document.write( "Question 1205768: It is claimed that a random sample of 49 tyres has mean life of 15200 km. This sampled was drawn from a population whose mean is 15150 km's and a standard deviation of 1200 km. Test the significance at 0.05 level. \n" ); document.write( "
Algebra.Com's Answer #842877 by Theo(13342)\"\" \"About 
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sample size is 49.
\n" ); document.write( "sample mean is 15200.
\n" ); document.write( "population mean is 15150.
\n" ); document.write( "population standard deviation is 1200.
\n" ); document.write( "confidence interval is .95\r
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\n" ); document.write( "\n" ); document.write( "standard error = standard deviation / sqrt(sample size) = 1200 / sqrt(49) = 171.4285714\r
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\n" ); document.write( "\n" ); document.write( "z-score formula is z = (x-m)/s
\n" ); document.write( "z = the z-score
\n" ); document.write( "x = the sample mean
\n" ); document.write( "m = the population mean
\n" ); document.write( "s = the standard error\r
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\n" ); document.write( "\n" ); document.write( "formula becomes:\r
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\n" ); document.write( "\n" ); document.write( "z = (15200 - 15150) / 171.4285714 = .291666667.\r
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\n" ); document.write( "\n" ); document.write( "area to the right of that z-score is equal to .3853.\r
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\n" ); document.write( "\n" ); document.write( "this is much greater than the critical p-value of .05/2 = .025.\r
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\n" ); document.write( "\n" ); document.write( "this indicates the results are not significant and that the sample mean is clearly within the confidence limits and therefore it is quite possible to get a sample mean of 15200 when the population mean is 15150.\r
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\n" ); document.write( "\n" ); document.write( "here are the results using an online calculator with the population mean of 15150 and the standard error of 171.4285714.\r
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