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

a

`0.5043 < p <0.9075`

b

`n = 24`

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 = (k)/(n )`

substituting values

`\r p = (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
`(\alpha )/(2)` from the normal distribution table, the value is
`Z_{(\alpha )/(2) } = 2.58`

The reason we are obtaining critical values of
`(\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

`(\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_{(\alpha )/(2) } * \sqrt{(\r p ( 1 - \r p))/(n) }`

substituting values

`MOE = 2.58 * \sqrt{( 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_{(\alpha )/(2) } * √(\r p (1- \r p )) }{(\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 ( 2.58 * √( 0.7059 (1- 0.7059)) )/((0.10 )/(2) )`

`n \ge 23.51`

=>
`n = 24`