Question

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.

Answer 1

Answer:

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

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:

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