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lara31 [8.8K]
3 years ago
13

What is the additive inverse of 5/11

Mathematics
1 answer:
jok3333 [9.3K]3 years ago
5 0

HERE IS UR ANSWER:-

-5/11

HOPE IT HELPED U BRO! ✌✌

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ME = 1.653 *\frac{7.9}{\sqrt{300}}= \pm 0.75

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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".

\bar X=34.5 represent the sample mean

\mu 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:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:

df=n-1=300-1=299

Since the Confidence is 0.90 or 90%, the value of \alpha=0.1 and \alpha/2 =0.05, 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 t_{\alpha/2}=1.653

And the margin of error is given by:

ME = 1.653 *\frac{7.9}{\sqrt{300}}= \pm 0.75

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