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Your in alaska panning for gold. Your scale is set to measure 7 oz at a time. Eight trips into town yield an average 7.3 oz with a standard deviation of s = 0.5 oz. Do you think your scale needs adjustments? 5 % level of significants.
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Test for mean involves standard error of the mean, SD/sqrt(number of samples) z = (7.3-7)/[0.5/sqrt(8)] = 0.6 sqrt(8) = 1.70 which is just above the 95%tile so, yes, the scale is off and needs adjustment. This used the normal distribution. Even better is Student's t distribution, which likely would have been less than 95%tile and thus not prove calibration needed. Use it if in your curriculum.
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