document.write( "Question 1083817: A new piece of equipment is suspected of having faulty temperature control the desired temperature is supposed to be 68 dergree and a random sample of 45 readings are collected and the mean of these is 69.178 with a S.E of 0.483. at the level significance 0.05 . what would your concluison have been around this hypothisis \n" ); document.write( "
Algebra.Com's Answer #697895 by Theo(13342) You can put this solution on YOUR website! desired temperature is 68 degrees.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "sample size is 45 \n" ); document.write( "mean of samples is 69.178 \n" ); document.write( "standard error of sample is .483\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "z-score of sample = (69.178 - 68) / .483 = 2.44\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "since the faulty readings could be high or low, a two tailed distribution is assumed.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "a significance level of .05 would be divided by 2 to get an alpha of .025 on each end.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "an alpha of .025 on each end gives you a critical z-score of plus or minus 1.96.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "the z-score of 2.44 is greater than the critical z-score, therefore the results of the test are statistically significant and you can conclude that the temperature readings are not within the prescribed limits.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "notice that you were given the standard error, and not the standard deviation of the sample.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "standard error is equal to standard deviation divided by square root of sample size.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "using this formula, the standard deviation of your sample would have been based on this formula to get:\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "se = sd / sqrt(ss)\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "se is standard error \n" ); document.write( "sd is standard deviation \n" ); document.write( "ss is sample size\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( ".483 = sd / sqrt(45)\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "solve for sd to get sd = .483 * sqrt(45) = 3.24.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "in order for the readings not to be considered out of limits, a test with a sample of 45 should have given you readings between what is shown below:\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \r\n" ); document.write( " \n" ); document.write( "\n" ); document.write( "you could calculate this using the z-score formula.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "that formula is z = (x-m)/se\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "at 95% two tailed confidence interval, the critical z-score is 1.96.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "to find the high side limits for 95% confidence interval, formula becomes:\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "1.96 = (x-68)/.483\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "solve for x to get x = 1.96 * .483 + 68 = 68.95\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "similar calculations using a z-score of -1.96 will get you the low side raw score limit.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |