The energy storage system is an important part of any electric car. An electric car company has decided to replace their batteries with a new type and want to examine wo types of batteries for its new vehicles: Lithium-ion and Lead-acid batteries. Due co the critical role of lifespan in the energy storage system, the company randomly selected 10 samples from each type and ran an accelerated life test to obtain the lifetime of each battery type. The measured lifespan is illustrated in the table below: Lithium-ion Lead-acid 365 370 366 358 369 362 364 364 363 357 365 355 364 366 365 359 369 355 362 356 (a) Test the hypothesis that the mean lifespan of the two battery types are equal. Use a = 0.01 and assume equal variances. (b) Find the P-value for this test.
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- The manufacturer of a sports car claims that the fuel injection system lasts 48 months before it needs to be replaced. A consumer group tests this claim by surveying a random sample of 10 owners who had the fuel injection system replaced. The ages of the cars at the time of replacement were (in months):Researchers studied 208 infants whose brains were temporarily deprived of oxygen due to complications at birth. When researchers detected oxygen deprivation, they randomly assigned babies to either usual care or to a whole-body cooling group. The goal was to see whether reducing body temperature for three days after birth increased the rate of survival without brain damage. What is the explanatory variable? A. Survival without brain damage B.Infants whose brains are temporarily deprived of oxygen C. Usual care or whole-body coolingIn this study, the protection of aquatic ecosystems is an important goal of Ville de Montréal. In particular, Thomas wants to see whether the contamination of creeks (ruisseaux in French) by Escherichia coli differs among the months of May, June, July, and August. Therefore, he randomly samples six creeks in different areas. For each creek, he collects water samples from May to August, and measures the amount of E. coli, expressed in colony forming units (CFU) per 100 mL of water. For each combination of a month and a creek, he thus obtains an E. coli contamination value, i.e., 24 observations in total. A preliminary statistical analysis produces the following results. 13-a) Does the mean E. coli contamination differ significantly among the four months? Justify your answer to this question with the result of a statistical test. Use α = 0.05. 13-b) Which pairs of months show significant differences in mean E. coli contamination? Explain your reasoning and justify your answer…
- A manufacturer is developing a nickel-metal hydride battery to be used in cellular telephones instead of a nickel-cadmium battery. The director of quality control decides to evaluate the newly developed battery by picking a random sample of employees with cellular phones and asking them to use both phones until the batteries run down. The results are given in the excel data file. At the 5% level of significance is there evidence that the new battery provides more minutes of talk time per battery charge? Show your work and give your answer in complete sentences in context of the problem. Data Table: Person Old Battery Type New Battery Type 1 72.39 89.95 2 75.25 64.60 3 82.10 71.25 4 58.98 87.07 5 67.07 103.11 6 35.98 76.46 7 75.37 83.92 8 66.00 83.42 9 73.14 84.30 10 67.14 92.48 11 83.06 93.48 12 57.00 59.11 13 65.92 92.27 14 76.12 61.81 15 85.21 72.83 16 80.73 42.17 17 60.65 83.89 18 36.63 94.41 19 77.98 70.30 20 75.40 77.72 21 84.70 84.35…We have a sample of 2,400 geriatric patients who are in an assisted living home, of which 1,200 participated in a new preventative Drug A. Rates of UTIs tend to be higher than average among this population. As part of a preventative and treatment intervention, we are examining the performance of several drugs: Preventative Drug (before the onset of UTI) Drug A: preventative UTI drug taken daily in hopes to prevent the growth of bacteria that causes UTIs Treatment Drugs (after the onset of UTI) Drug B: New antibiotic for treating UTIs Drug C: Conventional antibiotic for treating UTIs Information for how many patients took each drug or combination of drugs is summarized below in the two tables. Use these to answer questions a) -d) Table 1. Summary of performance of drug A: UTI rates among those taking and not taking drug A Did not take Drug A Did take Drug A Total UTI 759 887 1646 No UTI 441 312 753 Total 1200 1200 2400 Table 2. Summary of performance of drug B and C: recovery…A hospital conducted a study of the waiting time in its emergency room. The hospital has a main campus and three satellite locations. Management had a business objective of reducing waiting time for emergency room cases that did not require immediate attention. To study this, a random sample of 15 emergency room cases that did not require immediate attention at each location was selected on a particular day, and the waiting times (measured from check into when the patient was called into the clinic area) were collected and stored in the table. At the 0.05 level of significance, is there evidence of a difference in the mean waiting times in the four locations? Main Satellite 1 Satellite 2 Satellite 3 120.08 30.75 75.86 54.05 81.90 61.83 37.88 38.82 78.79 26.40 68.73 36.85 63.83 53.84 51.08 32.83 79.77 72.30 50.21 52.94 47.94 53.09 58.47 34.13 79.88 27.67 86.29 69.37 48.63 52.46 62.90 78.52 55.43 10.64 44.84 55.95 64.06 53.50 64.17 49.61 64.99 37.28 50.68…
