Construct A 95% Confidence Interval For The Population Propo

Construct a 95% confidence interval for the population proportion of people in New Hampshire with cholesterol levels over 200

You are told that a random sample of 150 people from Manchester, New Hampshire, have been given cholesterol tests, and 60 of these people had levels over the “safe” count of 200. Using Excel, construct a 95% confidence interval for the population proportion of people in New Hampshire with cholesterol levels over 200. This involves calculating a confidence interval for a proportion, which estimates the range within which the true proportion of the entire population is likely to fall, based on the sample data.

The formula for a confidence interval for a population proportion (p) is:

p̂ ± zcritical × sqrt([p̂(1 - p̂)] / n)

Where:

  • p̂ = sample proportion = number with high cholesterol / total sample size
  • zcritical = the z-value corresponding to the desired confidence level (for 95%, z ≈ 1.96)
  • n = sample size

Since this is about a population proportion, the standard deviation of the population is not needed, and the z-multiplier is used instead of the t-multiplier.

Paper For Above instruction

Constructing a confidence interval for a population proportion provides valuable insight into the likely range of the true proportion of individuals with a particular characteristic—in this case, high cholesterol levels—in a wider population. The sample data from Manchester, New Hampshire, indicating that 60 out of 150 people (p̂ = 0.4) have cholesterol levels over 200, forms the basis for estimating this range at a 95% confidence level.

Calculating the sample proportion (p̂) is straightforward:

p̂ = 60 / 150 = 0.4

Next, identifying the critical z-value for a 95% confidence level involves consulting standard normal distribution tables or using Excel functions. In Excel, the z-value can be obtained using the formula:

=NORMSINV(1 - (1 - confidence_level)/2)

For 95% confidence:

=NORMSINV(0.975) ≈ 1.96

This z-value reflects the number of standard deviations to capture 95% of the normal distribution centered around the sample proportion.

Calculating the standard error (SE) involves the formula:

SE = sqrt([p̂(1 - p̂)] / n) = sqrt[0.4 × 0.6 / 150] ≈ sqrt[0.24 / 150] ≈ sqrt[0.0016] ≈ 0.04

The margin of error (ME) is then:

ME = z × SE = 1.96 × 0.04 ≈ 0.0784

The confidence interval is therefore:

[p̂ - ME, p̂ + ME] = [0.4 - 0.0784, 0.4 + 0.0784] = [0.3216, 0.4784]

This means we are 95% confident that the true proportion of all people in Manchester, NH, with cholesterol levels over 200 falls between approximately 32.16% and 47.84%.

Constructing this interval in Excel demonstrates the practical application of statistical theory. The process involves calculating the sample proportion, the standard error, and then applying the z-multiplier to find the margin of error. Using Excel functions such as =NORMSINV(0.975) for the z-value simplifies the process and ensures precise results.

Understanding confidence intervals allows public health officials and clinicians to assess the prevalence of high cholesterol within a population accurately. Such estimates can lead to better targeted health interventions and resource allocation. They also underscore the importance of large sample sizes for more precise estimates, as the width of the confidence interval decreases with increasing n.

References

  • Agresti, A., & Coull, B. A. (1998). Approximate is better than "exact" for interval estimation of binomial proportions. The American Statistician, 52(2), 119-126.
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