Mathematics, 15.04.2020 19:58, csnider

A news report states that the 95% confidence interval for the mean number of daily calories consumed by participants in a medical study is (2080, 2260). Assume the population distribution for daily calories consumed is normally distributed and that the confidence interval was based on a simple random sample of 15 observations. Calculate the sample mean, the margin of error, and the sample standard deviation based on the stated confidence interval and the given sample size. Use the t distribution in any calculations and round non-integer results to 4 decimal places.

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answered: stormserena

$\bar X = \frac{2080+2260}{2}=2170$

The margin of error can be founded like this:

$ME = \frac{2260-2080}{2}= 90$

We know that the margin of error is given by:

$ME = t_{\alpha/2} \frac{s}{\sqrt{n}}$

The degrees of freedom are given by:

df = n-1= 15-1 = 14

And the critical value for a 95% of confidence and 14 degrees of freedom is $t_{\alpha/2} = 2.1448$ then we can solve for s like this:

$s= \frac{ME* \sqrt{n}}{t_{\alpha/2}}$

And replacing we got:

$s= \frac{90*\sqrt{15}}{2.1448}= 162.5180$

Step-by-step explanation:

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".

$\bar X$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n 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)

For this case we have the confidence interval given (2080,2260)

We can estimate the sample mean like this:

$\bar X = \frac{2080+2260}{2}=2170$

The margin of error can be founded like this:

$ME = \frac{2260-2080}{2}= 90$

We know that the margin of error is given by:

$ME = t_{\alpha/2} \frac{s}{\sqrt{n}}$

The degrees of freedom are given by:

df = n-1= 15-1 = 14

And the critical value for a 95% of confidence and 14 degrees of freedom is $t_{\alpha/2} = 2.1448$ then we can solve for s like this:

$s= \frac{ME* \sqrt{n}}{t_{\alpha/2}}$

And replacing we got:

$s= \frac{90*\sqrt{15}}{2.1448}= 162.5180$

answered: Guest

1. sometimes

2. never true

3. sometimes

step-by-step explanation:

these are pretty self explanatory if you know what supp right and com angles are

answered: Guest

when functions are first introduced, you will probably have some simplistic "functions" and relations to deal with, usually being just sets of points. these won't be terribly useful or interesting functions and relations, but your text wants you to get the idea of what the domain and range of a function are. small sets of points are generally the simplest sorts of relations, so your book starts with those.

step-by-step explanation:

state the domain and range of the following relation. is the relation a function?

{(2, –3), (4, 6), (3, –1), (6, 6), (2, 3)}

the above list of points, being a relationship between certain x's and certain y's, is a relation. the domain is all the x-values, and the range is all the y-values. to give the domain and the range, i just list the values without duplication:

domain: {2, 3, 4, 6}

range: {–3, –1, 3, 6}

(it is customary to list these values in numerical order, but it is not required. sets are called "unordered lists", so you can list the numbers in any order you feel like. just don't duplicate: technically, repetitions are okay in sets, but most instructors would count off for this.)

while the given set does indeed represent a relation (because x's and y's are being related to each other), the set they gave me contains two points with the same x-value: (2, –3) and (2, 3). since x = 2 gives me two possible destinations (that is, two possible y-values), then this relation is not a function.

note that all i had to do to check whether the relation was a function was to look for duplicate x-values. if you find any duplicate x-values, then the different y-values mean that you do not have a function. remember: for a relation to be a function, each x-value has to go to one, and only one, y-value.

state the domain and range of the following relation. is the relation a function?

{(–3, 5), (–2, 5), (–1, 5), (0, 5), (1, 5), (2, 5)}

i'll just list the x-values for the domain and the y-values for the range: