| 
 
 
| Question 980616:  Use the given degree of confidence and sample data to construct a confidence interval for the population mean μ. Assume
 that the population has a normal distribution.
 n = 10, x = 8.1, s = 4.8, 95% confidence
 Found 2 solutions by  stanbon, jim_thompson5910:
 Answer by stanbon(75887)
      (Show Source): 
You can put this solution on YOUR website!  Use the given degree of confidence and sample data to construct a confidence interval for the population mean μ. Assume that the population has a normal distribution.
 n = 10, x-bar = 8.1, s = 4.8, 95% confidence
 -----
 Confidence Interval:: x-bar - ME < u < x-bar + ME
 ------
 ME = 1.96*4.8/sqrt(10) = 2.975
 -------
 95%CI: 8.1-2.98 < u < 8.1+2.98
 -------
 Ans: 5.12 < u < 11.08
 ----------
 Cheers,
 Stan H.
 -------------
Answer by jim_thompson5910(35256)
      (Show Source): 
You can put this solution on YOUR website! xbar = 8.1 n = 10
 s = 4.8
 Since n = 10 makes n > 30 false, and because sigma is not known, we use a T distribution. The critical t value is t = 2.262 (at 95% confidence, df = n-1 = 10-1 = 9). Use a table to find this
 
 The confidence interval is of the form (L,U) where L is the lower limit and U is the upper limit.
 
 Lower Limit (L):
 L = xbar - t*s/sqrt(n)
 L = 8.1 - 2.262*4.8/sqrt(10)
 L = 8.1 - 2.262*1.51789327688082
 L = 8.1 - 3.43347459230442
 L = 4.66652540769558
 L = 4.67
 
 Upper Limit (U):
 U = xbar + t*s/sqrt(n)
 U = 8.1 + 2.262*4.8/sqrt(10)
 U = 8.1 + 2.262*1.51789327688082
 U = 8.1 + 3.43347459230442
 U = 11.5334745923044
 U = 11.53
 
 The 95% confidence interval for mu is (L,U) = (4.67, 11.53)
 
 Note: The confidence interval can also be stated as 4.67 < mu < 11.53. Some books will use plus/minus notation and say
  where 8.1 is the point estimate (xbar) and 3.43347 is the margin of error.
 | 
  
 | 
 |