Confidence Interval for Population Proportion

There were \(20\) cocoa pods sampled and \(8\) out of those were diseased. This gives you a population proportion of \(40\%\).

• Nope! What this does tell you is that “about” \(40\%\) of them are diseased.

So then, what does “about” mean in technical terms?

Well, it depends on how confident you want to be.

• The confidence interval for the population proportion gives you a range of values near \(40\%\) that you can say the actual percentage of diseased pods is in.
• The size of the interval will be smaller if you want to be more confident, and it will be larger if you are willing to be less confident.

In the cocoa example at the start of the article, \(20\) cocoa pods were sampled and \(8\) out of those were diseased.

In this context of this example, a success is a pod being diseased. So,

\[ \begin{align}\hat{p} &= \frac{\text{number of successes}}{\text{sample size}} \\&= \frac{8}{20} \\ &= 0.4.\end{align}\]

In the cocoa example at the start of the article, \(20\) cocoa pods were sampled and \(8\) out of those were diseased. What is the standard error of the population proportion?

Solution:

For this example, \(n = 20\) and you have already calculated that \(\hat{p} = 0.4\). Using the formula,

\[\begin{align}\sigma_{\hat{p}} & = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \\&= \sqrt{\frac{0.4(1-0.4)}{20}} \\&= \sqrt{0.12} \\&= 0.1095\end{align}\]

rounded to \(4\) decimal places.

In the cocoa example at the start of the article, \(20\) cocoa pods were sampled and \(8\) out of those were diseased. In the example from the “Standard Error” section, you have found that \( \hat{p} = 0.4 \) and the standard error is about \( 0.1095 \). Find the margin of error for the

1. \(90\%\),
2. \(95\%\), and
3. \(99\%\) confidence levels.

Solution:

Use the critical values at the \(90\%\), \(95\%\), and \(99\%\) confidence levels listed in the table above.

1. For the \(90\%\) confidence level, the critical value is \(1.645\), so\[\begin{align}\text{margin of error for } 90\% &= (\text{critical value})(\text{standard error}) \\&= (1.645)(0.1095) \\&\approx 0.18.\end{align}\]
2. Similarly, for the \(95\%\)confidence level, the critical value is \(1.96\), so\[\begin{align}\text{margin of error for } 95\% &= (\text{critical value})(\text{standard error}) \\&= (1.96)(0.1095) \\&\approx 0.21.\end{align}\]
3. Finally, for the \(99\%\)confidence level, the critical value is \(1.96\), so\[\begin{align}\text{margin of error for } 99\% &= (\text{critical value})(\text{standard error}) \\&= (2.58)(0.1095) \\&\approx 0.54.\end{align}\]

Let's put the results for the \(95\%\) confidence level into words.

• Remember that \(40\%\) of the pods were found to be diseased. What the margin of error tells you is that at the \(95\%\) confidence level, you can be \(95\%\) sure that for any random sample taken of the cocoa pods, that the percentage of diseased pods will be \(40\%\) with a \(21\%\) margin of error.

Comparing the results from the three confidence levels, notice that the margin of error goes up as the confidence level goes up! In other words, the more confident you want to be about the result, the larger your error might be.

In the example about cocoa pods, you don't know how the data is gathered. So, you can't tell whether the data is representative or not. If you do any statistical analysis based on this data, you will need to say something like:

No information is given about how the sample was selected. Therefore, the results are only valid if the sample selected was representative of the overall population.

On a cocoa farm, Indodo, the owner of the farm, sampled \(20\) cocoa pods and realized that \(8\) out of those were diseased. Determine if the sample size is large enough to find an appropriate confidence interval.

Solution:

Remember that the population proportion is \( \hat{p} = 0.4 \). Therefore,

\[ n \hat{p} = (20)(0.4) = 8 \]

and

\[ n (1 - \hat{p}) = (20)(0.6) = 12.\]

Since \( n \hat{p} < 10 \), this data does not meet the requirements for determining an appropriate confidence interval. That is why the margin of error in the previous example was so large!

