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:
Find the degrees of freedom (df):
[tex]df = n - 1 = 50 - 1 = 49[/tex]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].Calculate the standard error (SE):
[tex]SE = \frac{s}{\sqrt{n}} = \frac{13.86}{\sqrt{50}} \approx 1.96[/tex]Calculate the margin of error (ME):
[tex]ME = t_{\alpha/2, n-1} \times SE = 1.676 \times 1.96 \approx 3.29[/tex]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.