Question

An article reports that when each football helmet in a random sample of 34 suspension-type helmets was subjected to a certain impact test, 24 showed damage. Let p denote the proportion of all helmets of this type that would show damage tested in the prescribed manner.

Required:
a. Calculate a 99% Cl for p.
b. What sample size would be required for the width of a 99% Cl to beat most .10, irrespective of p ?

Answer 1

Answer:

a

    $0.5043 < p <0.9075$

b

   $n = 24$

Step-by-step explanation:

From the question we are told that

    The sample size is  n =  34

    The  number of damaged helmets is  x  =  24

Now the proportion of damaged helmets is mathematically represented as

      $\r p = \frac{k}{n }$

substituting values

     $\r p = \frac{24}{34 }$

    $\r p = 0.7059$

Given that the confidence level is 99% the level of significance can be evaluated as

     $\alpha = 100 - 99$

     $\alpha = 1$%

    $\alpha = 0.01$

Next we obtain the critical value of  $\frac{\alpha }{2}$  from the normal distribution table, the value is  $Z_{\frac{\alpha }{2} } = 2.58$

The reason we are obtaining critical values 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

   The margin of error is mathematically represented as

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

substituting values

      $MOE = 2.58 * \sqrt{\frac{ 0.7059 ( 1 - 0.7059)}{34} }$

     $MOE =0.2016$

The 99% confidence interval for p is mathematically represented as

      $p-MOE < p < p + MOE$

substituting values

       $0.7059 - 0.2016 < p <0.7059 + 0.2016$

      $0.5043 < p <0.9075$

The  sample size required for the width of a 99% Cl to beat most 0.10, irrespective of p ? is mathematically represented as  

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

Here  $\sigma = 0.10$  telling  us that the deviation from the sample proportion is  set to 0.10 irrespective of the value of  $\r p$

so the  sample size for this condition is

    $n \ge \frac{ 2.58 * \sqrt{ 0.7059 (1- 0.7059)} }{\frac{0.10 }{2} }$

    $n \ge 23.51$

=>   $n = 24$