Question

How do you construct a $ 90\% $ confidence interval for the population mean, $ \mu $? A sample of 15 randomly selected student grade point average is of $ 2.86 $ with a standard deviation of $ 0.78 $. Assume the population has a normal distribution.

Answer 1

Hint: Begin by mentioning the formula. Then we will evaluate all the required terms from the given question. Mention all the evaluated terms. Calculate the critical value. Then we evaluate the margin of error and finally, we will evaluate the confidence interval. Also, evaluate the upper and lower limit.

Complete step by step answer:
The formula for the confidence interval is given by:
 $ CI = \mathop x\limits^ - \pm z/{s}{{\sqrt n }} $
Now, we will evaluate all the required values of the terms and then mention them one by one.
Sample mean $ = \mathop x\limits^ - = 2.86 $
Sample standard deviation $ = s = 0.78 $
Sample size $ = n = 15 $
Significance level $ = \alpha = 1 - 0.9 = 0.1 $
Degrees of freedom for t-distribution, d.f. $ = n - 1 = 14 $
Critical value $ = {t_{\alpha /2,df}} = {t_{0.05/2,14}} = 1.761 $
Now, we will evaluate the margin of error. The margin of error is given by:
 $ E = {t_{\alpha /2,df}} \times /{{{s_x}}}{{\sqrt n }} $
Margin of error $ = $ $ E $ $ = {t_{\alpha /2,df}} \times /{{{s_x}}}{{\sqrt n }} = 1.761 \times /{{0.78}}{{\sqrt {15} }} $
 $
  E = 1.761 \times 0.201395 \\
  E = 0.354657 \\
  $
Limits of $ 90\% $ confidence interval are given by:
Lower limit:
  $
   = \mathop x\limits^ - - E \\
   = 2.86 - 0.354657 \\
   \approx 2.505 \\
  $
Upper limit:
 $
   = \mathop x\limits^ - + E \\
   = 2.86 - 0.354657 \\
   \approx 3.215 \\
  $
 $ 90\% $ confidence interval is:
 $
   = \mathop x\limits^ - \pm E \\
   = 2.86 \pm 0.354657 \\
   = 2.505343,\,3.214657 \\
  $
$\Rightarrow$ $ 90\% $ CI using t-dist:
$\Rightarrow$ $ 2.505 < \mu < 3.215 $
Additional Information: Confidence interval is a range of values we are fairly sure our true value lies in. We should use the standard deviation of the entire population, but in many cases, we won’t know it. We can use the standard deviation for the sample if we have enough observations. In the formula for confidence interval, $ CI = \mathop x\limits^ - \pm z/{s}{{\sqrt n }} $ , $ CI $ is the confidence level, $ \mathop x\limits^ - $ is the sample mean, $ z $ is the confidence level value, $ s $ is the standard deviation, $ n $ is the sample size.


Note:
Always mention the formula while solving any question which involves formulae. Evaluate all the terms from the question and then again mention the value of each term and its value. While rounding off any number make sure you follow all the steps for it. An approximation is a value or quantity that is nearly but not exactly correct. So, while performing approximation make sure you all the rules.