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.

a

   The Null hypothesis is  

   The defect did not exceed 0.01

b

   The  95%  confidence interval is  

   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  

The Null hypothesis is  

The Alternative hypothesis is  

Generally the proportion of defective calculators is mathematically represented as

substituting values

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

Now the test statistics is mathematically evaluated as

substituting values

Now comparing the values of  t to the value of we see that hence we fail to reject the null hypothesis

Generally the margin of error is mathematically represented as  

where   is the critical value of   which is obtained from the z-table.The  value is

The reason we are obtaining critical value of      instead of     is because    

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

   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

The  95%  confidence interval is mathematically represented as

substituting values

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

i think it would be 42cm