Question

The water works commission needs to know the mean household usage of water by the residents of a small town in gallons per day. Assume that the population standard deviation is 2.2 gallons. The mean water usage per family was found to be 15.8 gallons per day for a sample of 669 families. Construct the 90% confidence interval for the mean usage of water. Round your answers to one decimal place.

Answer 1

Answer:

The 90% confidence interval is   $15.66< \mu < 15.94$

Step-by-step explanation:

From the question we are told that

  The population standard deviation is  $\sigma = 2.2$

  The  sample mean is  $\= x = 15.8 \ gallon$

   The sample size is  n  =  669

   

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

      $\alpha = (100 - 90 ) \%$

=>   $\alpha = 0.10$

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

   $Z_{\frac{\alpha }{2} } =  1.645$

Generally the margin of error is mathematically represented as  

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

=>  $E = 1.645  *  \frac{ 2.2}{\sqrt{669} }$

=>  $E = 0.1399$

Generally 90% confidence interval is mathematically represented as  

      $\= x -E <  \mu <  \=x  +E$

=>    $15.8  - 0.1399 <\mu<  15.8  + 0.1399$

=>    $15.66< \mu < 15.94$