What is the approximate value of z for a 72 percent level of confidence when determining a proportion?

To determine the approximate value of z for a 72 percent level of confidence in the context of statistics, we begin by understanding what this confidence level represents.

A confidence level of 72% means that we are willing to accept a 28% chance of being incorrect in our estimate of a proportion. In statistical terms, this implies that there is a 72% probability that our true parameter will fall within the confidence interval we calculate.

To find the corresponding z-score, we can refer to the standard normal distribution (also known as the z-distribution). The z-score represents the number of standard deviations a data point is from the mean, and it can be used to determine confidence intervals.

Given a confidence level of 72%, we know that the area in the tails of the normal distribution is:

  • 100% – 72% = 28%

Since this area is split between the two tails of the distribution, we divide by 2, which gives us:

  • 28% / 2 = 14% (0.14)

To find the z-score that corresponds to an upper tail probability of 0.14, we can refer to a standard normal (z) table or use statistical software. Looking up the cumulative area of 0.86 (because 1 – 0.14 = 0.86) in the z-table reveals a z-score of approximately:

  • z ≈ 1.04

In summary, for a 72 percent level of confidence in estimating a proportion, the approximate value of z is 1.04. It is important to note that this value can be used to construct confidence intervals for proportions using the formula:

  • CI = p ± z * √(p(1-p)/n)

where p is the sample proportion and n is the sample size.

Leave a Comment