(b) Using critical value of 2.101, the 95% confidence interval for the slope parameter is(round off answers to tw decimal points): lower confidence interval upper confidence interval
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Q: What price do farmers get for their watermelon crops? In the third week of July, a random sample of…
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Q: What price do farmers get for their watermelon crops? In the third week of July, a random sample of…
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Q: What price do farmers get for their watermelon crops? In the third week of July, a random sample of…
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- Find the confidence intervalWhat 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…The prevalence rate of osteoporosis among women were measured in a survey, p(%) = 25.3 SE (p) = 1.8 calculate the upper bound of 95% confidence interval
- What price do farmers get for their watermelon crops? In the third week of July, a random sample of 42 farming regions gave a sample mean of = $6.88 per 100 pounds of watermelon. Assume that σ is known to be $1.94 per 100 pounds. (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 $ margin of error $ (b) Find the sample size necessary for a 90% confidence level with maximal error of estimate E = 0.33 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.…What is the formula for CONFIDENCE INTERVAL (CI) FOR μ1 – μ2 (TWO INDEPENDENT SAMPLES)?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 = $6.88 per 100 pounds of watermelon. Assume that σ is known to be $1.92 per 100 pounds. (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 $ margin of error $ (b) Find the sample size necessary for a 90% confidence level with maximal error of estimate E = 0.25 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…
- What price do farmers get for their watermelon crops? In the third week of July, a random sample of 45 farming regions gave a sample mean of = $6.88 per 100 pounds of watermelon. Assume that σ is known to be $1.98 per 100 pounds. (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 $ margin of error $ (b) Find the sample size necessary for a 90% confidence level with maximal error of estimate E = 0.45 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…What price do farmers get for their watermelon crops? In the third week of July, a random sample of 44 farming regions gave a sample mean of = $6.88 per 100 pounds of watermelon. Assume that σ is known to be $1.90 per 100 pounds. (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 $ ______margin of error $ ______ (b) Find the sample size necessary for a 90% confidence level with maximal error of estimate E = 0.25 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)?…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, 12.07
- 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 = $6.88 per 100 pounds of watermelon. Assume that σ is known to be $1.90 per 100 pounds. (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 $ margin of error $ (b) Find the sample size necessary for a 90% confidence level with maximal error of estimate E = 0.37 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…An SRS of 450 high school seniors gained an average of ?¯=20 points in their second attempt at the SAT Mathematics exam. Assume that the change in score has a Normal distribution with standard deviation ?=49 . (a) Find a 95% confidence interval for the mean change in score ? in the population of all high school seniors. (Enter your answers rounded to two decimal places.) lower bound of confidence interval: upper bound of confidence interval: (b) What is the margin of error for 95% ? (Enter your answer rounded to two decimal places.) margin of error: (c) Suppose we had an SRS of just 100 high school seniors. What would be the margin of error for 95% confidence? (Enter your answer rounded to three decimal places.) margin of error:what is The confidence interval ? Interpret the interval.