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Liono4ka [1.6K]
3 years ago
8

Examine the required sample size needed to be able to

Mathematics
1 answer:
Xelga [282]3 years ago
3 0

Answer:

n=(\frac{2.33(500)}{10})^2 =13572.25 \approx 13573

And if we use a sample level of n =2000 the margin of error would be higher as we can see here:

ME=2.33\frac{500}{\sqrt{2000}}=26.05  

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)

\sigma=500 represent the population standard deviation assumed

n represent the sample size  (variable of interest)

Confidence =98% or 0.98

ME = 10 represent the margin of error desired

The margin of error is given by this formula:

ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}    (a)

And on this case we have that ME =10 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

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

The critical value for 98% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.01;0;1)", and we got z_{\alpha/2}=2.33, replacing into formula (b) we got:

n=(\frac{2.33(500)}{10})^2 =13572.25 \approx 13573

So the answer for this case would be n=13573 rounded up to the nearest integer

And if we use a sample level of n =2000 the margin of error would be higher as we can see here:

ME=2.33\frac{500}{\sqrt{2000}}=26.05    

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