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confidence interval in statistics? help asap im desperate?
A random sample of 50 suspension helmets used by motorcycle riders and automobile race-car
drivers was subjected to an impact test, and on 18 of these helmets some damage was observed.
(a) Is the proportion of damaged helmets different from 0.3? Use a = 0.05, write down the null
and alternative hypotheses, the observed test statistic, the P-value and state your conclusion
in the context of the question. Check the assumptions required for applying this test.
(b) Find a 95% confidence interval on the true proportion of helmets that would show damage
from this test.
(c) Using the point estimate of p obtained from the preliminary sample of 50 helmets, how
many helmets must be tested to be 95% confident that the error in estimating the true value
of p is less than 0.02?
Let p be the true or population proportion of damaged helments used by motorcycle riders and automobile race-car.
drivers.
(a).
We wish to test
Ho(Null hypothesis): p = p0 = 0.3
versus
Ha(Alternative hypothesis): p≠(p0=0.3).
Since np0=50*0.3 = 15 and n*(1-p0)= 50*(1-0.3)=35 are both at least greater than 5, then n is considered to be large and hence the sampling distribution of all the possible sample proportions of damaged helments used by motorcycle riders and automobile race-car-
-^p will follow the z distribution.
Our sampled proportion of damaged helments used by motorcycle riders and automobile race-car.
= ^p
= 18/50
= 0.36
Based on our sample,
our observed test statistic
= z
= (^p-p0)/√[p0(1-p0)/n]
= (0.36-0.3)/√[0.3(1-0.3)/50]
= 0.93
|z| = |0.93| = 0.93
At alpha = 0.05 level of significance,
Our rejection point
= z(alpha/2)
= z(0.05/2)
= z0.025
= 1.96
Our p-value
= Twice the area on the right of the z test statistic of 0.93 under the standard normal curve
= 2*[P(z>0) - P(0
= 2*0.1762
= 0.3524
Conclusion: Because our test statistic of z=0.93 is smaller instead of greater than that of our rejection point of z(alpha/2)=z0.025 =1.96, then we fail to reject Ho in favor of Ha at the alpha = 0.05 level of significance, based on our sample. Also, since our two-tailed p-value of 0.3254 is greater instead of smaller than that of our chosen alpha value of 0.05, then we have no evidence that the proportion of damaged helmets is different from 0.3 and hence the claimed proportion of damaged helments of 0.3 is true.
(b).
Since n*^p= 50*0.36= 18 and n*(1-^p) = 50*(1-0.36) = 32 are both at least greater than 5, then n is considered to be large and hence the sampling distribution of all possible sample proportions of damaged helments-^p will follow the z distribution.
A 95% confidence interval on the true proportion of helmets that would show damage from this test
= [^p (+-)√[^p(1-^p)/(n-1)]
= [0.36 (+-)√[0.36(1-0.36)/(50-1)]
= [0.36 (+-)√0.2304/49]
= [0.2914,0.4286]
So, we are 95% confident that the true proportion of helmets that would show damage from this test should be any value lie between 0.2914 and 0.4286.
(c).
Let n be the number of helments that must be tested to be 95% confident that the error bound is less than 0.02.
Our given confidence level = 95%
100(1-alpha)% = 95%
1-alpha = 0.95
alpha = 0.05
Using ^p = 0.36 to estimate p, then
n= p(1-p)*[z(alpha/2)/B]^2
n = 0.36*(1-0.36)*[z(0.05/2)/0.02]^2
n = 0.36*(1-0.36)*[z0.025/0.02]^2
n = 0.2304*(1.96/0.02)^2
n ≈ 2213
So, around 2213 helments must be tested to be 95% confident that the error in estimating the true value of p is less than 0.02.
Hope this helps.
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