It is specified that a certain type of iron should contain 0.85 g of silicon per 100 g of iron (0.85%). The silicon content of each of 26 randomly selected iron specimens was determined, and the accompanying Minitab outputresulted from a test of the appropriate hypotheses.VariableNMeansil cont260.8890StDev0.1809SE MeanT0.03551.10P0.28(a) What hypotheses were tested?Hoi u = 0.85Ha: μ < 0.85Ο Ηγ: μ = 0.85Ha μ > 0.85Ο Ηγ: μ = 0.85Ha: μ ± 0.85Ho: μ > 0.85Ha: μ = 0.85a'Ηρ: μ < 0.85H₂: M == 0.85(b) What conclusion would be reached for a significance level of 0.05, and why?Since the P-value is ---Select--- ✓ the significance level of 0.05, we would ---Select--- ✓ the null hypothesis.What conclusion would be reached for a significance level of 0.10, and why?Since the P-value is ---Select--- ☑ the significance level of 0.10, we would ---Select--- ✓ the null hypothesis.

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Answered: It is specified that a certain type of…

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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Cindy is a server at a local restaurant who makes, on average, $58 in tips per daytime shift. Cindy thinks she will make more if she moves to the nighttime shift. Her manager will choose five random nighttime shifts for Cindy to work. Cindy plans to conduct a significance test on the hypotheses H0: mu = 58; Ha: mu > 58 at the alpha = 0.05 level, where is the true mean amount of tips Cindy will make working nighttime shifts. Are the conditions for performing the test met? No, the 10% condition is not met. No, the sample size is too small, and a Normal distribution cannot be assumed. Yes, all conditions are met for performing a significance test about the mean. Yes, only the random condition must be met in order to perform a significance test. No, there is no way the manager could choose a random sample of shifts to work.
The average height of 18-year-old American women is 64.2 inches. You wonder whether the mean height of this year's female graduates from your local high school is different from the national average. You measure an SRS of 78 female graduates and find that x⎯⎯⎯x¯ = 63.1 inches. What are your null and alternative hypotheses? Note: Type in "mu" as the substitute for μμ and "!=" for ≠≠. H0:Ha:
A sample of 12 radon detectors of a certain type was selected, and each was exposed to 100 pCi/L of radon. The resulting readings were as follows: 105.6 90.9 91.2 96.9 96.5 91.3 100.1 105.5 99.6 107.7 103.3 92.4 Does this data suggest that the population mean reading under these conditions differ from 100? (a) State the null and alternative hypotheses. (b) Calculate, correct to 2 decimal places, (i) the sample mean, x ; and (ii) the sample standard deviation, s. (c) Evaluate the appropriate test statistic. (d) Test these hypotheses at   0.05 level of significance.
A report in LTO stated that the average age of taxis in the Philippines is 12 years. An operations manager of a large taxi company selects a sample of 40 taxis and finds the average of the taxis is 11.2 years. The o of the population is 2.3 age years. At a = of the taxis in his company is less than the national average? 0.05, can it be concluded that the average age %3D
The data in NutritionStudy include information on nutrition and health habits of a sample of 315 people. One of the variables is VitaminUse, indicating whether a person takes a multivitamin pill regularly or occasionally or not at all. Use technology to test whether the data provide evidence that the proportion taking a vitamin pill regularly is different from 35 % given. Clearly state the null and alternative hypotheses. : : 42 : Pi : P2 :: 0.35 : 2 :: P1 : p2 :: :: r Ho: vs Hg: ::
A researcher noticed that today's adolescents seem to have higher rates of social anxiety than those of generations past. In 1990, the average level of social anxiety in adolescents was μ = 15.5. The researcher collects data by sampling 100 adolescent students from the tristate area and records their level of anxiety. Testing at an alpha level of a = .05, what will the researcher conclude about today's adolescents? a.) Step 1: State the Hypotheses H₂: H₁: b.) Step 2: Determine the critical region and draw it on the distribution above: df = c.) Step 3: Calculate the test statistic the sample mean is 16.5 and SS = 0.25 M= SM= tobs d.) Step 4: Make a decision. Circle one: Reject Null (Ho) or e.) State your conclusions about the research scenario in sentence form: tcritical = 72²= Fail to Reject Null (Ho) f.) Calculate an effect size in terms of the variance account for (2)? Do you characterize it as being small, medium or large?
