High School

A random sample of size 50 drawn from a population has a mean of 46.5 and a standard deviation of 13.86. What is a 90% confidence interval for the population mean?

Answer :

To calculate a 90% confidence interval for the population mean based on a random sample, we can use the formula for the confidence interval of the mean when the population standard deviation is unknown:

[tex]\text{Confidence Interval} = \bar{x} \pm \left( t_{\alpha/2, n-1} \times \frac{s}{\sqrt{n}} \right)[/tex]

Where:

  • [tex]\bar{x}[/tex] is the sample mean, which is 46.5 in this case.
  • [tex]s[/tex] is the sample standard deviation, which is 13.86.
  • [tex]n[/tex] is the sample size, which is 50.
  • [tex]t_{\alpha/2, n-1}[/tex] is the t-score that corresponds to the desired confidence level and degrees of freedom [tex]n-1[/tex].

Here's how to find the confidence interval step-by-step:

  1. Find the degrees of freedom (df):
    [tex]df = n - 1 = 50 - 1 = 49[/tex]

  2. Determine the t-score:
    For a 90% confidence interval with 49 degrees of freedom, you can look up the t-score in a t-distribution table or use statistical software. Typically, [tex]t_{0.05, 49} \approx 1.676[/tex].

  3. Calculate the standard error (SE):
    [tex]SE = \frac{s}{\sqrt{n}} = \frac{13.86}{\sqrt{50}} \approx 1.96[/tex]

  4. Calculate the margin of error (ME):
    [tex]ME = t_{\alpha/2, n-1} \times SE = 1.676 \times 1.96 \approx 3.29[/tex]

  5. Determine the confidence interval:
    [tex]\text{Lower limit} = \bar{x} - ME = 46.5 - 3.29 = 43.21[/tex]
    [tex]\text{Upper limit} = \bar{x} + ME = 46.5 + 3.29 = 49.79[/tex]

Therefore, the 90% confidence interval for the population mean is approximately (43.21, 49.79). This means we can be 90% confident that the true population mean falls within this interval.