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Old 08-06-2007   #1
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Default Specifically a statistics question?

Help, please take a look at this question and tell me where to start, what formula to use please.. A simple random sample of 800 persons is taken to estimate the percentage of independent voters among 55,000 eligible voters in a certain town. It turns out that 349 people in the sample are Independents. Find a 95% confidence interval for the percentage of Independence among the eligible voters in town. Also, test the null hypothesis that p (population proportion) is greater than .3.thanks
Old 08-06-2007   #2
Sarah C
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The following is my final answer. Sorry I have made calculation errors in the pervious versions of my answer.Confidence interval is used to estimate a population parameter. Confidence interval for the population mean or proportion is an interval constructed around the sample mean or proportion so that we are confident that this population contains the population mean or proportion. IN THIS CASE we are required to construct an interval around the sample proportion to be confident whether this interval contains the population proportion.Let p = population proportion of independent voters among 55000 eligible voters. n = sample size= 800 ^p = sample proportion = point estimate of true proportion p ^p = 349/800 = 0.43625 Since n^p = 800 * 0.43625 = 349and n(1-^p) = 800*[1-(0.43625)] = 451 are both at least greater than 5, then the sample size n= 800 is considered to be large in this case and we can use the z distribution to construct our confidence interval.100(1-∞)% = 95% 1 - ∞ = 0.95 -∞ = 0.95-1 = -0.05 ∞ = 0.05 = 5%where ∞ represents alpha. Then, a 95% confidence interval for p (the population proportion) is = [ ^p (+-) z(∞/2)*√^p(1-^p)/(n-1)]=[0.43625 (+-) z(0.05/2))*√0.43625(1-0.43625)/(800-1)]= [ 0.43625 (+-) z0. 025# *√(0.245935937/799)]= [0.43625 (+-) 1.96* 0.017544363]= [ 0.43625 (+-) 0.034386951]= [0.401863049, 0.470636951]# the value of z0.025= z(0.5-0.475) can be found in a table of areas under the standard normal curve.Hypothesis testing is basically a statistical procedure used to provide evidence in favor of some statement (called a hypothesis). In this case we are required to reject the above claim(null hypothesis) that p- population proportion of independent voters among 55000 eligible voters is greater than 0.3 and to prove our favor statement(alternative hypothesis) that p- population proportion of independent voters among 55000 eligible voters is less than 0.3 We form the following set of hypothesis to be tested:Null hypothesis -> is the statement that we want to reject when there is convincing sample evidence that it is false.Ho: P>P0 Ho: P>0.3 versusAlternative hypothesis-> is the statement that will be accepted only if there is convinving sample evidence that it is true.Ha: P≤P0 Ha: P≤0.3We can reject H0 in favor of Ha if z< -z∞. ^P = 0.43625Let po = p = 0.3 = CLAIMED true proportion of independent voters among the 50000 eligible voters.Since np0= 800*(0.3) = 240 and n(1-p0) = 800*(1-0.3) = 560,are both at least 5, then we can perform or do a z test (large sample hypothesis test about a population proportion) to test our set of hypothesis.The test static = z = (^p- p0)/√[P0(1-p0)/n] z = (0.43625 -0.3)/√[0.3(1-0.3)/800] z = 0.13625 / √(0.21/800) z = 0.13625/0.016201851 z = 8.40953296Because z = 8.40953296 is not less but much greater than -Z0.05 = -1.645, then we cannot reject Ho in favor of Ha at the ∞ = 0.05 =5% level of significance. Thus, in other words, this means we do NOT have strong evidence to reject the claim that the proportion of independent voters among the 55000 eligible voters is greater than 0.3.

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