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- In a sample of n=19 lichen specimens, the researchers found the mean and standard deviation of the amount of the radioactive element, cesium-137, that was present to be 0.009 and 0.006 microcurie per milliliter, respectively. Suppose the researchers want to increase the sample size in order to estimate the mean μ to within 0.002 microcurie per milliliter of its true value, using a 95% confidence interval.What price do farmers get for their seedless watermelon crops? In a recent month, a simple random sample of 31 farming regions gave a sample mean of $12.15 per 100 pounds of seedless watermelon. Assumer that σ=$1.46 per 100 pounds. Find the sample size necessary for a 90% confidence level with a margin of error E=0.61 for the mean price per 100 pounds of seedless watermelon. (Increase to the next whole number.)A. The average number of hours worked per week for college students is 27, and the standard deviation is 6. Assume the data is normally distributed. Determine the z-score for 35 hours. Determine the probability of someone working at least 35 hours. Determine the margin of error with 95% confidence if 64 people were surveyed. Determine the 95% confidence interval. B. A sample of 40 speedometer for “Pragia” is obtained in Tamale and each is calibrated to check for accuracy at 55kmh. The resulting sample average and sample standard deviation are 53.8 and 1.3 respectively. Does the sample information suggest that the true mean for the speedometers is not accurate at 55kmh? Use an alpha level of 0.01. C. A random sample of 30 households was selected as part of a study on electricity usage, and the number of kilowatt-hours (kWh) was recorded for each household in the sample for the March quarter of 2006. The average usage was found to be 375kWh. In a very large study in the March…
- what is the sampling error desired? In a sample of n=19 lichen specimens, the researchers found the mean and standard deviation of the amount of the radioactive element, cesium-137, that was present to be 0.009 and 0.006 microcurie per milliliter, respectively. Suppose the researchers want to increase the sample size in order to estimate the mean μ to within 0.002 microcurie per milliliter of its true value, using a 95% confidence interval. Complete parts a through c.Considering a sample of n = 9 water specimens selected for treatment by co-agulation, the sample mean arsenic concentration was 24.3 μg/L, and the samplestandard deviation was 4.1. Assume that the distribution of arsenic concentrationwas normal.(a) Calculate and interpret a 95% CI for true average arsenic concentration in allsuch water specimens.(b) Calculate a 95% upper confidence bound for the standard deviation of thearsenic concentration distribution(c) Assuming a new sample of n = 50 water specimens selected with the samplemean arsenic concentration of 23.5 μg/L and the sample standard deviation of0.75, calculate a 95% CI for true average arsenic concentration without assumingnormal distribution of arsenic concentration?A random sample of 16 adult male wolves from the Canadian Northwest Territories gave an average weight x1 = 97 lb with estimated sample standard deviation s1 = 6.3 lb. Another sample of 22 adult male wolves from Alaska gave an average weight x2 = 88 lb with estimated sample standard deviation s2 = 7.1 lb. (a) Let ?1 represent the population mean weight of adult male wolves from the Northwest Territories, and let ?2 represent the population mean weight of adult male wolves from Alaska. Find a 75% confidence interval for ?1 – ?2. (Round your answers to one decimal place.) lower limit upper limit (b) Examine the confidence interval and explain what it means in the context of this problem. Does the interval consist of numbers that are all positive? all negative? of different signs? At the 75% level of confidence, does it appear that the average weight of adult male wolves from the Northwest Territories is greater than that of the Alaska wolves? Because the interval…
