document.write( "Question 1201232: A soft drink manufacturer claims that its 120-milliliter bottles do not contain, on average, more than 32 calories. A random sample of 20 bottles of this soft drink, which were checked for calories, contained a mean of 33.3 calories with a standard deviation of 2.5 calories. Assume that the number of calories in the 120-milliliter bottles is normally distributed. Using a level of significance of 5%, is there a significant difference between the manufacturer’s claim and the findings of the study? t computed = 2.32\r
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Algebra.Com's Answer #835795 by Theo(13342)\"\" \"About 
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population mean is assumed to be 32 calories
\n" ); document.write( "sample size is 20
\n" ); document.write( "sample mean is 33.3
\n" ); document.write( "sample standard deviation is 2.5
\n" ); document.write( "standard error is standard deviation / square root of sample size = 2.5 / sqrt(20) = .5590169944.
\n" ); document.write( "test t-score is (x - m) / s = (33.3 - 32) / .5590169944 = 2.325510697
\n" ); document.write( "area to the right of that at 19 degrees of freedom is equal to .0156354548.
\n" ); document.write( "x is the sample mean
\n" ); document.write( "m is the assumed population mean
\n" ); document.write( "s is the standard error.
\n" ); document.write( "at .05 two tailed level of significance, this is less than .025, making the results of the test significant.
\n" ); document.write( "if you use critical t-score, that would be equal to 2.093024022.
\n" ); document.write( "since that is less than the test t-score of 2.32....., the results are, once again, significant, as they should be, because the critical t-score and the critical level of significance always give results that are consistent with each other.\r
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