document.write( "Question 1135587: The manager of a hotel room report that the mean number of rooms rented out per night is at least 212 based on his observation for the past several months. One of the corporate official observed an over estimation, where a sample of 150 nights produce a mean of 201.3 rooms and a standard deviation of 45.5 rooms.. if the result suggest the manager has inflated his report he will be given a warning, at 5% level. what is the managers Fate? \n" ); document.write( "
Algebra.Com's Answer #753221 by Theo(13342)\"\" \"About 
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population mean is assumed to be 212 rooms rented out per night.
\n" ); document.write( "a sample of 150 nights gave a sample mean of 201.3 rooms rented out per night with a standard deviation of 45.5.\r
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\n" ); document.write( "\n" ); document.write( "you want to test this sample mean against the population mean to see if having a sample mean this low is due to chance variation in sample means or whether the probability is so low that it could mean that the population mean may be overstated.\r
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\n" ); document.write( "\n" ); document.write( "since you don't know what the population standard deviation is, then use the sample standard deviation and test with the t-score rather than the z-score.\r
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\n" ); document.write( "\n" ); document.write( "the degrees of freedom are equal to the sample size minus 1, making them equal to 149.\r
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\n" ); document.write( "\n" ); document.write( "the sample size is 150.\r
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\n" ); document.write( "\n" ); document.write( "the standard error is equal to the standard deviation divided by the square root of the sample size.\r
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\n" ); document.write( "\n" ); document.write( "that makes the standard error equal to 45.5 / sqrt(150) = 3.715059443.\r
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\n" ); document.write( "\n" ); document.write( "the t-score with 149 degrees of freedom is equal to the (sample mean minus the population mean) / the standard error.\r
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\n" ); document.write( "\n" ); document.write( "you get t(149) = (201.3 - 212) / 3.715059443 = -2.880169258.\r
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\n" ); document.write( "\n" ); document.write( "use a t-score calculator to find the area under the normal distribution curve to the left of this t-score with 149 degrees of freedom to get:\r
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\n" ); document.write( "\n" ); document.write( "t-score alpha = .0022805084\r
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\n" ); document.write( "\n" ); document.write( "one sided critical t-score alpha is .05.\r
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\n" ); document.write( "\n" ); document.write( "your sample alpha is way below this, so the probability that you would get a sample mean of 201.3 is more then likely not due to random variations in samplemeans of size 150.\r
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\n" ); document.write( "\n" ); document.write( "this mean that the population mean of 212 rooms a night is over-stated and so the manager who gave the report should be given a warning.\r
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\n" ); document.write( "\n" ); document.write( "if you're checking against the critical t-score, that is equal to -1.655...\r
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\n" ); document.write( "\n" ); document.write( "your sample t-score is way beyond that as well.\r
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\n" ); document.write( "\n" ); document.write( "checking your sample t-score against the critical t-score is one way of determing if your sample mean is within limits.\r
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\n" ); document.write( "\n" ); document.write( "checking your sample alpha against the critical t-score alpha is another way of determing if your sample mean is within limits.\r
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\n" ); document.write( "\n" ); document.write( "both tests will be consistent, i.e. if one if out of limits, the other one is out of limits as well.\r
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\n" ); document.write( "\n" ); document.write( "i'm pretty sure this is correct.\r
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\n" ); document.write( "\n" ); document.write( "t-score calculator can be found at https://stattrek.com/online-calculator/t-distribution.aspx\r
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\n" ); document.write( "\n" ); document.write( "critical t-score table can be found at http://www.sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf
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