document.write( "Question 1084046: An article in a journal reports that 23% of American fathers take no responsibility for childcare. A researcher claims that the figure is higher for fathers in the town of Littleton. A random sample of 200 fathers from Littleton yielded 54 who did not help with childcare. Find the P value for a test of the researcher’s claim.\r
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document.write( "a. 0.0038
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document.write( "b. 0.0019
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document.write( "c. 0.0529
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document.write( "d. 0.0901
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Algebra.Com's Answer #698465 by Theo(13342)![]() ![]() You can put this solution on YOUR website! you are comparing proportions.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "the population proportion is .23 (given)\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "the sample proportion is .27 (54/200 = .27)\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "the standard error of your test is sqrt(.23 * .77 / 200) = .029757352\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "your z-score is (.27 - .23) / .029757352 = 1.344205625.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "round this to 1.34 and look it up in the z-score table and you will find that your alpha is equal to 1 - .9099 = .0901\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( ".0901 is your solution.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "when you calculate the standard error of the test, you need to use the population proportion rather than the sample proportion.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "that gets you the correct answer.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "you also needed to round the z-score to 2 digits.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "the general formula is:\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "pmp = population mean proportion \n" ); document.write( "pmq = 1 minus population mean proportion \n" ); document.write( "smp = sample mean proportion \n" ); document.write( "se = standard error of the test. \n" ); document.write( "ss = sample size\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "the formula for se is:\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "se = sqrt(pmp * pmq / ss)\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "the formula for z-score is:\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "z = (smp - pmp) / se\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "in your problem:\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "pmp = .23 \n" ); document.write( "smp = 54/200 = .27 \n" ); document.write( "pmq = 1 - .23 = .77\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "you first need to calculate the standard error of the test.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "that formula is se = sqrt(pmp * pmq / ss) which becomes se = sqrt(.23 * .77 / 200) which becomes .029757352\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "once you find the standard error of the test, you then find the z-score.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "that formula is z = (smp - pmp) / se which becomes z = (.27 - .23) / .029757352 which becomes 1.344205625.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "here you had to round the z-score to 2 decimal digits which gave you z = 1.34\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "if you didn't round it, and used a calculator, you would have gotten something different from .0901 but close.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "using the z-score table, 1.34 z-score gave you .9099 proportion of the normal distribution curve to the left of that z-score.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "your alpha is the proportion of the distribution curve to the right of that z-score, which you obtain by subtracting .9099 from 1 to get .0901.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "visually, it looks like this:\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " ![]() \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "which is 1 minus what looks like this:\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " ![]() \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "in the first picture i was looking for the area of the normal distribution curve above the z-core of 1.34 which is the same as looking for the area to the right of the z-score.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "in the second picture i was looking for the area of the normal distribution curve below the z-score of 1.34 which is the same as looking for the area to the left of the z-score which is what the tables would normally show you.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "to get the area to the right of the z-score, i had to subtract .9099 from 1 to get .0901.\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); 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