Assuming that Indodo is doing a random sample, and he finds that \( \hat{p} = 0.4 \) every time, how large does his sample need to be to say that the sample is large enough?

Solution:

1. You are trying to find the sample size \(n\) that gives you both\[ n \hat{p} \geq 10 \]and\[ n (1 - \hat{p}) \geq 10. \]
2. Using \( \hat{p} = 0.4 \), that means you need\[ \begin{align}0.4 n &\geq 10 \\n &\geq \frac{10}{0.4} \\n &\geq 25.\end{align} \]and\[ \begin{align}n (1 - 0.4) &\geq 10 \\n (0.6) &\geq 10 \\n &\geq \frac{10}{0.6} \\n &\geq 17.\end{align} \]

Choosing the larger of the two values for \(n\), Indodo needs to sample at least \(25\) pods to make sure the sample is large enough to find an appropriate confidence interval.

Indodo has decided to do another sample of cocoa pods on his farm. He samples \(100\) of them, and finds that \(25\) of them are diseased. Based on that data, what can you learn about the proportion of cocoa pods that are diseased on the entire farm?

Solution:

1. E: Estimate– Explain what population characteristic you plan to estimate.
   • You will estimate the value of \( \hat{p} \), the proportion of cocoa pods on the farm that are diseased.

2. M: Method – Decide which statistical inference method you want to use.

   • Because:

     • the question is asking you for an estimation,

     • the situation involves using sample data,

     • the type of data involved is one categorical variable, and

     • there is only one sample,

     • you can use the confidence intervals for a population proportion method.

   • Because a confidence level is associated with the confidence interval, you need to specify a confidence level for the problem. A confidence level isn't given, so use a confidence level of \(95\%\).

3. C: Check – There are \(2\) conditions that must be met in order to use the confidence interval of a population proportion:

   1. The data must be truly representative of the population.

      • It isn't specifically stated how Indodo picked out the pods for the sample, so you don't know that the data is representative.
        • That means you will need to assume that the data is representative of the population and include a statement about not having this information when you present the results.

   2. The sample size must be large enough.For this example, a success is defined as finding a diseased pod. Since he sampled \(100\) pods, \( n = 100 \) and\[ \begin{align}\hat{p} &= \frac{ \text{number of successes} }{ \text{sample size} } \\&= \frac{25}{100} = 0.25.\end{align} \]So,\[ \begin{align}n \hat{p} &= (100)(0.25) \\&= 25 \geq 10\end{align} \]and\[ \begin{align}n (1 - \hat{p}) &= 100 (1 - 0.25) \\&= 100 (0.75) \\&= 75 \geq 10.\end{align} \] Because both checks for the required sample size are greater than or equal to \(10\), the sample size is large enough.

4. C: Calculate– Use the formula to calculate the confidence interval.
   1. The sample size is \( n = 100 \).
   2. The population proportion is \( \hat{p} = 0.25 \).
   3. The confidence level is \( 95\% \), so the critical value of the confidence level is \( \text{critical value} = 1.96 \).
   4. Find the standard error.\[ \begin{align}\text{standard error } &= \sqrt{ \frac{ \hat{p} (1 - \hat{p}) }{n} } \\&= \sqrt{ \frac{0.25 (1 - 0.25)}{100} } \\&= \sqrt{0.001875} \\&\approx 0.0433\end{align} \]
   5. Find the margin of error.Using the critical value for the \(95\%\) confidence level, the margin of error is:\[ \begin{align}\text{margin of error for } 95\% &= (\text{critical value})(\text{standard error}) \\&= (1.96) (0.0433) \\&\approx 0.084.\end{align} \]
   6. Find the confidence interval.Now you can construct the confidence interval using:\[ \hat{p} \pm \text{margin of error} = 0.25 \pm 0.085, \]so the interval is:\[ (0.25 - 0.085, 0.25 + 0.085 ) = (0.165, 0.335). \]
5. C: Communicate – Answer the question asked in the problem, stating what you learned from the data and addressing any potential risks or shortcomings.
   1. First, a statement about the confidence level.
      • The method used to construct the confidence interval will ensure that the actual population proportion is contained in the confidence interval about \(95\%\) of the time.
      • No information is given about how the sample was selected. Therefore, the results are only valid if the sample selected was representative of the overall population.
   2. Then, a statement about the results with regard to the actual problem.
      • If the sample was selected reasonably, you can be \(95\%\) confident that the actual proportion of diseased cocoa pods is somewhere between \(0.165\) and \(0.335\).
      • In terms of percentages, you can be \(95\%\) confident that the actual percentage of diseased cocoa pods is somewhere between \(16.5\%\) and \(33.5\%\).

