Question

A recent study of 26 city residents showed that the time they had lived at their present address has a mean of 10.3 years with the standard deviation of 3 years. Find the 95% confidence interval of the true mean. Assume that the variable is approximately normally distributed.

Answer 1

Answer:

The 95% confidence interval is  $9.15< \mu < 11.45$

Step-by-step explanation:

From the question we are told that

     The sample size is  n =  26

      The mean is  $\= x = 10.3$

       The standard deviation is  $s = 3$

From the question we are told the confidence level is  95% , hence the level of significance is    

      $\alpha = (100 - 95 ) \%$

=>   $\alpha = 0.05$

Generally from the normal distribution table the critical value  of  $\frac{\alpha }{2}$ is  

   $Z_{\frac{\alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as  

      $E = Z_{\frac{\alpha }{2} } * \frac{\sigma }{\sqrt{n} }$

=>   $E = 1.96 * \frac{3}{\sqrt{26} }$

=>   $E = 1.1532$

Generally 95% confidence interval is mathematically represented as  

      $\= x -E < \mu < \=x +E$

=>   $10.3 -1.1532 < \mu < 10.3 + 1.1532$

=>    $9.15< \mu < 11.45$