Construct a 90% Confidence Interval based on the following data: 45, 55, 67, 45, 68, 79, 98, 87, 84, 82.
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- In a survey of 2847 adults, 1408 say they have started paying bills online in the last year. Construct a 99% confidence interval for the population proportion. Interpret the results.Twenty-four fast-food restaurant hourly employees were surveyed to see how many hours they worked per week, on average. The following data was obtained: 34, 45, 43, 41, 49, 48, 29, 36, 50, 40, 40, 49, 36, 54, 31, 41, 42, 37, 47, 29, 47, 36, 47, 48 Assuming the population standard deviation is a = 6, construct a 87% confidence interval for the mean number of hours per week that the population of fast-food restaurant employees work. I = a 2 NINA || Margin of Error: E = We are 87% confident that fast-food restaurant hourly employees, on average, work between hours and hours per week.In a survey of 3342 adults, 1489 say they have started paying bills online in the last year. Construct a 99% confidence interval for the population proportion. Interpret the results.
- Of 346 items tested, 22 are found to be defective. Construct the 98% confidence interval for the proportion of allsuch items that are defective.In a survey of 3426 adults, 1487 say they have started paying bills online in the last year. Construct a 99% confidence interval for the population proportion. Interpret the results.A sanatorium conducted a study on 3,500 women over 35 years of age, of whom 220 reported having suffered kidney pain. Determine the 95% and 99% confidence intervals.
- A second-semester junior at a large high school wants to estimate the number of colleges a senior applies to on average before deciding where they will attend. The junior randomly selects 19 seniors and asks them how many colleges they applied to and records the number. Use the data below to construct a 99% confidence interval to estimate the true mean number of colleges seniors at this school apply to before deciding about which college to attend. 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5, 6, 7, 7, 8, 9 3. Do. Use technology to find the mean and standard deviation of the sample. Round each answer to 2 decimal places. Use those values and t* from the table to calculate the confidence interval. Round the endpoints to 1 decimal place. 99% confidence interval is (, )In a survey of 3196 adults, 1405 say they have started paying bills online in the last year. Construct a 99% confidence interval for the population proportion. Interpret the results.The number of points held by a sample of the NHL's highest scorers for both the Eastern Conference and the Western Conference is shown below. Find a 95% confidence interval for the difference of the means. Eastern Conference: 83, 60, 75, 58,78, 59, 70, 58,62, 61, 59 Western Conference: 77, 59, 72, 58,37, 57, 66, 55,61,65
- In a survey of 2588 adults, 1466 say they have started paying bills online in the last year. Construct a 99% confidence interval for the population proportion. Interpret the results.The data represents a sample of the number of home fires started by fault electricalwiring for the past eight years: 20, 16, 27, 19, 18, 23, 24, and 25. Find the 90%confidence interval for the mean number of home fires started by faulty electricalwiring each yearIn a survey of 3246 adults, 1418 say they have started paying bills online in the last year. Construct a 99% confidence interval for the population proportion. Interpret the results.