Interval Estimation for Population Mean(In case when Population Variance is Unknown and Large Samples)

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In previous section, we have explained the pattern 1, which is the Interval Estimation of population average \mu when the population variance \sigma^2 is a prior knowledge. Although this pattern is the simplest one, which we can first introduce for logical consistency, the applicability conditions may not be enough realistic. The latter is because it is unlikely that we can know the population variance \sigma^2 in advance, as this system feature belongs to the population set and not to the sample set.

 

Then, the next question is whether we can estimate population average only by using information of sample set, excluding population information as prior knowledge.

 

Luckily, the answer to this question is affirmative in condition that the sample size n is enough large (sample size n is more than 30 elements). When the sample size n is enough large, we can estimate population mean \mu by slightly modifying the previously introduced pattern 1.

 

The required modification is simple. Basically, what we have to do is just replace the population standard deviation \sigma in the previous formula (pattern 1) by the sample standard deviation s. Here, the sample standard deviation s is defined to be the standard deviation of the sample randomly picked up from the population set:

s^2 = \frac{1}{n} \sum_{i=1}^n (X_i - \bar{X})^2

 

Then, we can obtain the formula for the estimation of population mean for the case when the population variance is unkown and the sample size is enough large:

\displaystyle P(\bar{X} - 1.96\times \frac{s}{\sqrt{n}} < \mu<\bar{X}+1.96\times\frac{s}{\sqrt{n}})=0.95

 

Here, the parameters in the equation are denoted as follows:

\mu : The mean population(This is what we want to know!)

s : The sample standard deviation(This is easy to compute from the sample set.)

\bar{X} : The sample mean (This is easy to compute from the sample set.)

n : The sample size (We of course know from the starting point. )

P(w)=0.95 : This denotes that the probability to become the state of w is 0.95 (95%).

 

As it can seen in the formula, we can compute the population mean \mu by using only the information from the sample. Hence, the information from the population itself is not necessary at all. !

 

This derived equation can be widely applied in many distinct context.

 

(Exercise) Suppose that we want to know the average hand grip strength of all the 20 years-old male in one country. First, we randomly pick up one hundred people from the whole population and investigate their hand grip strength. The result indicates that this sample average is 40kg and the corresponding sample standard deviation is 5kg. Then, please estimate the average hand grip strength of all the 20 years-male in the country with the confidence level 95%.

(Answer) By inserting the input data of the excercise into the above formula, we obtain:

\displaystyle P(40 - 1.96\times \frac{5}{\sqrt{100}} < \mu<40+1.96\times\frac{5}{\sqrt{100}})=0.95
Then, after a simple calculation, we get:

\displaystyle P(39.02 < \mu< 40.98)=0.95

 

The result of the computation shows that the average hand grip strength \mu of all the 20 years-male in the country is between 39.02(kg) < \mu< 40.98(kg) with the confidence level 95%.

 
*We also explain this page in the following video.

 

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