You are studying relocation patterns of U.S. adults aged \(21\) years or older who moved back home or in with friends during the previous year. You conducted a survey of \(843\) U.S. adults age \(21\) or older, and \(62\) of them reported that in the previous year they had moved in with friends or relatives. Based on these data, what can you learn about the proportion of all U.S. adults aged \(21\) years or older who moved in with friends or relatives during the previous year?

Solution:

1. E: Estimate – Explain what population characteristic you plan to estimate.

   • You are estimating the value of \( \hat{p} \), the proportion of U.S. adults age \(21\) or older who have moved in with friends or relatives in the past year.

2. M: Method – Decide which statistical inference method you want to use.

   • Because:

     • the question is asking you for an estimation,

     • the situation involves using sample data,

     • the type of data involved is one categorical variable, and

     • there is only one sample,you can use the confidence intervals for a population proportion method.

   • Because a confidence level is associated with the confidence interval, you need to specify a confidence level for the problem. A confidence level isn't given, so use a confidence level of \(95\%\).

3. C: Check – There are \(2\) conditions that must be met in order to use the confidence interval of a population proportion:

   1. The data must be truly representative of the population.

      • It isn't specifically stated how the sample was selected, so you don't know that the data is representative.
        • That means you will need to assume that the data is representative of the population and include a statement about not having this information when you present the results.

   2. The sample size must be large enough.

      • For this example, a success is defined as an adult moving in with friends or relatives. Since the sample included \(843\) adults, \( n = 843 \) and\[ \begin{align}\hat{p} &= \frac{ \text{number of successes} }{ \text{sample size} } \\&= \frac{62}{843} \approx 0.0735.\end{align} \]So,\[ \begin{align}n \hat{p} &= (843)(0.0735) \\&= 62 \geq 10\end{align} \]and\[ \begin{align}n (1 - \hat{p}) &= 843 (1 - 0.0735) \\&= 843 (0.9265) \\&= 781.0395 \geq 10.\end{align} \] Because both checks for the required sample size are greater than or equal to \(10\), the sample size is large enough.

4. C: Calculate – Use the formula to calculate the confidence interval.

   1. The sample size is \( n = 843 \).
   2. The population proportion is \( \hat{p} = 0.0735 \).
   3. The confidence level is \( 95\% \), so the critical value of the confidence level is \( \text{critical value} = 1.96 \).
   4. The confidence interval is:\[ \begin{align}\hat{p} &\pm (\text{critical value}) \sqrt{ \frac{ \hat{p} (1 - \hat{p}) }{n} } \\0.0735 &\pm 1.96 \sqrt{ \frac{(0.0735) (1 - 0.0735)}{843} } \\0.0735 &\pm 1.96 \sqrt{0.00008078} \\0.0735 &\pm 0.0176\end{align} \] Written in interval notation, you have:\[ \begin{align}\text{confidence interval} &= (0.0735 - 0.0176, 0.0735 + 0.0176) \\&= (0.0559, 0.0911)\end{align} \]

5. C: Communicate – Answer the question asked in the problem, stating what you learned from the data and addressing any potential risks or shortcomings.

   • Confidence interval:If the sample was selected such that it truly represents the population, you can be \(95\%\) confident that the actual proportion of U.S. adults aged \(21\) years or older who moved back home or in with friends during the previous year is somewhere between \(0.0559\) and \(0.0911\).

