Mathematics, 22.04.2020 00:54 struckedblazing

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: 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? Question 16 options: ±12.996 ounces ±0.75 ounce ±0.456 ounce z = 1.645

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Answer from: nessuhbae6722

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

±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

Answer from: rileigh2302

The approximate margin of error in the estimate is ±0.75 ounces.

Step-by-step explanation:

The question is incomplete:

The following results were recorded: xbar=34.5 ounces, s=7.9 ounces.

The sample size is n=300.

We will use the sample standard deviation to estimate the population standard deviation, so we will use the t-statistic.

To develop a confidence interval, we first have to calculate the degrees of freedom, and then look up in a t-students distribution table the critical value for a 90% confidence interval.

The degrees of freedom are:

df=n-1=300-1=299

The critical value for a 90% CI is t=1.65.

Now, we can calculate the margin of error of the confidence interval as:

$E=t\cdot s/\sqrt{n}=1.65*7.9/\sqrt{300}=13.035/17.32=0.75$

The lower and upper bounds of the confidence interval will be:

$LL=\bar x-t\cdot s/\sqrt{n}=34.5-0.75=33.75\\\\UL=\bar x+t\cdot s/\sqrt{n}=34.5+0.75=35.25$

The confidence interval is (33.75, 35.25)

The Wisconsin Dairy Association is interested in estimating the mean weekly consumption of milk for

Answer from: Quest

answer: 100

step-by-step explanation:

Answer from: Quest

18.50+s=30

100% correct no other way you can put it

question answered by

(jacemorris04)

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