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An officer states that the average fine levied b
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<p>Null Hypothesis: u<= 200.</p><p>Alternative Hypothesis: u>200</p><p>We assume the XXXX Hypothesis to be true and XXX to either reject it or fail XX reject it.</p><p>To perform this test, we need the standard deviation of XXX XXXXXX means XXXXX is XXXXX by XXX XXXXXXX:</p><p>(XXXXXX XXXXXXXX XXXXXXXXX)/sqrt(XXXXXX XXXX) = 16/XXXX(XX) = X.06</p><p>XXXX we calculate XXX t-XXXXX: (XXXXXX mean - XXXXXXXXXX mean)/(XXXXXXXX deviation XX sample mean) =</p><p>(210 - 200)/5.06 = 1.XXX</p><p>XX need XX XXXX XXX XXXXXXXXXXX of XXXXXXXXX a t-score of X.XXX or XXXXXX, so XX XXX XXXXXX a table XX p-values for a t-XXXXXXXXXXXX or a calculator's tcdf() function.</p><p>For XXXXXX XX XXXXX methods, we need the XXXXXXX XX freedom XXXXX equals sample XXXX - X = XX - 1 = 9.</p><p>XXXXX a table, just XXXX XXX p-XXXXX XXX the t-XXXXX and degrees XX freedom.</p><p>For a XXXXXXXXXX (such as XX-XX), use function XXXX(1.XXX, X*10^XX,X) = 0.XXXX = p-XXXXX.</p><p>Since XXX p-XXXXX, X.0398, is XXXX than the significance level XX X.XX, XX XXX XXXXXX XXX null. XXXXX is XXXXXXXXXXX evidence that XXX average fine XX higher than $XXX.</p><p>Note: XX cannot XXXXXX a XXXXXXXXXX, only reject or fail to XXXXXX. XXXX this helps :)</p>">