   • Confidence level:The method you used to determine the interval estimate is successful in capturing the actual value of the population proportion approximately \(95\%\) of the time.No information is given about how the sample was selected. Therefore, the results are only valid if the sample selected was representative of the overall population.

In a study involving \(10,000\) parents, \(40\%\) of parents between the ages of \(18\) to \(34\) years created a social media account for their babies. Assuming this population is representative, determine the confidence intervals for a population proportion with \(90\%\), \(95\%\), and \(99\%\) confidence levels.

Solution:

1. E: Estimate – Explain what population characteristic you plan to estimate.

   • You are estimating the value of \( \hat{p} \), the proportion of parents between the ages of \(18\) to \(34\) years who created a social media account for their babies.

2. M: Method – Decide which statistical inference method you want to use.

   • Because:

     • the question is asking you for an estimation,

     • the situation involves using sample data,

     • the type of data involved is one categorical variable, and

     • there is only one sample,you can use the confidence intervals for a population proportion method.

   • Confidence levels of \(90\%\), \(95\%\), and \(99\%\) were specified.

3. C: Check – There are \(2\) conditions that must be met in order to use the confidence interval of a population proportion:

   1. The data must be truly representative of the population.

      • You were told to assume the sample population is representative.
        • That means you will need to include a statement about how the results are only valid if the sample selected was representative of the overall population when you present the results.

   2. The sample size must be large enough.

      • For this example, a success is defined as a parent creating a social media profile for their baby. Since the sample included \(10000\) parents, \( n = 10000 \) and\[ \begin{align}\hat{p} &= \frac{ \text{number of successes} }{ \text{sample size} } \\&= \frac{(10000)(0.40)}{10000} = 0.4.\end{align} \]So,\[ \begin{align}n \hat{p} &= (10000)(0.4) \\&= 4000 \geq 10\end{align} \]and\[ \begin{align}n (1 - \hat{p}) &= 10000 (1 - 0.4) \\&= 10000 (0.6) \\&= 6000 \geq 10.\end{align} \] Because both checks for the required sample size are greater than or equal to \(10\), the sample size is large enough.

4. C: Calculate – Use the formula to calculate the confidence interval.

   1. The sample size is \( n = 10000 \).
   2. The population proportion is \( \hat{p} = 0.4 \).
   3. The confidence levels are:
      • \( 90\% \), and the critical value of \( 90\% \) is \( \text{critical value} = 1.645 \).
      • \( 95\% \), and the critical value of \( 95\% \) is \( \text{critical value} = 1.96 \).
      • \( 99\% \), and the critical value of \( 99\% \) is \( \text{critical value} = 2.58 \).
   4. The confidence intervals are:
      • For a confidence level of \( 90\% \):\[ \begin{align}\hat{p} &\pm (\text{critical value}) \sqrt{ \frac{ \hat{p} (1 - \hat{p}) }{n} } \\0.4 &\pm 1.645 \sqrt{ \frac{(0.4) (1 - 0.4)}{10000} } \\0.4 &\pm 1.645 \sqrt{0.000024} \\0.4 &\pm 0.008\end{align} \] Written in interval notation, you have:\[ \begin{align}\text{confidence interval} &= (0.4 - 0.008, 0.4 + 0.008) \\&= (0.392, 0.408)\end{align} \]
      • For a confidence level of \( 95\% \):\[ \begin{align}\hat{p} &\pm (\text{critical value}) \sqrt{ \frac{ \hat{p} (1 - \hat{p}) }{n} } \\0.4 &\pm 1.96 \sqrt{ \frac{(0.4) (1 - 0.4)}{10000} } \\0.4 &\pm 1.96 \sqrt{0.000024} \\0.4 &\pm 0.0096\end{align} \] Written in interval notation, you have:\[ \begin{align}\text{confidence interval} &= (0.4 - 0.0096, 0.4 + 0.0096) \\&= (0.3904, 0.4096)\end{align} \]
      • For a confidence level of \( 99\% \):\[ \begin{align}\hat{p} &\pm (\text{critical value}) \sqrt{ \frac{ \hat{p} (1 - \hat{p}) }{n} } \\0.4 &\pm 2.58 \sqrt{ \frac{(0.4) (1 - 0.4)}{10000} } \\0.4 &\pm 2.58 \sqrt{0.000024} \\0.4 &\pm 0.0126\end{align} \] Written in interval notation, you have:\[ \begin{align}\text{confidence interval} &= (0.4 - 0.0126, 0.4 + 0.0126) \\&= (0.3874, 0.4126)\end{align} \]

