Suppose the investigators had made a rough guess of 330 for the value of s before collecting data.
What sample size would be necessary to obtain an interval width of 50 ml for a confidence level of 95%?

Respuesta :

Answer:

[tex]n=(\frac{1.960(330)}{25})^2 =669.36 \approx 670[/tex]

So the answer for this case would be n=670 rounded up to the nearest integer

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".

[tex]\bar X[/tex] represent the sample mean for the sample  

[tex]\mu[/tex] population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

Solution to the problem

The margin of error is given by this formula:

[tex] ME=z_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]    (a)

And on this case sicne we want a confidence interval with a wisth of 50 ml then we have that ME =50/2=25 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

[tex]n=(\frac{z_{\alpha/2} \sigma}{ME})^2[/tex]   (b)

We can use an estimator of the population deviation the sample deviation [tex]\hat \sigma = s = 330[/tex]

The critical value for 95% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.025;0;1)", and we got [tex]z_{\alpha/2}=1.960[/tex], replacing into formula (b) we got:

[tex]n=(\frac{1.960(330)}{25})^2 =669.36 \approx 670[/tex]

So the answer for this case would be n=670 rounded up to the nearest integer