Question

15. A manufacturer of electronic calculators is interested in estimating the fraction of defective units produced. A random sample of 800 calculators contains 10 defectives. a. Formulate and test the hypothesis to determine if the fraction defective exceeds 0.01. Use 0.05 significance level. b. Calculate a 95% CI for this problem. Does the CI agreed with your result on (a) explain.

Answer 1

Answer:

a

   The Null hypothesis is  $H_o : p = 0.01$

   The defect did not exceed 0.01

b

   The  95%  confidence interval is   $0.004801 < p < 0.020199$

   Yes the CI agrees with the result in a  because the value 0.01 fall within the CI

Step-by-step explanation:

From the question we are told that

     The sample size is  n = 800

      The number of defective calculators is  k =  10

       The population is  $p = 0.01$

The Null hypothesis is  $H_o : p = 0.01$

The Alternative hypothesis is  $H_a : P> 0.01$

Generally the proportion of defective calculators is mathematically represented as

         $\r p = \frac{k}{n}$

substituting values

          $\r p = \frac{10}{800}$

          $\r p = 0.0125$

Next is to obtain the critical value of  $\alpha$ from the z-table.The  value is  

          $Z_{\alpha } = 1.645$

Now the test statistics is mathematically evaluated as

        $t = \frac{\r p - p }{ \sqrt{ \frac{p (1- p )}{n} } }$

substituting values

          $t = \frac{ 0.0125 - 0.01 }{ \sqrt{ \frac{0.01 (1- 0.01 )}{800} } }$

           $t = 0.71067$

Now comparing the values of  t to the value of $Z_{\alpha }$ we see that $t < Z_{\alpha }$ hence we fail to reject the null hypothesis

Generally the margin of error is mathematically represented as  

       $E = Z_{\frac{\alpha }{2} } * \sqrt{\frac{\r p (1-\r p )}{n} }$

where  $Z_{\frac{\alpha }{2} }$ is the critical value of  $\frac{\alpha }{2}$ which is obtained from the z-table.The  value is

              $Z_{\frac{\alpha }{2} } = Z_{\frac{0.05 }{2} } = 1.96$

The reason we are obtaining critical value of    $\frac{\alpha }{2}$  instead of    $\alpha$ is because    $\alpha$

represents the area under the normal curve where the confidence level interval (   $1- \alpha$ ) did not cover which include both the left and right tail while  

 $\frac{\alpha }{2}$  is just the area of one tail which what we required to calculate the margin of error .

NOTE: We can also obtain the value using critical value calculator (math dot armstrong dot edu)

So

      $E = 1.96 * \sqrt{\frac{ 0.0125 (1-0.0125 )}{800} }$

     $E = 0.007699$

The  95%  confidence interval is mathematically represented as

       $\r p - E < p < \r p - E$

substituting values

      $0.0125 - 0.007699 < p < 0.0125 + 0.007699$

      $0.004801 < p < 0.020199$

Now given the p = 0.01 is within this interval then the CI  agrees with answer gotten in a