified analyte concentrations. SRM 1951c is intended in part for “evaluating the accuracy of clinical procedures for the determination of total cholesterol . . . in human serum.” The certified total cholesterol is 6.244 ± 0.072 mmol/L. For the sake of this question, let’s treat 6.244 mmol/L as its “true” value (μ). We will use this standard to test a method with an intrinsic variability (for an experienced analyst) of σ = 0.131 mmol/L. **Please show written work! It is difficult to understand "text" work/answers with mathematical/statistical problems** 1. For what value of z does exactly 95% of the area under a Gaussian curve lie between ± z? Use this multiplier to find the range of measured cholesterol concentrations that would include 95% of the trials from the “parent population” (μ and σ) given above. 2. Calculate z for a trial in which the…
The National Institute of Standards and Technology (NIST) offers a wide variety of “standard reference materials” (SRMs) with accurately specified analyte concentrations. SRM 1951c is intended in part for “evaluating the accuracy of clinical procedures for the determination of total cholesterol . . . in human serum.” The certified total cholesterol is 6.244 ± 0.072 mmol/L. For the sake of this question, let’s treat 6.244 mmol/L as its “true” value (μ). We will use this standard to test a method with an intrinsic variability (for an experienced analyst) of σ = 0.131 mmol/L. In three trials, you obtained the following cholesterol concentrations (all in mmol/L): 6.476, 6.590, 6.338. 1. Showing all work, calculate a 95% confidence interval for your measured cholesterol concentration and write it as an intermediate result (with guard digits). Does your measurement differ from μ at 95% confidence? 2. Repeat part 1 for 90% confidence and explain any change in your conclusion. 3. Another…
Sulfur compounds cause "off-odors" in wine, and winemakers want to know the odor threshold – the lowest concentration of a compound that the human nose can detect. The odor threshold for dimethyl sulfide (DMS) in trained wine tasters is about 25 micrograms per liter of wine (µg/L). The untrained noses of consumers may be less sensitive, however. A set of DMS odor threshold data for 10 untrained students was analyzed. How sensitive are the untrained noses of students? You want to estimate the mean DMS odor threshold among all students, and you would be satisfied to estimate the mean to within ±0.25 with 95% confidence. The standard deviation of the odor threshold for untrained noses is known to be o = 8 micrograms per liter of wine. How large an SRS of untrained students do you need? Give your answer rounded up to the nearest whole number. n = Incorrect
One operation of a mill is to cut pieces of steel into parts that will later be used as the frame for front seats in an automobile. The steel is cut with a diamond saw and requires the resulting parts to be within 10.005 inch of the length specified by the automobile company. Data are collected from a sample of 50 steel parts and are shown in the following table. The measurement reported is the difference in inches between the actual length of the steel part, as measured by a laser measurement device, and the specified length of the steel part. For example, the first value, -0.003, represents a steel part that is 0.003 inch shorter than the specified length. Complete parts a through c Click the icon to view the data table. a. Construct a frequency distribution Difference in Length -0.005 but less than -0.003: -0.003but less than -0.001 -0.001but less than 0.001 0.001but less than 0.003 0.003but less than 0.005 Frequency Difference Between Actual and Specified Lengths 0.002 0 -0.003…
Listed below are pulse rates (beats per minute) from samples of adult males and females. Does there appear to be adifference? Find the coefficient of variation for each of the two samples; then compare the variation. A. The coefficient of variation for the male pulse rates iS (Type an integer or decimal rounded to one decimal place as needed.) B. The coefficient of variation for the female pulse rates is (Type an integer or decimal rounded to one decimal place as needed.)
To investigate water quality, the Columbus Dispatch took water specimens at 16 Ohio State Park swimming areas in central Ohio. Those specimens were taken to laboratories and tested for E. coli, which are bacteria that can cause serious gastrointestinal problems. For reference, if a 100-milliliter specimen (about 3.3 ounces) of water contains more than 130 E. coli bacteria, it is considered unsafe. The E. coli levels per 100 milliliters found by the laboratories are presented in the table: 291.0 10.9 47.0 86.0 44.0 18.9 1.0 50.0 190.4 45.7 28.5 18.9 16.0 34.0 8.6 9.6 To access the complete data set, click the link for your preferred software format: Excel Minitab JMP SPSS TI R Mac-TXT PC-TXT CSV CrunchIt! Find the mean E. coli level. Mean E.coli level (rounded to two decimal places) per 100 ml
Two analysts were given a standard reference material (SRM) and were asked to determine its copper concentration (in mg/kg) using the EPA standard method. Each analyst was given sufficient time so they could produce as many accurate results as possible. Their results (mg/kg) were as follows: Analyst A 45.2, 47.3, 51.2, 50.4, 52.2, 48.7 (Standard error s = 2.62) Analyst B 49.4, 50.3, 51.6, 52.1, 50.9 (Standard error s = 1.06) (a) For Analyst A, is 52.2 mg/kg a possible outlier? (b) Calculate the RSD for each student; which analyst is more precise? (c) If the known concentration of copper in SRM is 49.5 mg/kg, which analyst is more accurate (without considering the deletion of outliers if any)?
1. Let μ denote the true average level of radioactivity of the water in a nuclear plant. Five (5) picocuries per liter is considered the dividing line between safe and unsafe water. a. State the null and alternate hypothesis you would test if you worked at the plant andexplain why you chose your alternate hypothesis.b. In the context of this situation, describe what happens if there is a type I error and type II error.
A sociologist wants to determine if the life expectancy of people in Africa is less than the life expectancy of people in Asia. The data obtained is shown in the table below. Africa Asia = 63.3 yr. 1 X,=65.2 yr. 2 o, = 9.1 yr. = 7.3 yr. n1 = 120 = 150