(a) Find the critical z* for a level 96.5% confidence interval. z*= (b) Find the critical z* for a level 58% confidence interval. z*=
Q: Allen's hummingbird (Selasphorus Sasin or these birds is x = 3.15 grams. Based on previous studies,…
A: Hi! Thank you for the question, As per the honor code, we are allowed to answer three sub-parts at a…
Q: Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther.+ Suppose a small…
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Q: Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther.t Suppose a small…
A: Note: Hi there! Thank you for posting the question. As there are multiple sub parts, according to…
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(a) Find the critical z* for a level 96.5% confidence interval.
z*=
(b) Find the critical z* for a level 58% confidence interval.
z*=
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- What price do farmers get for their watermelon crops? In the third week of July, a random sample of 40 farming regions gave a sample mean of x = Assume that o is known to be $1.92 per 100 pounds. $6.88 per 100 pounds of watermelon. (a) Find a 90% confidence interval for the population mean price (per 100 pounds) that farmers in this region get for their watermelon crop (in dollars). What is the margin of error (in dollars)? (For each answer, enter a number. Round your answers to two decimal places.) lower limit $ upper limit 2$ margin of error $ (b) Find the sample size necessary for a 90% confidence level with maximal error of estimate E = 0.27 for the mean price per 100 pounds of watermelon. (Enter a number. Round up to the nearest whole number.) farming regions (c) A farm brings 15 tons of watermelon to market. Find a 90% confidence interval for the population mean cash value of this crop (in dollars). What is the margin of error (in dollars)? Hint: 1 ton is 2000 pounds. (For each…Assume that last year in a particular state there were 211 children out of 1990 who were diagnosed with Autism Spectrum Disorder. Nationally, 1 out of 88 children are diagnosed with ASD. It is believed that the incident of ASD is more common in that state than nationally. Calculate a 96% confidence interval for the percentage of children in that state diagnosed with ASD.P: Parameter What is the correct parameter symbol for this problem? What is the wording of the parameter in the context of this problem? A: Assumptions Since information was collected from each object, what conditions do we need to check? Check all that apply. n≥30n≥30 or normal population. N≥20nN≥20n n(pˆ)≥10n(p̂)≥10 σσ is known. σσ is unknown. n(1−pˆ)≥10n(1-p̂)≥10 Check those assumptions: 1. nˆpnp^= which is 2. n(1−ˆp)n(1-p^)= which is 3. NN= which is If no N is given in the problem, use 1000000N: Name the procedure The…A sample of 66 night-school students' ages is obtained in order to estimate the mean age of night-school students. x = 24.7 years. The population variance is 18. (a) Give a point estimate for u. (Give your answer correct to one decimal place.) (b) Find the 95% confidence interval for u. (Give your answer correct to two decimal places.) Lower Limit Upper Limit (c) Find the 99% confidence interval for u. (Give your answer correct to two decimal places.) Lower Limit Upper Limit
- Suppose we are making predictions of the dependent variable y for specific values of the independent variable x using a simple linear regression model holding the confidence level constant. Let Width (C.I) = the width of the confidence interval for the average value y for a given value of x, and Width (P.I) = the width of the prediction interval for a single value y for a given value of x. Which of the following statements is true? Width (C.I) = 0.5 Width (P.I) Width (C.I) = Width (P.I) Width (C.I) > Width (P.I) Width (C.I) < Width (P.I)Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther.t Suppose a small group of 20 Allen's hummingbirds has been under study in Arizona. The average weight for these birds is x = 3.15 grams. Based on previous studies, we can assume that the weights of Allen's hummingbirds have a normal distribution, with o = 0.32 gram. (a) Find an 80% confidence interval for the average weights of Allen's hummingbirds in the study region. What is the margin of error? (Round your answers to two decimal places.) lower limit upper limit margin of error (b) What conditions are necessary for your calculations? (Select all that apply.) O normal distribution of weights Oo is known O n is large O o is unknown O uniform distribution of weights (c) Interpret your results in the context of this problem. O The probability that this interval contains the true average weight of Allen's hummingbirds is 0.20. O we are 80% confident that the true average weight of Allen's hummingbirds falls…The data below were collected in a clinical trial to compare a new drug to a placebo for its effectiveness in lowering total serum cholesterol. Generate a 95% confidence interval for the difference in mean total cholesterol levels between treatments. Generate a 95% confidence interval for the proportion of all patients with total cholesterol < 200. New Drug Placebo Total Sample (n=75) (n=75) (n=150) Mean (SD) Total Serum Cholesterol 185.0 (24.5) 204.3 (21.8) 194.7 (23.2) % Patients with Total Cholesterol < 200 78.00% 65.00% 71.50% a. 95% CI ( , )
- In a large scale study of energy conservation in single family homes, 20 homes were randomly selected from homes built in a housing development in southern England. Ten of the houses, randomly selected from the 20 houses selected for the study, were constructed with standard levels of insulation (70 mm of roof insulation and 50 mm of insulation in the walls). The other 10 houses were constructed with extra insulation (120 mm of roof insulation and 100 mm of insulation in the walls). Each house was heated with a gas furnace and energy consumption was monitored for eight years. The data shown below give annual gas consumption in MWh for each of the 20 houses. Standard Insulation Extra Insulation 13.8 15.1 17.8 13.9 18.0 15.9 17.3 17.2 16.9 15.2 19.9 13.8 13.6 11.3 17.6 13.2 15.9 18.8 12.3 14.0 Sample means Sample Standard Deviation 16.31 14.84 2.38 2.12A. True or False? A researcher compares the BMI of two populations of individuals from different socioeconomic backgrounds and finds that the proportion of individuals considered obese and from a poor background is .37 (n = 112), and the proportion of individuals considered obese and from an affluent background is .27 (n = 87); a 95% confidence interval for the risk difference for obesity among members of the poor population compared to the affluent population is (-0.03, 0.23). B. Why is it true or false?What critical value of t* should be used for a 99.5% confidence interval for the population mean based upon a random sample of 18 observations? Find the t-table here. t* = 2.878 t* = 2.898 t* = 3.197 t* = 3.222
- The sample mean for the fill weighs of 100 boxes is x = 12.05. The population variance of the fill weighs is known to be 0.100. Find a 95% confidence interval for the population mean u fill weighs of the boxes. Use z = 1.96 12.03, 12.70 -12.30, 12.70 -12.03, 12.07 O 12.30, 12.07 12.03, 12.07Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther.t Suppose a small group of 12 Allen's hummingbirds has been under study in Arizona. The average weight for these birds is x = 3.15 grams. Based on previous studies, we can assume that the weights of Allen's hummingbirds have a normal distribution, with a = 0.22 gram. When finding an 80% confidence interval, what is the critical value for confidence level? (Give your answer to two decimal places.) (a) Find an 80% confidence interval for the average weights of Allen's hummingbirds in the study region. What is the margin of error? (Round your answers to two decimal places.) lower limit upper limit margin of error (b) What conditions are necessary for your calculations? (Select all that apply.) Oo is known Oo is unknown O uniform distribution of weights O normal distribution of weights On is large (c) Interpret your results in the context of this problem. We are 80% confident that the true average weight of…
