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Dafna1 [17]
3 years ago
10

Tamara's dosage of medication has recently decreased 20 milligrams per day. What is the total decrease in Tamara's medication at

the end of the week. Show your work.
Mathematics
2 answers:
Leviafan [203]3 years ago
6 0
7x20 = 140 making that the total amount of decrease per week. Hope this helps!
Vika [28.1K]3 years ago
3 0
7x20=140 making it decrease  per week.
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The Wisconsin Dairy Association is interested in estimating the mean weekly consumption of milk for adults over the age of 18 in
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Answer:

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

±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".

\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

±0.75 ounce

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