When estimating a population proportion, p, we must have the same conditions that were used in finding bias and precision, and one new condition as well:

 

The Central Limit Theorem (CLT) for Sample Proportions:

If we take a random sample from a population, and if the sample size is large and the population size much larger than the sample size, then the sampling distribution of  <img class="Wirisformula" style="max-width: none; vertical-align: -8px;" role="math" src="data:;base64,<!–MathML: p^–>@font-face{font-family:’stix_italic7f6ffaa6bb0b408017b6′;src:url(data:font/truetype;charset=utf-8;base64″ alt=”p with hat on top” width=”10″ height=”26″ align=”middle” data-mathml=”«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mover»«mi»p«/mi»«mo»^«/mo»«/mover»«/math»” /> is approximately

<img class="Wirisformula" style="max-width: none; vertical-align: -21px;" role="math" src="data:;base64,<!–MathML: p^~Np” alt=”p with hat on top tilde N open parentheses p comma square root of fraction numerator p open parentheses 1 minus p close parentheses over denominator n end fraction end root close parentheses” width=”156″ height=”55″ align=”middle” data-mathml=”«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mover»«mi»p«/mi»«mo»^«/mo»«/mover»«mo»~«/mo»«mi»N«/mi»«mfenced»«mrow»«mi»p«/mi»«mo»,«/mo»«msqrt»«mfrac»«mrow»«mi»p«/mi»«mfenced»«mrow»«mn»1«/mn»«mo»-«/mo»«mi»p«/mi»«/mrow»«/mfenced»«/mrow»«mi»n«/mi»«/mfrac»«/msqrt»«/mrow»«/mfenced»«/math»” />

If you don’t know the value of p, then you can substitute the value of   <img class="Wirisformula" style="max-width: none; vertical-align: -8px;" role="math" src="data:;base64,<!–MathML: p^–>@font-face{font-family:’stix_italic7f6ffaa6bb0b408017b6′;src:url(data:font/truetype;charset=utf-8;base64″ alt=”p with hat on top” width=”10″ height=”26″ align=”middle” data-mathml=”«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mover»«mi»p«/mi»«mo»^«/mo»«/mover»«/math»” /> to calculate the estimated standard error.

Example 1

A research center surveyed a random 1428 adults in the US and asked whether they own a dog or a cat. The survey found that 357 of them own a cat. What is the approximate probability that the proportion in our sample will be bigger than 29%?

Solution:

<img class="Wirisformula" style="max-width: none; vertical-align: -20px;" role="math" src="data:;base64,<!–MathML: p^=3571428=0.25–>@font-face{font-family:’stix_italic7f6ffaa6bb0b408017b6′;src:url(data:font/truetype;charset=utf-8;base64″ alt=”p with hat on top equals 357 over 1428 equals 0.25″ width=”110″ height=”49″ align=”middle” data-mathml=”«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mover»«mi»p«/mi»«mo»^«/mo»«/mover»«mo»=«/mo»«mfrac»«mn»357«/mn»«mn»1428«/mn»«/mfrac»«mo»=«/mo»«mn»0«/mn»«mo».«/mo»«mn»25«/mn»«/math»” />

25% of the sample own a cat.

Since the sample is large enough, we can use the Central Limit Theorem for sample proportions. By the CLT, we can calculate the standard deviation using the formula:

Here, since we do not know the population proportion, we substitute the sample proportion in the formula instead.

<img class="Wirisformula" style="max-width: none; vertical-align: -21px;" role="math" src="data:;base64,<!–MathML: p1-pn=0.251-0.251428≈0.0115–>@font-face{font-family:’stix_italic7d3c6b5bd3e761d87a24′;src:url(data:font/truetype;charset=utf-8;base64″ alt=”square root of fraction numerator p open parentheses 1 minus p close parentheses over denominator n end fraction end root equals square root of fraction numerator 0.25 open parentheses 1 minus 0.25 close parentheses over denominator 1428 end fraction end root almost equal to 0.0115″ width=”294″ height=”55″ align=”middle” data-mathml=”«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«msqrt»«mfrac»«mrow»«mi»p«/mi»«mfenced»«mrow»«mn»1«/mn»«mo»-«/mo»«mi»p«/mi»«/mrow»«/mfenced»«/mrow»«mi»n«/mi»«/mfrac»«/msqrt»«mo»=«/mo»«msqrt»«mfrac»«mrow»«mn»0«/mn»«mo».«/mo»«mn»25«/mn»«mfenced»«mrow»«mn»1«/mn»«mo»-«/mo»«mn»0«/mn»«mo».«/mo»«mn»25«/mn»«/mrow»«/mfenced»«/mrow»«mn»1428«/mn»«/mfrac»«/msqrt»«mo»§#8776;«/mo»«mn»0«/mn»«mo».«/mo»«mn»0115«/mn»«/math»” />

Therefore, the sample proportion has a normal distribution with mean of 0.25 and the standard deviation of 0.0115.

