A sample of 50 suspension helmets by motorcyclists and car racing used car drivers subjected to an impact assessment and was observed 18 of these helmets some damage. (a) Is the percentage of damaged helmets different from 0.3? Use a = 0.05, write the null and alternative Hypotheses, the observed test statistic, the P-value and give you your degree in connection with the issue. Check the conditions for the application of this test is required. (B) Find a 95% confidence interval about the true proportion of helmets, would hurt the show from this test. (c) The estimated value of p from the provisional Sample of 50 helmets, how many helmets have come to be tested is 95% sure that the error in estimating the true value of p less than 0.02?
Let p be the true proportion of the population or helment damaged by motorcyclists and car racing-car use. Drivers. (a). We want to test Ho (null hypothesis): p = p0 = 0.3 versus Ha (alternative hypothesis): p ≠ (p 0 = 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 as large and thus the sampling distribution of all possible sample proportions of damaged helment used by motorcycle riders and car racing-car - ^ p is the z-distribution follow. Our sample proportion of helment be damaged by motor cyclists and racing car. = ^ P = 18/50 = 0.36 in 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 significant, our rejection point = z (alpha / 2) = z (0.05 / 2) = = 1.96 Our z0.025 p-value = twice the area on the right side of the z test statistic of 0.93 below the normal standard curve = 2 * [P (z> 0) - P (0
= 2*(0.5-0.3238)
= 2*0.1762
= 0.3524
Conclusion: Because our test Statistics from z = 0.93 is smaller rather than larger than our rejection point z (alpha / 2) = z0.025 = 1.96, then we are not in favor of Ho Ha at the alpha = 0.05 reject level of importance based on our sample. Also, because our two-sided p-value of 0.3254 is larger rather than smaller than those we selected alpha value of 0.05, then we have no evidence that the proportion of damaged helmet from 0.3 to ensure that the various alleged proportion of damaged helment of 0.3 is true. (b). Since s ^ p * = 50 * 0.36 = 18 and n * (1 - ^ p) = 50 * (1-0.36) = 32 are both at least greater than 5, then n is as large and thus the sampling distribution all possible sample proportions of damaged helment-^ p will follow the z-distribution. A 95% confidence interval for the true proportion of helmets would harm arising from this Test = ^ [p show (+ -) √ [^ p (1 - ^ p) / (n-1)] = [0.36 (+ -) √ [0.36 (1-0.36) / (50 -- 1)] = [0.36 (+ -) √ 0.2304/49] = [0.2914,0.4286] Thus we are confident that 95% show the true proportion of helmets would harm resulting from this test, all its value range from 0.2914 and 0.4286. (c). Let n be the number of helment that need to be tested, 95% believe that the error bound less than 0.02. Notice given confidence level = 95% 100 (1-alpha)% = 95% 1-alpha = 0.95, alpha = 0.05 with 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 to 2213, therefore helment need to be tested, 95% believe that the error in estimating the true value of p less than 0.02. Hope this helps.
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