- Do lizards play a role in spreading plant seeds? Some research carried out in South Africa would suggest so. Researchers on a study collected 400 seeds of a particular type of fig, 100 of which were from each treatment: lizard dung, bird dung, rock hyrax dung, and uneaten figs. They planted these seeds in batches of 5, and for each group of 5 they recorded how many of the seeds germinated. This resulted in 20 observations for each treatment. The treatment means and standard deviations are given in the accompanying table. Treatment n x s Uneaten figs 20 2.50 0.30 Lizard dung 20 2.30 0.33 Bird dung 20 1.70 0.34 Hyrax dung 20 1.45 0.28 (a) Construct the appropriate ANOVA table. (Use technology. Round your answers to three decimal places.) Source ofvariation Degrees offreedom Sum ofsquares Meansquares F Ratio F Prob Between Groups Within Group Total 79 Test the hypothesis that there is no difference between mean number of seeds…A researcher is interested in the effect of the pandemic on social media use among college students. The researcher hypothesizes that college students spend more time on social media than they did before the pandemic. To test his hypothesis, he randomly selected 50 college students and asked how many hours do they spend on social media per day before and after the pandemic. What is the research question? Do college students spend more time on social media that they did before the pandemic? What is/are the variable(s)? What graphs can be used to display the variable(s) visually? Variables- social media & college students. The graph that could be used is What statistical test can be used to answer the research question?Turtles are one of the longest existing creatures in the world. Due to several environmental factors, many species of turtles are already endangered. A wildlife biologist would like to study the effect of two randomly selected substrates during incubation (A1 = sand, A2 = vermiculite) and temperature (B1 = 25°C, B2 = 30°C) on the number of days before a turtle egg is hatched. The experiment was laid out in CRD with five replications.
- Cadmium, a heavy metal, is toxic to animals. Mushrooms, however, are able to absorb and accumulate cadmium at high concentrations. The Czech and Slovak governments have set a safety limit for cadmium in dry vegetables at 0.5 part per million (ppm). M. Melgar et al. measured the cadmium levels in a random sample of the edible mushroom Boletus pinicola and published the results in the paper “Influence of Some Factors in Toxicity and Accumulation of Cd from Edible Wild Macrofungi in NW Spain” (Journal of Environmental Science and Health, Vol. B33(4), pp. 439–455). A hypothesis test is to be performed to decide whether the mean cadmium level in Boletus pinicola mushrooms is greater than the government’s recommended limit. Hypothesis tests are proposed. For each hypothesis test,a. determine the null hypothesis.b. determine the alternative hypothesis.c. classify the hypothesis test as two tailed, left tailed, or right tailed.Tardigrades, or water bears, are a type of micro-animal famous for their resilience. In examining the effects of radiation on organisms, an expert claimed that the amount of gamma radiation needed to sterilize a colony of tardigrades no longer has a mean of 900 Gy (grays). (For comparison, humans cannot withstand more than 10 Gy .) A study was conducted on a sample of 27 randomly selected tardigrade colonies, finding that the amount of gamma radiation needed to sterilize a colony had a sample mean of 907 Gy , with a sample standard deviation of 17 Gy . Assume that the population of amounts of gamma radiation needed to sterilize a colony of tardigrades is approximately normally distributed. Complete the parts below to perform a hypothesis test to see if there is enough evidence, at the 0.05 level of significance, to support the claim that μ , the mean amount of gamma radiation needed to sterilize a colony of tardigrades, is not equal to 900…Tardigrades, or water bears, are a type of micro-animal famous for their resilience. In examining the effects of radiation on organisms, an expert claimed that the amount of gamma radiation needed to sterilize a colony of tardigrades no longer has a mean of 900 Gy (grays). (For comparison, humans cannot withstand more than 10 Gy .) A study was conducted on a sample of 27 randomly selected tardigrade colonies, finding that the amount of gamma radiation needed to sterilize a colony had a sample mean of 907 Gy , with a sample standard deviation of 17 Gy . Assume that the population of amounts of gamma radiation needed to sterilize a colony of tardigrades is approximately normally distributed. Complete the parts below to perform a hypothesis test to see if there is enough evidence, at the 0.05 level of significance, to support the claim that μ , the mean amount of gamma radiation needed to sterilize a colony of tardigrades, is not equal to 900…