  <img class="Wirisformula" style="max-width: none; vertical-align: -8px;" role="math" src="data:;base64,<!–MathML: p^~N0.25″ alt=”p with hat on top tilde N open parentheses 0.25 comma space 0.0115 close parentheses” width=”145″ height=”26″ align=”middle” data-mathml=”«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mover»«mi»p«/mi»«mo»^«/mo»«/mover»«mo»~«/mo»«mi»N«/mi»«mfenced»«mrow»«mn»0«/mn»«mo».«/mo»«mn»25«/mn»«mo»,«/mo»«mo»§#160;«/mo»«mn»0«/mn»«mo».«/mo»«mn»0115«/mn»«/mrow»«/mfenced»«/math»” />

Now, we can use the normal calculator to calculate the probability that the proportion in our sample will be bigger than 29%. We need to find

<img class="Wirisformula" style="max-width: none; vertical-align: -8px;" role="math" src="data:;base64,<!–MathML: PX≥0.29=?–>@font-face{font-family:’stix_italic92a147196e7a6d476769′;src:url(data:font/truetype;charset=utf-8;base64″ alt=”P open parentheses X greater or equal than 0.29 close parentheses equals ?” width=”111″ height=”24″ align=”middle” data-mathml=”«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»P«/mi»«mfenced»«mrow»«mi»X«/mi»«mo»§#8805;«/mo»«mn»0«/mn»«mo».«/mo»«mn»29«/mn»«/mrow»«/mfenced»«mo»=«/mo»«mo»?«/mo»«/math»” />

 

 

The probability that the proportion in our sample will be bigger than 29% is 0.0003.

 

 

Example 2.

According to a candy company, packages of a certin candy contain 15% orange candies. Find the approximate probability that the random sample of 500 candies will contain 18% or more orange candies.

Solution:

According to the candy company, the population proportion is 0.15.

 

<img class="Wirisformula" style="max-width: none; vertical-align: -8px;" role="math" src="data:;base64,<!–MathML: p=0.15–>@font-face{font-family:’stix_italic7f6ffaa6bb0b408017b6′;src:url(data:font/truetype;charset=utf-8;base64″ alt=”p equals 0.15″ width=”55″ height=”24″ align=”middle” data-mathml=”«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»p«/mi»«mo»=«/mo»«mn»0«/mn»«mo».«/mo»«mn»15«/mn»«/math»” />

Since the sample size 500 is large enough, we can use the Central Limit Theorem for sample proportions. We first calculate the standard deviation by the formula:

 

<img class="Wirisformula" style="max-width: none; vertical-align: -21px;" role="math" src="data:;base64,<!–MathML: p1-pn=0.151-0.15500≈0.0160–>@font-face{font-family:’stix_italic7d3c6b5bd3e761d87a24′;src:url(data:font/truetype;charset=utf-8;base64″ alt=”square root of fraction numerator p open parentheses 1 minus p close parentheses over denominator n end fraction end root equals square root of fraction numerator 0.15 open parentheses 1 minus 0.15 close parentheses over denominator 500 end fraction end root almost equal to 0.0160″ width=”294″ height=”55″ align=”middle” data-mathml=”«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«msqrt»«mfrac»«mrow»«mi»p«/mi»«mfenced»«mrow»«mn»1«/mn»«mo»-«/mo»«mi»p«/mi»«/mrow»«/mfenced»«/mrow»«mi»n«/mi»«/mfrac»«/msqrt»«mo»=«/mo»«msqrt»«mfrac»«mrow»«mn»0«/mn»«mo».«/mo»«mn»15«/mn»«mfenced»«mrow»«mn»1«/mn»«mo»-«/mo»«mn»0«/mn»«mo».«/mo»«mn»15«/mn»«/mrow»«/mfenced»«/mrow»«mn»500«/mn»«/mfrac»«/msqrt»«mo»§#8776;«/mo»«mn»0«/mn»«mo».«/mo»«mn»0160«/mn»«/math»” />

Therefore the sample proportion has a normal distribution with mean of 0.15 and a standard deviation of 0.016.

<img class="Wirisformula" style="max-width: none; vertical-align: -8px;" role="math" src="data:;base64,<!–MathML: p^~N0.15″ alt=”p with hat on top tilde N open parentheses 0.15 comma space 0.0160 close parentheses” width=”145″ height=”26″ align=”middle” data-mathml=”«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mover»«mi»p«/mi»«mo»^«/mo»«/mover»«mo»~«/mo»«mi»N«/mi»«mfenced»«mrow»«mn»0«/mn»«mo».«/mo»«mn»15«/mn»«mo»,«/mo»«mo»§#160;«/mo»«mn»0«/mn»«mo».«/mo»«mn»0160«/mn»«/mrow»«/mfenced»«/math»” />

Now, we can use the normal calculator to find the probability that the sample will contain 18% or more orange candies. We need to find:

<img class="Wirisformula" style="max-width: none; vertical-align: -8px;" role="math" src="data:;base64,<!–MathML: P0.18≤X=?–>@font-face{font-family:’stix_italic92a147196e7a6d476769′;src:url(data:font/truetype;charset=utf-8;base64″ alt=”P open parentheses 0.18 less or equal than X close parentheses equals ?” width=”111″ height=”24″ align=”middle” data-mathml=”«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»P«/mi»«mfenced»«mrow»«mn»0«/mn»«mo».«/mo»«mn»18«/mn»«mo»§#8804;«/mo»«mi»X«/mi»«/mrow»«/mfenced»«mo»=«/mo»«mo»?«/mo»«/math»” />

So, the approximate probability that the random sample of 500 candies will contain 18% or more orange candies is 0.0304. In another words, there is almost a 3% chance that a sample of 500 candies will contain 18% orange candies.