Express 620 ft in yards and feet Thanks
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Answer:
±0.75 ounce
Step-by-step explanation:
Assuming this complete question: The Wisconsin Dairy Association is interested in estimating the mean weekly consumption of milk for adults over the age of 18 in that state. To do this, they have selected a random sample of 300 people from the designated population. The following results were recorded: xbar=34.5 ounces, s=7.9 ounces Given this information, if the leaders wish to estimate the mean milk consumption with 90 percent confidence, what is the approximate margin of error in the estimate?
Previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
represent the sample mean
population mean (variable of interest)
s=7.9 represent the sample standard deviation
n=300 represent the sample size
Solution to the problem
The confidence interval for the mean is given by the following formula:
(1)
In order to calculate the critical value we need to find first the degrees of freedom, given by:
Since the Confidence is 0.90 or 90%, the value of and
, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,199)".And we see that
And the margin of error is given by:
±0.75 ounce