0.234 Calcium is essential to tree growth. In 1990, the concentration of calcium in precipitation in Chautauqua, New York, mg was 0.11 milligram per liter A random sample of 8 precipitation dates in 2018 results in the following data: L 0.065 0.262 = 0.120 What are the null and alternative hypotheses? Ho: P 0.11 H₁: P # 0.11 (Type integers or decimals. Do not round.) Find the test statistic. 0.108 to = (Round to two decimal places as needed.) Points: 0 of 1 0.070 ... A normal probability plot suggests the data could come from a population that is normally distributed. A boxplot does not show any outliers. Does the sample evidence suggest that calcium concentrations have changed since 1990? Use the a = 0.1 level of significance. 0.183 Save 0.313
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- The hemoglobin count (HC) in grams per 100 milliliters of whole blood is approximately normally distributed with population mean 14 for healthy adult women. Suppose a female patient has had 16 laboratory blood tests during the past year. Her average HC is 16.94 with s = 1.05. Is %3D there evidence that her average HC is not 14? (a) State the null and alternative hypotheses: (Type "mu" for the symbol u, e.g. mu >1 for the mean is greater than 1, mu <1 for the mean is less than 1, mu not = 1 for the mean is not equal to 1) Ho : На (b) Find the test statistic (give to at least 3 decimal places), t =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…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:
- 1.56 Longleaf pine trees. The Wade Tract in Thomas County, Georgia, is an old-growth forest of longleaf pine trees (Pinus palustris) that has survived in a relatively undisturbed state since before the settlement of the area by Europeans. A study collected data on 584 of these trees. One of the variables measured was the diameter at breast height (DBH). This is the diameter of the tree at 4.5 feet, and the units are centimeters (cm). Only trees with DBH greater than 1.5 cm were sampled. Here are the diameters of a random sample of 40 of these trees: PINES 27 10.5 13.3 26.0 18.3 52.2 9.2 26.1 17.6 40.5 31.8 47.2 11.4 2.7 69.3 44.4 16.9 35.7 5.4 44.2 2.2 4.3 7.8 38.1 2.2 11.4 51.5 4.9 39.7 32.6 51.8 43.6 2.3 44.6 31.5 40.3 22.3 43.3 37.5 29.1 27.9 (a) Find the five-number summary for these data. (b) Make a boxplot. (c) Make a histogram. (d) Write a short summary of the major features of this distribution. Do you prefer the boxplot or the histogram for these data?According to a report from the United States Environmental Protection Agency, burning one gallon of gasoline typically emits about 8.9 kg of CO2. A fuel company wants to test a new type of gasoline designed to have lower CO2 emissions. Their hypotheses are Ho : µ = 8.9 kg and H, : µ < 8.9 kg. where u is the mean amount of CO, emitted by burning one gallon of this new gasoline. Suppose that Ho is actually true. Which situation below would have the lowest probability of a Type I error? Choose 1 answer: n = 20 and a = 0.01 n = 50 and a = 0.05 n = 75 and a = 0.10An article includes the accompanying data on compression strength (lb) for a sample of 12-oz aluminum cans filled with strawberry drink and another sample filled with cola. Beverage Sample Size Sample Mean Sample SD Strawberry Drink 10 535 24 Cola 10 559 15 Does the data suggest that the extra carbonation of cola results in a higher average compression strength? Base your answer on a P-value. (Use ? = 0.05.) State the relevant hypotheses. (Use ?1 for the strawberry drink and ?2 for the cola.) H0: ?1 − ?2 = 0Ha: ?1 − ?2 ≥ 0H0: ?1 − ?2 = 0Ha: ?1 − ?2 ≠ 0 H0: ?1 − ?2 = 0Ha: ?1 − ?2 < 0H0: ?1 − ?2 = 0Ha: ?1 − ?2 > 0 Calculate the test statistic and determine the P-value. (Round your test statistic to one decimal place and your P-value to three decimal places.) t = P-value = State the conclusion in the problem context. Reject H0. The data suggests that cola has a higher average compression strength than the strawberry drink.Reject H0. The…
- A chi-square test for homogeneity is conducted on three populations and one categorical variable that has three values. Computation of the chi-square statistic yields Χ2 = 2.05. What is the p-value between? a.p > 0.25 b.0.05 < p < 0.1 c.0.02 < p < 0.025 d.0.01 < p < 0.02 e.0.005 < p < 0.01According to the U.S. Climate Data website, the average monthly rainfall totals (in inches) in San Diego, California for the months of July through December are shown in the Expected column of the table below. Assuming the significance level is p=0.95, what is the chi-square (χ2) critical value for one 6-month period? Does the data match the model? Month Observed(Inches) Expected(Inches) July 0.07 0.04 August 0.05 0.04 September 0.13 0.16 October 0.59 0.55 November 0.89 1.02 December 1.28 1.54 a The chi-square critical value is 1.15 and the data matches the model. b The chi-square critical value is 1.64 and the data matches the model. c The chi-square critical value is 1.15 and the data does not match the model. d The chi-square critical value is 1.64 and the data does not match the model.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?
- Based on the following statistical statement, χ2(4, N = 148) = 11.76, p = .02, is there a significant association?ol 106. Use the two scenarios presented in Figure 7.12 to answer the following question about the two sources of variability in ANOVA. 3.5 3.0 2.5 2.0 1.5 (i) HA: P₁ P2 ‡ P3 (iii) HA not all proportions are equal to this po 1.0 Figure 7.12: Within vs Between Variability in ANOVA onnisth Scenario A L (d) Recall that, O 2 H 3 2.05 2.10 2.15 2.20 2.25 2.30 2.35 (ii) Ho: M₁ M₂ = μ3 sans are (iv) HA: not all means are equal 720 qeaoluet alt sot mol 4 mobor 2924 (6) 2 3 4 gob za aim onbala (1) di batt (9) w-qodt fonqtotal bac o2 (6) (Troy ote12. (0) (a) Which scenario provides stronger evidence that the difference in factor level means represents true differences in the population means? Explain why in your own words. Scenario B 1 = stogol 05 O 1 oftaltete fest out of sulav gris I (b) Which scenario exhibits more variability within each factor level? Is this measured by SSError or SSFactor? Agulow (c) Which scenario exhibits more variability between each factor level? Is this measured by…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:H0: μμ = 181(Population mean weight of men in 2014 is same as that in 1990) Ha:Ha: μμ > 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 I Error because H0H0 was incorrectly rejected. Type II Error because H0H0 was incorrectly not 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.