- A simple random sample of 29 filtered 100-mm cigarettes is obtained from a normally distributed population, and the tar content of each cigarette is measured. The sample has a standard deviation of 0.22 mg. Use a 0.05 significance level to test the claim that the tar content of filtered 100-mm cigarettes has a standard deviation different from 0.35 mg, which is the standard deviation for unfiltered king-size cigarettes. Complete parts (a) through (d) below. a. What are the null and alternative hypotheses? A. H0: sigmaσnot equals≠0.350.35 mg Upper H 1 : sigma equals 0.35H1: σ=0.35 mg B. H0: sigmaσequals=0.350.35 mg Upper H 1 : sigma less than 0.35H1: σ<0.35 mg C. H0: sigmaσgreater than>0.350.35 mg Upper H 1 : sigma less than or equals 0.35H1: σ≤0.35 mg D. H0: sigmaσequals=0.350.35 mg Upper H 1 : sigma not equals 0.35H1: σ≠0.35 mg Your answer is correct. b. Find the test statistic. chi squaredχ2equals=11.063 (Round to…A simple random sample of 29 filtered 100-mm cigarettes is obtained from a normally distributed population, and the tar content of each cigarette is measured. The sample has a standard deviation of 0.22 mg. Use a 0.05 significance level to test the claim that the tar content of filtered 100-mm cigarettes has a standard deviation different from 0.35 mg, which is the standard deviation for unfiltered king-size cigarettes. Complete parts (a) through (d) below. a. What are the null and alternative hypotheses? A. H0: sigmaσnot equals≠0.350.35 mg Upper H 1 : sigma equals 0.35H1: σ=0.35 mg B. H0: sigmaσequals=0.350.35 mg Upper H 1 : sigma less than 0.35H1: σ<0.35 mg C. H0: sigmaσgreater than>0.350.35 mg Upper H 1 : sigma less than or equals 0.35H1: σ≤0.35 mg D. H0: sigmaσequals=0.350.35 mg Upper H 1 : sigma not equals 0.35H1: σ≠0.35 mg Your answer is correct. b. Find the test statistic. chi squaredχ2equals=?? (Round to three…Construct a 90% confidence interval for the true average cost of a 50000 kilometre car service in Sandton , if it was found that a sample of 20 different garages in the area charged an average of R1000 for a service. Assume a population standard deviation of R90.26
- A simple random sample of 26 filtered 100-mm cigarettes is obtained from a normally distributed population, and the tar content of each cigarette is measured. The sample has a standard deviation of 0.18 mg. Use a 0.01 significance level to test the claim that the tar content of filtered 100-mm cigarettes has a standard deviation different from 0.25 mg, which is the standard deviation for unfiltered king-size cigarettes. Complete parts (a) through (d) below. a. What are the null and alternative hypotheses? A. H0: σ≠0.25 mg H1: σ=0.25 mg B. H0: σ=0.25 mg H1: σ≠0.25 mg C. H0: σ=0.25 mg H1: σ<0.25 mg D. H0: σ>0.25 mg H1: σ≤0.25 mg b. Find the test statistic. χ2=_____ (Round to three decimal places as needed.) c. Find the P-value of the test statistic. The P-value of the test statistic is ______ (Round to three decimal places as needed.) d. State the conclusion. (Fail to reject OR Reject…A random sample of 16 adult male wolves from the Canadian Northwest Territories gave an average weight x1 = 96 lb with estimated sample standard deviation s1 = 6.7 lb. Another sample of 22 adult male wolves from Alaska gave an average weight x2 = 88 lb with estimated sample standard deviation s2 = 7.5 lb. (a) Let μ1 represent the population mean weight of adult male wolves from the Northwest Territories, and let μ2 represent the population mean weight of adult male wolves from Alaska. Find a 75% confidence interval for μ1 – μ2. (Round your answers to one decimal place.) lower limit upper limitA simple random sample of 30 filtered 100-mm cigarettes is obtained from a normally distributed population, and the tar content of each cigarette is measured. The sample has a standard deviation of 0.22mg. Use a 0.05 significance level to test the claim that the tar content of filtered 100-mm cigarettes has a standard deviation different from 0.35mg, which is the standard deviation for unfiltered king-size cigarettes. Complete parts (a) through (d) below. a. What are the null and alternative hypotheses?