Mathematics: 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 ?

C no capStep-by-step explanation:...

An article reports that when each football helmet

Ответ:

lilbit217

30.01.2022

Mathematics

0.5043 < p

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

0.5043 < p

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

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