Question

The mean and standard deviation of a random sample of n measurements are equal to and ​, respectively. a. Find a ​% confidence interval for if n. b. Find a ​% confidence interval for if n. c. Find the widths of the confidence intervals found in parts a and b. What is the effect on the width of a confidence interval of quadrupling the sample size while holding the confidence coefficient​ fixed?

Answer 1

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

$33.55 < \mu < 35.5$

b

$34.03 < \mu < 34.969$

c

Generally the width at  n =  49 is mathematically represented as

   $w = 2 * E$

    $w = 2 * 0.952$

     $w = 1.904$

Generally the width at  n =  196 is mathematically represented as

   $w = 2 * E$

    $w = 2 * 0.4687$

     $w = 0.9374$

d

The correct option is E

Step-by-step explanation:

From the question we are told  that

   The sample mean is  $\= x = 34.5$

    The standard deviation is  $s = 3.4$

Generally given that the confidence level is 95% then 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  

   $Z_{\frac{\alpha }{2} } = 1.96$

Considering question a

From the question  n  =  49

Generally the margin of error is mathematically represented as  

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

=>   $E = 1.96* \frac{ 3.4 }{\sqrt{49} }$

=>   $E = 0.952$

Generally 95% confidence interval is mathematically represented as  

      $\= x -E < p < \=x +E$

      $34.5 -0.952 < p < 34.5 + 0.952$

=>    $33.55 < \mu < 35.5$

Considering question b

From the question  n  =  196

Generally the margin of error is mathematically represented as  

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

=>   $E = 1.96* \frac{ 3.4 }{\sqrt{196} }$

=>   $E = 0.4687$

Generally 95% confidence interval is mathematically represented as  

      $\= x -E < p < \=x +E$

      $34.5 -0.4687 < p < 34.5 +0.4687$

=>   $34.03 < \mu < 34.969$

Considering question c

Generally the width at  n =  49 is mathematically represented as

   $w = 2 * E$

    $w = 2 * 0.952$

     $w = 1.904$

Generally the width at  n =  196 is mathematically represented as

   $w = 2 * E$

    $w = 2 * 0.4687$

     $w = 0.9374$

Now when the sample size is quadrupled i.e from n = 49 to  n =  196  

The width of the  confidence interval  decrease by 2 from 1.904 to  0.9374