How many voters should be sampled for a 99% confidence interval? Round up to the nearest whole number.
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- In a test of the effectiveness of garlic for lowering cholesterol, 25 patients were treated with raw garlic. The changes in their levels of LDL cholesterol (in mg/dL) had a mean of 0.4 and a standard deviation of 21.0. a. Find the margin of error E that corresponds to 98% confidence level. Round E to five decimal places. b. Find the 98% confidence interval estimate of the population mean. Follow rounding rules and use appropriate notation. c. What does the confidence interval suggest about the effectiveness of garlic in reducing LDL cholesterol?A random sample of 10 items is taken from a normal population. The sample had a mean of 82 and astandard deviation is 26. Which is the appropriate 99% confidence interval for the population mean?Create a 85% confidence interval for a population proportion from a sample of size 103 with sample proportion ˆpp^ = 4110341103. ( , ) [three decimal accuracy] [three decimal accuracy]
- Data from a survey of a representative sample was used to estimate that 33% of computer users in 2011 had tried to get on a Wi-Fi network that was not their own in order to save money. Suppose you decide to conduct a survey to estimate this proportion for the current year. What is the required sample size if you want to estimate this proportion with a margin of error of 0.05? (Assume a 95% confidence level.) Calculate the required sample size first using 0.33 as a preliminary estimate of p. (Round your answer up to the nearest whole number.) n = ?? Calculate the required sample size using the conservative value of 0.5. (Round your answer up to the nearest whole number.) n=??Given that the sample proportion of Americans who are dissatisfied with the quality of education is 0.53 and the sample size is 1012. a. Verify that the sample is large enough to use the normal formula to find a confidence interval for the proportion of Americans who are dissatisfied with the quality of education students receive in kindergarten through grade 12. b. Construct a 90% confidence interval for the proportion of U.S. adults who are dissatisfied with the quality of education students receive in kindergarten through grade 12. Use three decimal places in your margin of error. c. Provide an interpretation of your interval in the context of this data situation.The mean weight for college students is 175 pounds. A sample of 30 students at a college is taken, and the mean of the sample is 180 pounds with standard deviation 20 pounds. A company wants to see if the weights of the students at the college are statistically different from the national average at the 95% confidence level. Find the p-value in the table for the table value of t closest to the calculated value of the test statistic (round to 2 decimal places).
- A pack of whey powder is marked to have 30 grams. It was determined that the average mean weight content is 27.5 grams and a standard deviation of 0.9 grams. The bakery randomly selected 36 packs from the stock room. What is the standard error? Construct a 95% confidence interval, provide the margin of error Construct a 95% one-sided confidence bound, provide the margin of error Assuming that the population is normally distributed, test to see if the mean of the population is at least 30 grams. Use 0.01 level of significance. Applying the same hypotheses, use a 0.075 level of significance.The operations manager at a light emitting diode (LED) light bulb factory needs to estimatethe mean life of a large shipment of LEDs. The manufacturer’s specifications are that thestandard deviation is 1,500 hours. A random sample of 64 LEDs indicated a sample mean lifeof 49,875 hours.Construct a 95% confidence interval estimate for the population mean life of LED light bulbsin this shipment.Ans: CI = ______________________ Paste your working here:The freshness of produce at a super-store is rated on a scale of 1 to 5 with 5 being very fresh. Froma random sample of 49 customers, the average score was 3.8 with a standard deviation of 0.7. a) Obtain an 80% confidence interval for the population mean - the mean score for thedistribution of all possible customers. Specify and check all conditions needed for theproblem. b) Does the interval obtained in Part(a) contain the population mean score for the customers?Why? c) Does the interval obtained in Part(a) contain the sample mean score for the customers in thedata? Why? d.) In a long series of repeated experiments, with new random samples of 49 customers eachday, what proportion of the resulting confidence intervals will contain the true populationmean? Explain your reasoning.
- TOTAL PLASMA VOLUME IS IMPORTANT IN DETERMINING THE REQUIRED PLASMA COMPONENT IN BLOOD REPLACEMENT THERAPY FOR A PERSON UNDERGOING SURGERY. PLASMA VOLUME IS INFLUENCED BY THE OVERALL HEALTH AND PHYSICAL ACTIVITY OF AN INDIVIDUAL. SUPPOSE THAT A RANDOM SAMPLE OF 45 MALE FIREFIGHTERS ARE TESTED AND THAT THEY HAVE A PLASMA VOLUME SAMPLE MEAN OF 37.5ML/KG WITH A STANDARD DEVIATION OF 7.50ML/KG. 1) Find a 95% confidence interval for the mean blood plasma volume in male firefighters. Interpretthe interval. 2) The Chief Scientist at the lab wants the estimation of the mean blood plasma volume to be correctto within 1 unit of the true mean with 95% probability. How many additional firefighters have tobe sampled in order to achieve this estimation accuracy?A random sample of 100 observation produced a sample mean 22.3 and given population standard deviation 6. (b) find a 90% confidence interval for μ. Round to 2 decimal places.A look at all Ontario households using natural gas heating in 2012 shows that the mean monthly consumption of natural gas is 107.73 ft3, with a population standard deviation of 13.9a ft3. For a sample of 6a households, the sample mean found was 105.6a ft3. Use this information to find value, the margin of error E, and to construct a 93% confidence interval for the mean monthly consumption for the population of all Ontario households. Is this sample a good estimation of the population mean monthly consumption of natural gas for all Ontario households? (a=1)
