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07.09.2021 • 
Mathematics

All of our measurements exceed 47 milliliters. Set up a t-test of the null hypothesis that: (i) the correct measurement is 47 milliliters, (ii) the different measurements are due only to chance measurement error , (iii) and that measurement error is Normally distributed. In each part, answer correct to at least three decimal places. First find the average and SD of the data. Average of the data: SD of the data (remember, it's a small sample): If the null is right, the average of five measurements has an expected value of milliliters. What is the difference between the observed value and the expected value? milliliters For comparison, how far off from the expected value is the average of 5 measurements likely to be, if the null is right? milliliters (Estimate the standard error.) There are 5 measurements. Assuming the null is right, measurements vary only by measurement error that is Normally distributed . The standardized average (t-statistic) of 5 measurements follows a Student t-distribution with degrees of freedom. This standardized average here is observed to be . The P-value is approximately . (Probability calculator required.)

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