Thus we can write as. Find a 90 confidence interval for the population mean cash value of this crop in dollars.
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X 300 688.
. What is the margin of error. 1 ton is 2000 pounds. Find a 90 confidence interval for the population mean cash value of this crop.
Find a 90 confidence interval for the population mean cash value of this crop. Round your answers to two decimal places. In the third week of July a random sample of 42 farming regions gave a sample mean of 688 per 100 pounds of watermelon.
Critical z-value for significance level of 01 is z 1645. No standard deviation value is stated-----. 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. Find a 90 confidence interval for the population mean cash value of this crop. Given E 029 Z 005 1645 σ 192.
Assume that σ is known to be 192 per 100 pounds. Lower limit 191602 upper limit 221198 margin of. Find a 90 confidence interval for the population mean cash value of this crop.
What is the margin of error. C A farm brings 15 tons of watermelon to market. Up to 25 cash back b Sample size.
Find a 90 confidence interval for the population mean cash value of this crop. Assume that σ is known to be 192 per 100 pounds. 1 ton is 2000 pounds.
C A farm brings 15 tons of watermelon to market. What is the margin of error. What is the margin of error.
Assume that σ is known to be 186 per 100 pounds. Farming regions c A farm brings 15 tons of watermelon to market. σ 198 300.
CWe are told that the farm brings 15 tons of watermelon to market. 1 ton is 2000 pounds. In the third week of July a random sample of 41 farming regions gave a sample mean of x 688 per 100 pounds of watermelon.
1 ton is 2000 pounds. Farming regions c A farm brings 15 tons of watermelon to market. We must convert 15 tons to pounds before finding the confidence interval.
C A farm brings 15 tons of watermelon to market. Find a 90 confidence interval for the population mean cash value of this crop. The detailed solution is given below.
C A farm brings 15 tons of watermelon to market. No unit price sees to be available. In the third week of July a random sample of 42 farming regions gave a sample mean of x 688 per 100 pounds of watermelon.
Hint 1 ton is 2000. 1 ton is 2000 pounds. First note that the sample mean is in units of dollars per 100 pounds of watermelon.
1 ton is 2000 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.
Round your answers to two decimal places time left. Starting with the lower value convert this interval to be in terms of 30000. In the third week of July a random sample of 44 farming regions game a sample mean of.
What is the margin of error. _____ farming regions c A farm brings 15 tons of watermelon to market. C A farm brings 15 tons of watermelon to market.
A farm brings 15 tons of watermelon to market. What is the margin of error. Find a 90 confidence interval for the population mean cash value of this crop.
Now margin of error is. Find a 90 confidence interval for the population mean cash value of this crop. C A farm brings 15 tons of watermelon to market.
What is the margin of error in dollars. Round your answers to two decimal places -----It appears that the sample mean is 15 tons. What is the margin of error in dollars.
1 ton is 2000 pounds. 1 ton is 2000 pounds. For each answer enter a number.
Find a 90 confidence interval for the population mean cash value of this crop in dollars. Find a 90 confidence interval for the population mean cash value of this cropWhat is the margin of error. C A farm brings 15 tons of watermelon to market.
Round your answers to two decimal places. Assume that σ known to be 192 per 100 pounds. We know that n zσE2.
What is the margin of error. Also 1 ton 2000 pounds. Up to 25 cash back c a farm brings 15 tons of watermelon to market.
C A farm brings 15 tons of watermelon to market. 1 ton is 2000 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.
15 tons 15 2000 pounds 30000 pounds. 1 ton is 2000 pounds. Solution for What price do farmers get for their watermelon crops.
Round your answers to two decimal places I have posted this more than once and everyone keeps ignoring the last part. 1 ton is 2000 pounds. Round your answers to two decimal places lower limit upper limit margin of error.
In the third week of July a random sample of 42 farming regions gave a sample mean of x 688 per 100 pounds of watermelon. Find a 90 confidence interval for the population mean cash value of this crop. What is the.
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. Round your answers to two decimal places You want to conduct a survey to determine the proportion of people who favor a proposed tax policy. Now this is per 100 pounds.
B Find the sample size necessary for a 90 confidence level with maximal error of estimate E 035 for the mean price per 100 pounds of watermelon. Find the sample size necessary for a 9o confidence level with maximal error of estimate E03 for the mean price per 100 pounds of watermelon. For each answer enter a number.
What is the margin of error. First note that the sample mean is in units of dollars per 100 pounds of watermelon. Round your answers to two decimal places Lower limit Upper limit Margin of error.
2000 Ibs 1 ton 15 tons x 30000 Ibs From part a we know that the confidence Interval for 100 pounds of watermelon is 638 to 738. 30000100 300 pounds. Find a 90 confidence interval for the population mean cash value of this crop in dollars.
Round up to the nearest whole number 4 farming regions c A farm brings 15 tons of watermelon to market.
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