00:01
Okay, so i see that you need help with this question, and it says, 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.
00:22
Assume that the standard deviation is known to be $2 per 100 pounds.
00:31
Find the 90 % confidence interval for the population mean price that the farmers in the region get for their watermelon crop.
00:39
What is the margin of error? so the first thing that you're going to do is you are going to take 6 .88 plus or minus 90 % confidence interval is 1 .645 times your standard deviation of two over the square root of 45.
01:04
So 6 .88, oops, not yet.
01:08
So two divided by the square root of 45, and you have 6 .88 plus or minus 1 .645 times 0 .2, and i'm rounding to two decimal places, so i'll just go out three for right now, 298, and then times 1 .645, and so i have 6 .88 plus or minus 0 .49, and then my margin of error is 0 .49, and then my confidence interval is 6 .88 minus 0 .49, and that's 6 .39 is my lower limit, and then if i add them together, that's 7 .37.
01:59
Then for the next one, find the sample size necessary for 90 % confidence interval with a maximum, maximal error estimate of 0 .43 for the mean price per 100 pounds of watermelon...