According to the CDC, the average weight of men in 1990 was 181 pounds. Consider a claim that the average weight of men increased by 2014. (https://abcnews.go.com/Health/average-weight-american-men-15-pounds-20-years/story?id=41100782) Use these hypotheses: H0: μ = 181(Population mean weight of men in 2014 is same as that in 1990) Ha: μ > 181(Population mean weight of men in 2014 has increased) Based on sample data, you fail to reject the null hypothesis and compared to the unknown truth if your decision was wrong, then you would have made: Type II Error because H0H0 was incorrectly not rejected. Type I Error because H0H0 was incorrectly rejected. Based on sample data, you fail to reject the null hypothesis. Then your conclusion would be: The sample data support the claim that the average weight of men increased. There is not sufficient sample evidence to support the claim that the average weight of men increased.
In a sample of 44 men, 35 said that they had less leisure time today than they had 10 years ago. In a sample of 44 women, 28 women said that they had less leisure time today than they had 10 years ago. At a= .01, is there a difference in the proportions?Use p1 for the proportion of men with less leisure time. State the hypotheses and identify the claim H0: H1: one tailed or two tailed Find the critical value(s). Round the answer to at least three decimal places and ensure your degrees of freedom are correct before proceeding. If there is more than one critical value, seperate them with commas. Critical value(s)= Compute the test value. Round sample proportions to three decimal places and z score values to two decimal places. z= Reject or do not reject the null hypothesis Is there enough evidence to support the claim Find the 99% confidence interval for the difference of the two proportions. Round the answers to three decimal places ( ) < p1-p2 < ( )
A golf association requires that golf balls have a diameter that is 1.68 inches. To determine if golf balls conform to the standard, a random sample of golf balls was selected. Their diameters are shown in the accompanying data table. Do the golf balls conform to the standards? Use the α = 0.05 level of significance. Click the icon to view the data table. First determine the appropriate hypotheses. Ho: H₁: (Type integers or decimals. Do not round.) Data Table Golf Ball Diameter (inches) 1.682 1.685 1.684 1.685 Print 1.677 1.678 1.684 1.682 1.681 1.686 1.673 1.675 Done O - X
A presidential candidate's aide estimates that, among all college students, the proportion p who intend to vote in the upcoming election is at least 60%. If 139 out of a random sample of 235 college students expressed an intent to vote, can we reject the alde's estimate at the 0.05 level of significance? Perform a one-tailed test. Then complete the parts below. Carry your intermediate computations to three or more decimal places and round your answers as specified in the table. (If necessary, consult a list of formulas.) (a) state the null hypothesis H, and the alternative hypothesis H1. Họ :0 H :0
A study was conducted to examine if children with autism spectrum disorder (ASD) had higher prenatal exposure to air pollution, specifically particulate matter < 2.5 g in diameter (PM2.5). Researchers obtained birth records of all children born in Los Angeles between 2000 and 2008 and linked these to the Department of Developmental Services records to determine if any of those subjects had been diagnosed with ASD or not. They used the birth addresses given in the birth records to determine the average daily PM2.5 for the third trimester for each child. The standard deviation for PM2.5 among ASD subjects was found to be 34.6 and for non-ASD subjects was 16.8. Assume PM2.5 is normally distributed. 4a. What was the study design? * Randomized Clinical Trial (RCT) * Case Report * Nested Case-Control Study * Case-Control Study * cross-sectional study Cohort Study 4B. What are the null and alternative hypotheses? 4c. What type of statistical test would you use to analyze the…
According to a university wildlife ecology and conservation researcher, the average level of mercury uptake in wading birds in a certain region has declined over the om past several years. Ten years ago, the average level was 16.5 parts per million. Complete parts a through c. ent a. Give the null and alternative hypotheses for testing whether the average level today is less than 16.5 ppm. (Type integers or decimals. Do not round.) al Ho: ent ctu b. Describe a Type I error for this test. Complete the sentence below. Sh Conclude that the mean mercury uptake in wading birds in the region today is Gra parts per million, when in fact, the mean mercury uptake is actually parts per million. (Type integers or decimals. Do not round.) nch c. Describe a Type II error for this test. ets Complete the sentence below. Conte Conclude that the mean mercury uptake in wading birds in the region today is parts per million, when in fact, the mean mercury uptake is actually parts per million. er 1 N /Tune…

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