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Ch 8.3 Confidence Interval for Population Proportion

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    15923
  • Ch 8.3  Confidence Interval for population proportion:

    Terms: Population proportion: p

                  Sample proportion: \( \hat{p} \) = (x/n)

                  Number of success: x;  sample size: n

                  Confidence Level: C-level

                  Significant Level: α = 1 – C-level  (probability of unlikely)

                  EBP: Error Bound for proportion, margin of error

    A) To Estimate p:

    \( \fbox {1) point estimate: \( \hat{p} = x/n \) }  \)

    \( \fbox{ 2) Interval estimate: \( \hat{p} - E \text{ to } \hat{p} + E \) } \)

    \( E(EBP) = z_ {\alpha /2} \sqrt{\frac{\hat{p}\hat{q}}{n}} \)

    Note: E is the approximated value of sampling error of sampling distribution of \( \hat{p} \), which has a normal distribution when x and n-x  ≥ 5.

    Use Statdisk online calculator to find the confidence interval.

    - identify C-level, sample size n and success = x.

    - https://www.statdisk.com/# , Analysis/ Confidence Intervals/Proportion one sample/

    output: E (EBP) and   lower < p < upper

     

     \( z_ {\alpha /2} \) is the critical value with C-level in the middle.

     critical valueTo find  \( z_ {\alpha /2} \)

    Use online Inverse Normal calculator, set area = \( \alpha /2 \), mean = 0, sd =1,

                                    click above, recalculate.

    Explanation:

    X = number of success is a binomial distribution with mean = np and sd = \( \sqrt{npq} \). 

    When np and nq ≥ 5, distribution of X is normal with mean = np and sd = \( \sqrt{npq} \) .

    So  distribution of  \( \hat{p} = \frac{x}{n} \) is normal with mean = p and SD =  \( \sqrt{\frac{\hat{p}\hat{q}}{n}} \) .

     At a given C-level, the maximum error of \( \hat{p} \) and p is E where  \( E(EBP) = z_ {\alpha /2} \sqrt{\frac{\hat{p}\hat{q}}{n}} \)

    Confidence interval

    Note: the requirement for this confidence interval is not n > 30 but np and nq ≥  5.

     

    Interpret a 95% Confidence Interval:

    We are 95% confidence that the interval from ____ to ____   actual contain the true value of the population proportion of the category of interest.

    The Confidence interval also shows that 95% of all confidence Intervals contain the true value of p.

    Make conclusion from confidence interval:

    1) Any value in the confidence Interval can be p.

    2) If the whole interval > a, we can conclude p > a

    3) If the whole interval < a, we can conclude p < a.

    4) When two confidence interval overlap, we can conclude that the two p may be the same. We cannot conclude one of the p is higher.

     

     Ex1. A research is conducted to determine how many household use Netflix to stream videos. A random sample of 500 households show that 442 households use Netflix.

    a) Use a 90% confidence level to compute a confidence interval estimate of true proportion of households using Netflix.

    use Statdisk Analysis/Confidence Intervals/Proportion one sample : Clevel = 0.9, n = 500, x = 442,

    confidence interval output

    b) Find the critical value. 

    Use Inverse Normal calculator, since clevel = 90%,  \( \alpha = 0.1,  \alpha /2  = 0.05 \), set area = 0.05, mean = 0, sd = 1,

    click above, recalculate.   

     \( z_ {\alpha /2} \) = 1.645

    c) Interpret the confidence interval in non-technical term.

     We estimate with 90% confidence that the true proportion of all households that use Netflix is between 86.0% to 90.8%

     

    d) Can we conclude with 90% confidence that more than 80% of households use Netflix?

      Since the interval contains 86.0% to 90.8%, all values are more than 80%, so we can conclude that.

     

    Ex2: A Gallup poll of 1487 adults showed that 43% of the respondents have Facebook pages.

    a) Find the number in the sample who have Facebook pages.   \( x = n(\hat{p}) \) = 1487 (0.43) = 639

    c) Find the interval estimate of p at 95% confidence level  and margin of error E.

     Use statdisk with Clevel = 0.95, n = 1487, x = 639

     Interval estimate is  0.405 < p < 0.455, E = 0.025

    d) Write a non-technical interpretation of the above

    We are 95% confidence that the true percent of adults who have Facebook pages are between 40.5% to 45.5%.

    e) Can we claim that less than 60% of all adults have Facebook pages?

    Since the whole interval is less than 0.6, yes, we can conclude less than 60% of all adults have Facebook pages.

     

    B) Determine sample size for a desired E

    Since   \( E(EBP) = z_ {\alpha /2} \sqrt{\frac{\hat{p}\hat{q}}{n}} \)

    \( n = \frac{(z_ {\alpha/2})^2\hat{p}\hat{q}}{E^2}  \) when an estimate \( \hat{p} \) is known.

    \( n = \frac{(z_ {\alpha/2})^2\cdot 0.25}{E^2}  \)     when no estimate \( \hat{p} \)  is known.  

     

    Use statdisk / Analysis/Sample size Determination/ Estimate proportion to find sample size for a given error E.

    Input C-Level,  Desired E,  Estimate of p = \( \hat{p} = \frac{x}{n} \), evaluate.

     

    Ex3. What sample size should be used if we want the keep the margin of error within 2.1% when estimating a proportion at a 90% confidence interval. use \(\hat{p} = 0.43 \) as an old estimation of p.

    C-level = 0.9,  E = 0.021,  estimate of p = 0.43, Evaluate

    Sample size p = 1504.

     

    b) If no previous study has been done , what sample size will be needed. (do not use p-hat = 0.43)

    Use statdisk / Analysis/Sample size Determination/ Estimate proportion

    C-level = 0.9,  E = 0.021,  estimate of p = blank, evaluate

    sample size p = 1534 

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