5. C: Communicate – Answer the question asked in the problem, stating what you learned from the data and addressing any potential risks or shortcomings.

   • Confidence interval:Assuming the sample is truly representative of the population, for a \(90\%\) confidence level, the actual value should be within \(39.2\%\) and \(40.8\%\). For a \(95\%\) confidence level, the actual value should be within \(39.04\%\) and \(40.96\%\). Meanwhile, for a \(99\%\) confidence level, the actual value is expected to fall within \(38.74\%\) and \(41.26\%\).What can you make out of the above result, which applies three confidence levels?

     • You can tell that the \(90\%\) confidence level has the smallest range of \(1.6\%\) (from \(40.8\% - 39.2\%\)), followed by \(95\%\) with a range of \(2\%\), and lastly, \(99\%\) with a range of \(2.6\%\).

     • A smaller interval size means that you are closer to the actual value; however, a lower confidence level means a reduced assurance of the accuracy or precision that the actual value is found there. Now you see, in as much as you wish to have more assurance (as given by the \(99\%\) confidence level), you would prefer that the interval is as small as possible to narrow the interval closer to the actual value (as given by the \(90\%\) confidence level). Hence, it is REASONABLY assuring to rely on the \(95\%\) confidence level.

   • Confidence level:The method you used to determine the interval estimate is successful in capturing the actual value of the population proportion approximately \(90\%\), \(95\%\), or \(99\%\) of the time, depending on which interval you choose to consider.You were told to assume the sample was truly representative of the population. Therefore, the results are only valid if the sample selected was actually representative of the overall population.

Mary and her twin sister Elizabeth embarked separately on a random survey in the same area involving the support to build a pilot school. Mary's confidence intervals for the population proportion are \((0.34, 0.41)\), and those of Elizabeth are \((0.37, 0.39)\).

1. What explanation can be given to the difference in the confidence intervals, even within the same area?
2. Whose confidence interval is of higher precision?
3. Assuming both had a \(95\%\) confidence level, determine which result was derived from a smaller sample size with reasons.
4. Assuming both had used the same sample size, determine who would have used a higher confidence level and give your justification.

Solution:

1. Although both individuals had worked in the same sample area, there are several factors which may affect the uniformity of their results.
   1. Firstly, they could have worked with different sample sizes. Remember that a larger sample size would mean a lower margin of error. Hence, the interval difference would be smaller.
   2. Another factor is the confidence level. If a higher confidence level is used, even with the same population size, the boundary between the intervals would be wider, meaning less precision. However, a lower confidence level would give a smaller boundary between intervals, meaning more precision but at the expense of the level of assurance.
2. From the intervals, the size of the boundary would be calculated to determine the degree of precision.In Mary's case:\[ 0.41 - 0.34 = 0.07 \]In Elizabeth's case:\[ 0.39 - 0.37 = 0.02 \]
   • Knowing that a smaller boundary size means a higher precision, you can say that Elizabeth's result is more precise.
3. If both conducted their survey with a \(95\%\) confidence level, then the sample size becomes the sole basis upon which precision is determined. A larger sample size would mean more precision because of the smaller margin of error. Since Elizabeth's result has more precision, it means she worked with a larger sample size than Mary.
   • Therefore, Mary's sample size was smaller.
4. If both conducted their survey with the same sample size, the confidence level becomes the sole basis in determining precision. With Elizabeth's result being more precise, it means a lower confidence limit was used.
   • Therefore, Mary would have used a higher confidence level.